Linear Momentum
1
A 2 kg ball is moving with a velocity of 5 m/s to the right. What is the linear momentum of the ball?
Practice
A
10 kg·m/s to the right
B
10 kg·m/s to the left
C
2.5 kg·m/s to the right
D
7 kg·m/s to the right
✅ Show Answer
✔ Correct Answer:
A
(10 kg·m/s to the right)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Velocity of the ball, v = 5 m/s to the right
Linear momentum is given by:
p = mv
Substituting the given values:
p = 2 × 5
p = 10 kg·m/s
The direction of momentum is the same as the direction of velocity.
Since the ball is moving to the right, its momentum is also directed to the right.
Therefore:
The linear momentum of the ball is 10 kg·m/s to the right.
Mass of the ball, m = 2 kg
Velocity of the ball, v = 5 m/s to the right
Linear momentum is given by:
p = mv
Substituting the given values:
p = 2 × 5
p = 10 kg·m/s
The direction of momentum is the same as the direction of velocity.
Since the ball is moving to the right, its momentum is also directed to the right.
Therefore:
The linear momentum of the ball is 10 kg·m/s to the right.
2
Two objects have the same velocity of 4 m/s to the right. Object A has a mass of 2 kg and Object B has a mass of 6 kg. Which object has greater linear momentum?
Practice
A
Object A has greater momentum.
B
Object B has greater momentum.
C
Both objects have the same momentum.
D
Neither object has momentum.
✅ Show Answer
✔ Correct Answer:
B
(Object B has greater momentum.)
💡 Explanation
Given:
Velocity of both objects, v = 4 m/s to the right
Mass of Object A, m_A = 2 kg
Mass of Object B, m_B = 6 kg
Linear momentum is given by:
p = mv
Momentum of Object A:
p_A = 2 × 4
p_A = 8 kg·m/s
Momentum of Object B:
p_B = 6 × 4
p_B = 24 kg·m/s
Since 24 kg·m/s is greater than 8 kg·m/s, Object B has greater linear momentum.
Therefore:
Object B has greater momentum.
Velocity of both objects, v = 4 m/s to the right
Mass of Object A, m_A = 2 kg
Mass of Object B, m_B = 6 kg
Linear momentum is given by:
p = mv
Momentum of Object A:
p_A = 2 × 4
p_A = 8 kg·m/s
Momentum of Object B:
p_B = 6 × 4
p_B = 24 kg·m/s
Since 24 kg·m/s is greater than 8 kg·m/s, Object B has greater linear momentum.
Therefore:
Object B has greater momentum.
🎯 Conclusion
If velocity is constant, linear momentum is directly proportional to mass.
3
A 3 kg object is moving with a velocity of 4 m/s to the right. What will be its momentum if its velocity is increased to 8 m/s in the same direction?
Practice
A
12 kg·m/s
B
24 kg·m/s
C
8 kg·m/s
D
6 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(24 kg·m/s)
💡 Explanation
Given:
Mass of the object, m = 3 kg
Initial velocity, v_i = 4 m/s
Final velocity, v_f = 8 m/s
Linear momentum is given by:
p = mv
Since we are asked for the momentum after the velocity is increased:
p_f = mv_f
p_f = 3 × 8
p_f = 24 kg·m/s
Therefore:
The final momentum of the object is 24 kg·m/s to the right.
Mass of the object, m = 3 kg
Initial velocity, v_i = 4 m/s
Final velocity, v_f = 8 m/s
Linear momentum is given by:
p = mv
Since we are asked for the momentum after the velocity is increased:
p_f = mv_f
p_f = 3 × 8
p_f = 24 kg·m/s
Therefore:
The final momentum of the object is 24 kg·m/s to the right.
🎯 Conclusion
If mass is constant, linear momentum is directly proportional to velocity.
4
A 4 kg ball is moving at 6 m/s to the right. It is brought to rest. What is the change in its linear momentum?
Practice
A
24 kg·m/s to the right
B
24 kg·m/s to the left
C
0 kg·m/s
D
10 kg·m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(24 kg·m/s to the left)
💡 Explanation
Given:
Mass of the ball, m = 4 kg
Initial velocity, v_i = 6 m/s to the right
Final velocity, v_f = 0 m/s
Take the right direction as positive.
Initial momentum:
p_i = mv_i
p_i = 4 × 6
p_i = +24 kg·m/s
Final momentum:
p_f = mv_f
p_f = 4 × 0
p_f = 0 kg·m/s
Change in momentum is given by:
Δp = p_f - p_i
Therefore:
Δp = 0 - 24
Δp = -24 kg·m/s
The negative sign indicates that the change in momentum is opposite to the initial direction of motion.
Therefore:
The change in momentum is 24 kg·m/s to the left.
Mass of the ball, m = 4 kg
Initial velocity, v_i = 6 m/s to the right
Final velocity, v_f = 0 m/s
Take the right direction as positive.
Initial momentum:
p_i = mv_i
p_i = 4 × 6
p_i = +24 kg·m/s
Final momentum:
p_f = mv_f
p_f = 4 × 0
p_f = 0 kg·m/s
Change in momentum is given by:
Δp = p_f - p_i
Therefore:
Δp = 0 - 24
Δp = -24 kg·m/s
The negative sign indicates that the change in momentum is opposite to the initial direction of motion.
Therefore:
The change in momentum is 24 kg·m/s to the left.
🎯 Conclusion
When an object moving in one direction is brought to rest, its change in momentum is opposite to its initial direction of motion.
5
A 2 kg ball is moving at 5 m/s to the right. It reverses its direction and moves at 3 m/s to the left. What is the change in its linear momentum?
Practice
A
4 kg·m/s to the left
B
16 kg·m/s to the left
C
10 kg·m/s to the right
D
4 kg·m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(16 kg·m/s to the left)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 5 m/s to the right
Final velocity, vₓ = 3 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +5 m/s
vₓ = -3 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-3)
pₓ = -6 kg·m/s
Change in momentum is given by:
Δp = pₓ − pᵢ
Therefore:
Δp = -6 − (+10)
Δp = -16 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Therefore:
The change in momentum is 16 kg·m/s to the left.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 5 m/s to the right
Final velocity, vₓ = 3 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +5 m/s
vₓ = -3 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-3)
pₓ = -6 kg·m/s
Change in momentum is given by:
Δp = pₓ − pᵢ
Therefore:
Δp = -6 − (+10)
Δp = -16 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Therefore:
The change in momentum is 16 kg·m/s to the left.
🎯 Conclusion
When an object reverses its direction, the initial and final momenta are in opposite directions, so their signed values must be used to find the change in momentum.
6
A 5 kg object is initially at rest. A constant force of 10 N acts on it for 2 s. What is the impulse delivered to the object?
Practice
A
5 N·s
B
10 N·s
C
20 N·s
D
50 N·s
✅ Show Answer
✔ Correct Answer:
C
(20 N·s)
💡 Explanation
Given:
Mass of the object, m = 5 kg
Initial velocity, vᵢ = 0 m/s
Force, F = 10 N
Time for which the force acts, t = 2 s
Impulse is given by:
J = F × t
Substituting the given values:
J = 10 × 2
J = 20 N·s
Therefore:
The impulse delivered to the object is 20 N·s.
Mass of the object, m = 5 kg
Initial velocity, vᵢ = 0 m/s
Force, F = 10 N
Time for which the force acts, t = 2 s
Impulse is given by:
J = F × t
Substituting the given values:
J = 10 × 2
J = 20 N·s
Therefore:
The impulse delivered to the object is 20 N·s.
🎯 Conclusion
For a constant force, impulse is directly proportional to the time for which the force acts.
7
A 4 kg object is initially at rest. A force of 8 N acts on it for 3 s. What is the change in its linear momentum?
Practice
A
11 kg·m/s
B
24 kg·m/s
C
32 kg·m/s
D
96 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(24 kg·m/s)
💡 Explanation
Given:
Mass of the object, m = 4 kg
Initial velocity, vᵢ = 0 m/s
Force, F = 8 N
Time for which the force acts, t = 3 s
Impulse-momentum theorem states:
J = Δp
For a constant force:
J = F × t
Therefore:
Δp = F × t
Δp = 8 × 3
Δp = 24 kg·m/s
Therefore:
The change in linear momentum is 24 kg·m/s.
Mass of the object, m = 4 kg
Initial velocity, vᵢ = 0 m/s
Force, F = 8 N
Time for which the force acts, t = 3 s
Impulse-momentum theorem states:
J = Δp
For a constant force:
J = F × t
Therefore:
Δp = F × t
Δp = 8 × 3
Δp = 24 kg·m/s
Therefore:
The change in linear momentum is 24 kg·m/s.
🎯 Conclusion
The change in linear momentum is equal to the impulse acting on the object.
8
A force of 20 N acts on an object for 0.5 s. What impulse is produced by the force?
Practice
A
10 N·s
B
20 N·s
C
40 N·s
D
0.025 N·s
✅ Show Answer
✔ Correct Answer:
A
(10 N·s)
💡 Explanation
Given:
Force, F = 20 N
Time for which the force acts, t = 0.5 s
Impulse is given by:
J = F × t
Substituting the given values:
J = 20 × 0.5
J = 10 N·s
Therefore:
The impulse produced by the force is 10 N·s.
Force, F = 20 N
Time for which the force acts, t = 0.5 s
Impulse is given by:
J = F × t
Substituting the given values:
J = 20 × 0.5
J = 10 N·s
Therefore:
The impulse produced by the force is 10 N·s.
🎯 Conclusion
For a constant force, impulse is equal to the product of force and the time for which it acts.
9
Two forces act on the same object. Force A is 10 N and acts for 2 s, while Force B is 20 N and acts for 1 s. Which statement about the impulses produced by the two forces is correct?
Practice
A
Force A produces twice the impulse of Force B.
B
Force B produces twice the impulse of Force A.
C
Both forces produce the same impulse.
D
The impulse cannot be determined.
✅ Show Answer
✔ Correct Answer:
C
(Both forces produce the same impulse.)
💡 Explanation
Given:
For Force A:
F₁ = 10 N
t₁ = 2 s
For Force B:
F₂ = 20 N
t₂ = 1 s
Impulse is given by:
J = F × t
Impulse produced by Force A:
J₁ = F₁ × t₁
J₁ = 10 × 2
J₁ = 20 N·s
Impulse produced by Force B:
J₂ = F₂ × t₂
J₂ = 20 × 1
J₂ = 20 N·s
Therefore:
J₁ = J₂ = 20 N·s
Both forces produce the same impulse.
For Force A:
F₁ = 10 N
t₁ = 2 s
For Force B:
F₂ = 20 N
t₂ = 1 s
Impulse is given by:
J = F × t
Impulse produced by Force A:
J₁ = F₁ × t₁
J₁ = 10 × 2
J₁ = 20 N·s
Impulse produced by Force B:
J₂ = F₂ × t₂
J₂ = 20 × 1
J₂ = 20 N·s
Therefore:
J₁ = J₂ = 20 N·s
Both forces produce the same impulse.
🎯 Conclusion
For a constant force, impulse depends on the product of force and time; different force-time combinations can produce the same impulse.
10
A 2 kg ball is moving at 6 m/s to the right. It hits a wall and rebounds at 4 m/s to the left. What is the magnitude of the change in its linear momentum?
Practice
A
4 kg·m/s
B
8 kg·m/s
C
20 kg·m/s
D
12 kg·m/s
✅ Show Answer
✔ Correct Answer:
C
(20 kg·m/s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 6 m/s to the right
Final velocity, vₓ = 4 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +6 m/s
vₓ = -4 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 6
pᵢ = +12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-4)
pₓ = -8 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -8 − (+12)
Δp = -20 kg·m/s
The question asks for the magnitude of the change in momentum.
|Δp| = 20 kg·m/s
Therefore:
The magnitude of the change in linear momentum is 20 kg·m/s.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 6 m/s to the right
Final velocity, vₓ = 4 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +6 m/s
vₓ = -4 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 6
pᵢ = +12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-4)
pₓ = -8 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -8 − (+12)
Δp = -20 kg·m/s
The question asks for the magnitude of the change in momentum.
|Δp| = 20 kg·m/s
Therefore:
The magnitude of the change in linear momentum is 20 kg·m/s.
🎯 Conclusion
When an object reverses direction, the change in momentum is found using the signed initial and final velocities; the magnitude can be obtained from |Δp|.
11
A 3 kg ball moving at 4 m/s to the right is brought to rest by a constant force acting for 0.6 s. What is the magnitude of the average force acting on the ball?
Practice
A
5 N
B
10 N
C
15 N
D
20 N
✅ Show Answer
✔ Correct Answer:
D
(20 N)
💡 Explanation
Given:
Mass of the ball, m = 3 kg
Initial velocity, vᵢ = 4 m/s to the right
Final velocity, vₓ = 0 m/s
Time for which the force acts, t = 0.6 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 4
pᵢ = 12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 0
pₓ = 0 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 12
Δp = -12 kg·m/s
The magnitude of the impulse is:
|J| = |Δp| = 12 N·s
For an average force:
|J| = Fₐᵥₑ × t
Therefore:
Fₐᵥₑ = |J| / t
Fₐᵥₑ = 12 / 0.6
Fₐᵥₑ = 20 N
Therefore:
The magnitude of the average force is 20 N.
Mass of the ball, m = 3 kg
Initial velocity, vᵢ = 4 m/s to the right
Final velocity, vₓ = 0 m/s
Time for which the force acts, t = 0.6 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 4
pᵢ = 12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 0
pₓ = 0 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 12
Δp = -12 kg·m/s
The magnitude of the impulse is:
|J| = |Δp| = 12 N·s
For an average force:
|J| = Fₐᵥₑ × t
Therefore:
Fₐᵥₑ = |J| / t
Fₐᵥₑ = 12 / 0.6
Fₐᵥₑ = 20 N
Therefore:
The magnitude of the average force is 20 N.
🎯 Conclusion
For a given change in momentum, the average force is inversely proportional to the time over which the change occurs.
12
A 4 kg cart moving at 5 m/s to the right collides with a stationary 2 kg cart. After the collision, the 4 kg cart moves at 2 m/s to the right. What is the velocity of the 2 kg cart after the collision?
Practice
A
1 m/s to the right
B
6 m/s to the right
C
4 m/s to the right
D
2 m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(6 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = 5 m/s to the right
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Final velocity of the first cart, v₁ₓ = 2 m/s to the right
Final velocity of the second cart, v₂ₓ = ?
Take the right direction as positive.
According to the law of conservation of linear momentum:
Total initial momentum = Total final momentum
Therefore:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
Substituting the given values:
(4)(5) + (2)(0) = (4)(2) + (2)v₂ₓ
20 = 8 + 2v₂ₓ
12 = 2v₂ₓ
v₂ₓ = 6 m/s
The positive sign indicates that the second cart moves to the right.
Therefore:
The velocity of the second cart after the collision is 6 m/s to the right.
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = 5 m/s to the right
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Final velocity of the first cart, v₁ₓ = 2 m/s to the right
Final velocity of the second cart, v₂ₓ = ?
Take the right direction as positive.
According to the law of conservation of linear momentum:
Total initial momentum = Total final momentum
Therefore:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
Substituting the given values:
(4)(5) + (2)(0) = (4)(2) + (2)v₂ₓ
20 = 8 + 2v₂ₓ
12 = 2v₂ₓ
v₂ₓ = 6 m/s
The positive sign indicates that the second cart moves to the right.
Therefore:
The velocity of the second cart after the collision is 6 m/s to the right.
🎯 Conclusion
In an isolated system, the total linear momentum before a collision is equal to the total linear momentum after the collision.
13
A 6 kg cart is moving at 4 m/s to the right. What is its linear momentum?
Practice
A
10 kg·m/s to the right
B
24 kg·m/s to the right
C
24 kg·m/s to the left
D
2.5 kg·m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(24 kg·m/s to the right)
💡 Explanation
Given:
Mass of the cart, m = 6 kg
Velocity, v = 4 m/s to the right
Linear momentum is given by:
p = mv
Therefore:
p = 6 × 4
p = 24 kg·m/s
The momentum is in the same direction as the velocity.
Therefore:
The linear momentum is 24 kg·m/s to the right.
Mass of the cart, m = 6 kg
Velocity, v = 4 m/s to the right
Linear momentum is given by:
p = mv
Therefore:
p = 6 × 4
p = 24 kg·m/s
The momentum is in the same direction as the velocity.
Therefore:
The linear momentum is 24 kg·m/s to the right.
🎯 Conclusion
Linear momentum is directly proportional to mass when velocity is constant.
14
A 5 kg object is moving at 3 m/s to the left. Taking the right direction as positive, what is its linear momentum?
Practice
A
+15 kg·m/s
B
-15 kg·m/s
C
+8 kg·m/s
D
-8 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(-15 kg·m/s)
💡 Explanation
Given:
Mass of the object, m = 5 kg
Velocity, v = 3 m/s to the left
Take the right direction as positive.
Therefore, the velocity is:
v = -3 m/s
Linear momentum is given by:
p = mv
Therefore:
p = 5 × (-3)
p = -15 kg·m/s
The negative sign indicates that the momentum is directed to the left.
Mass of the object, m = 5 kg
Velocity, v = 3 m/s to the left
Take the right direction as positive.
Therefore, the velocity is:
v = -3 m/s
Linear momentum is given by:
p = mv
Therefore:
p = 5 × (-3)
p = -15 kg·m/s
The negative sign indicates that the momentum is directed to the left.
🎯 Conclusion
The direction of linear momentum is the same as the direction of velocity.
15
A 4 kg object changes its velocity from 2 m/s to 5 m/s in the same direction. What is the change in its linear momentum?
Practice
A
8 kg·m/s
B
12 kg·m/s
C
20 kg·m/s
D
28 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(12 kg·m/s)
💡 Explanation
Given:
Mass of the object, m = 4 kg
Initial velocity, vᵢ = 2 m/s
Final velocity, vₓ = 5 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 2
pᵢ = 8 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 5
pₓ = 20 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 20 − 8
Δp = 12 kg·m/s
Therefore:
The change in momentum is 12 kg·m/s.
Mass of the object, m = 4 kg
Initial velocity, vᵢ = 2 m/s
Final velocity, vₓ = 5 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 2
pᵢ = 8 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 5
pₓ = 20 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 20 − 8
Δp = 12 kg·m/s
Therefore:
The change in momentum is 12 kg·m/s.
🎯 Conclusion
When mass is constant, the change in momentum depends directly on the change in velocity.
16
A 3 kg object moving at 8 m/s to the right slows down to 2 m/s in the same direction. What is the change in its linear momentum?
Practice
A
18 kg·m/s to the right
B
18 kg·m/s to the left
C
30 kg·m/s to the left
D
6 kg·m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(18 kg·m/s to the left)
💡 Explanation
Given:
Mass of the object, m = 3 kg
Initial velocity, vᵢ = 8 m/s to the right
Final velocity, vₓ = 2 m/s to the right
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 8
pᵢ = +24 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 2
pₓ = +6 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 6 − 24
Δp = -18 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Mass of the object, m = 3 kg
Initial velocity, vᵢ = 8 m/s to the right
Final velocity, vₓ = 2 m/s to the right
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 8
pᵢ = +24 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 2
pₓ = +6 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 6 − 24
Δp = -18 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
🎯 Conclusion
When an object slows down while moving in one direction, its change in momentum is opposite to its direction of motion.
17
A 2 kg ball moving at 7 m/s to the right is brought to rest in 0.5 s. What is the magnitude of the average force acting on the ball?
Practice
A
7 N
B
14 N
C
28 N
D
35 N
✅ Show Answer
✔ Correct Answer:
C
(28 N)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 7 m/s to the right
Final velocity, vₓ = 0 m/s
Time, t = 0.5 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 7
pᵢ = 14 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 14
Δp = -14 kg·m/s
Magnitude of the impulse:
|J| = |Δp| = 14 N·s
Average force is given by:
|F| = |J| / t
Therefore:
|F| = 14 / 0.5
|F| = 28 N
Therefore:
The magnitude of the average force is 28 N.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 7 m/s to the right
Final velocity, vₓ = 0 m/s
Time, t = 0.5 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 7
pᵢ = 14 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 14
Δp = -14 kg·m/s
Magnitude of the impulse:
|J| = |Δp| = 14 N·s
Average force is given by:
|F| = |J| / t
Therefore:
|F| = 14 / 0.5
|F| = 28 N
Therefore:
The magnitude of the average force is 28 N.
🎯 Conclusion
For a fixed change in momentum, the average force increases when the time interval decreases.
18
A force of 15 N acts on an object for 4 s. What is the impulse produced by the force?
Practice
A
19 N·s
B
30 N·s
C
60 N·s
D
75 N·s
✅ Show Answer
✔ Correct Answer:
C
(60 N·s)
💡 Explanation
Given:
Force, F = 15 N
Time, t = 4 s
Impulse is given by:
J = F × t
Therefore:
J = 15 × 4
J = 60 N·s
Therefore:
The impulse produced by the force is 60 N·s.
Force, F = 15 N
Time, t = 4 s
Impulse is given by:
J = F × t
Therefore:
J = 15 × 4
J = 60 N·s
Therefore:
The impulse produced by the force is 60 N·s.
🎯 Conclusion
For a constant force, impulse increases directly with the time for which the force acts.
19
A 2 kg ball moving at 5 m/s to the right rebounds from a wall at 3 m/s to the left. What is the magnitude of the impulse acting on the ball?
Practice
A
4 N·s
B
10 N·s
C
16 N·s
D
6 N·s
✅ Show Answer
✔ Correct Answer:
C
(16 N·s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 5 m/s to the right
Final velocity, vₓ = 3 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +5 m/s
vₓ = -3 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-3)
pₓ = -6 kg·m/s
Impulse is equal to the change in momentum:
J = Δp
Therefore:
Δp = pₓ − pᵢ
Δp = -6 − (+10)
Δp = -16 kg·m/s
Magnitude of impulse:
|J| = 16 N·s
Therefore:
The magnitude of the impulse is 16 N·s.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 5 m/s to the right
Final velocity, vₓ = 3 m/s to the left
Take the right direction as positive.
Therefore:
vᵢ = +5 m/s
vₓ = -3 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-3)
pₓ = -6 kg·m/s
Impulse is equal to the change in momentum:
J = Δp
Therefore:
Δp = pₓ − pᵢ
Δp = -6 − (+10)
Δp = -16 kg·m/s
Magnitude of impulse:
|J| = 16 N·s
Therefore:
The magnitude of the impulse is 16 N·s.
🎯 Conclusion
Impulse is equal to the change in momentum, and a reversal of direction can produce a large change in momentum.
20
A 3 kg cart moving at 4 m/s to the right collides with a stationary 1 kg cart. If they stick together, what is their common velocity after the collision?
Practice
A
1 m/s to the right
B
2 m/s to the right
C
3 m/s to the right
D
4 m/s to the right
✅ Show Answer
✔ Correct Answer:
C
(3 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 3 kg
Initial velocity of the first cart, v₁ᵢ = 4 m/s to the right
Mass of the second cart, m₂ = 1 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Since the carts stick together, they have a common final velocity, vₓ.
According to conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting the values:
(3)(4) + (1)(0) = (3 + 1)vₓ
12 = 4vₓ
vₓ = 3 m/s
Therefore:
The common velocity after the collision is 3 m/s to the right.
Mass of the first cart, m₁ = 3 kg
Initial velocity of the first cart, v₁ᵢ = 4 m/s to the right
Mass of the second cart, m₂ = 1 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Since the carts stick together, they have a common final velocity, vₓ.
According to conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting the values:
(3)(4) + (1)(0) = (3 + 1)vₓ
12 = 4vₓ
vₓ = 3 m/s
Therefore:
The common velocity after the collision is 3 m/s to the right.
🎯 Conclusion
In a perfectly inelastic collision, the objects stick together and move with a common velocity while total momentum is conserved.
21
A 4 kg cart moving at 3 m/s to the right collides with a 2 kg cart moving at 1 m/s to the left. If the carts stick together, what is their common velocity after the collision?
Practice
A
1 m/s to the right
B
1.67 m/s to the right
C
2 m/s to the right
D
2.33 m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(1.67 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = +3 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = -1 m/s
Take the right direction as positive.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting the values:
(4)(3) + (2)(-1) = (4 + 2)vₓ
12 − 2 = 6vₓ
10 = 6vₓ
vₓ = 10/6
vₓ = 1.67 m/s
The positive sign means the common velocity is toward the right.
Therefore:
The carts move together at approximately 1.67 m/s to the right.
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = +3 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = -1 m/s
Take the right direction as positive.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting the values:
(4)(3) + (2)(-1) = (4 + 2)vₓ
12 − 2 = 6vₓ
10 = 6vₓ
vₓ = 10/6
vₓ = 1.67 m/s
The positive sign means the common velocity is toward the right.
Therefore:
The carts move together at approximately 1.67 m/s to the right.
🎯 Conclusion
In an isolated collision, the direction of the final motion depends on the net initial momentum of the system.
22
A 5 kg object is moving at 6 m/s to the right. What is its linear momentum?
Practice
A
11 kg·m/s to the right
B
30 kg·m/s to the right
C
30 kg·m/s to the left
D
1.2 kg·m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(30 kg·m/s to the right)
💡 Explanation
Given:
Mass of the object, m = 5 kg
Velocity, v = 6 m/s to the right
Linear momentum is given by:
p = mv
Therefore:
p = 5 × 6
p = 30 kg·m/s
Therefore:
The linear momentum is 30 kg·m/s to the right.
Mass of the object, m = 5 kg
Velocity, v = 6 m/s to the right
Linear momentum is given by:
p = mv
Therefore:
p = 5 × 6
p = 30 kg·m/s
Therefore:
The linear momentum is 30 kg·m/s to the right.
🎯 Conclusion
Linear momentum is the product of mass and velocity.
23
A 4 kg object moving at 8 m/s to the right is slowed down to 3 m/s in the same direction. What is the change in its linear momentum?
Practice
A
20 kg·m/s to the right
B
20 kg·m/s to the left
C
44 kg·m/s to the left
D
12 kg·m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(20 kg·m/s to the left)
💡 Explanation
Given:
Mass, m = 4 kg
Initial velocity, vᵢ = 8 m/s to the right
Final velocity, vₓ = 3 m/s to the right
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 8
pᵢ = 32 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 3
pₓ = 12 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 12 − 32
Δp = -20 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Therefore:
The change in momentum is 20 kg·m/s to the left.
Mass, m = 4 kg
Initial velocity, vᵢ = 8 m/s to the right
Final velocity, vₓ = 3 m/s to the right
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 8
pᵢ = 32 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 3
pₓ = 12 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 12 − 32
Δp = -20 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Therefore:
The change in momentum is 20 kg·m/s to the left.
🎯 Conclusion
When an object slows down, its change in momentum is opposite to its direction of motion.
24
A 3 kg ball moving at 10 m/s to the right is brought to rest in 0.5 s. What is the magnitude of the average force acting on the ball?
Practice
A
15 N
B
30 N
C
60 N
D
90 N
✅ Show Answer
✔ Correct Answer:
C
(60 N)
💡 Explanation
Given:
Mass, m = 3 kg
Initial velocity, vᵢ = 10 m/s
Final velocity, vₓ = 0 m/s
Time, t = 0.5 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 10
pᵢ = 30 kg·m/s
Final momentum:
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 30
Δp = -30 kg·m/s
Magnitude of impulse:
|J| = 30 N·s
Average force:
|F| = |J| / t
|F| = 30 / 0.5
|F| = 60 N
Therefore:
The magnitude of the average force is 60 N.
Mass, m = 3 kg
Initial velocity, vᵢ = 10 m/s
Final velocity, vₓ = 0 m/s
Time, t = 0.5 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 10
pᵢ = 30 kg·m/s
Final momentum:
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 30
Δp = -30 kg·m/s
Magnitude of impulse:
|J| = 30 N·s
Average force:
|F| = |J| / t
|F| = 30 / 0.5
|F| = 60 N
Therefore:
The magnitude of the average force is 60 N.
🎯 Conclusion
For a given change in momentum, average force equals the magnitude of the change in momentum divided by the time interval.
25
A 0.5 kg ball moving at 12 m/s to the right rebounds from a wall at 8 m/s to the left. What is the magnitude of the change in its linear momentum?
Practice
A
2 kg·m/s
B
4 kg·m/s
C
10 kg·m/s
D
20 kg·m/s
✅ Show Answer
✔ Correct Answer:
C
(10 kg·m/s)
💡 Explanation
Given:
Mass, m = 0.5 kg
Initial velocity, vᵢ = +12 m/s
Final velocity, vₓ = -8 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 0.5 × 12
pᵢ = +6 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0.5 × (-8)
pₓ = -4 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -4 − 6
Δp = -10 kg·m/s
Magnitude:
|Δp| = 10 kg·m/s
Therefore:
The magnitude of the change in momentum is 10 kg·m/s.
Mass, m = 0.5 kg
Initial velocity, vᵢ = +12 m/s
Final velocity, vₓ = -8 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 0.5 × 12
pᵢ = +6 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0.5 × (-8)
pₓ = -4 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -4 − 6
Δp = -10 kg·m/s
Magnitude:
|Δp| = 10 kg·m/s
Therefore:
The magnitude of the change in momentum is 10 kg·m/s.
🎯 Conclusion
When velocity reverses direction, the signed initial and final momenta must be used to calculate the change in momentum.
26
A constant force of 25 N acts on an object for 0.8 s. What is the impulse delivered to the object?
Practice
A
10 N·s
B
20 N·s
C
25 N·s
D
31.25 N·s
✅ Show Answer
✔ Correct Answer:
B
(20 N·s)
💡 Explanation
Given:
Force, F = 25 N
Time, t = 0.8 s
Impulse is given by:
J = F × t
Therefore:
J = 25 × 0.8
J = 20 N·s
Therefore:
The impulse delivered to the object is 20 N·s.
Force, F = 25 N
Time, t = 0.8 s
Impulse is given by:
J = F × t
Therefore:
J = 25 × 0.8
J = 20 N·s
Therefore:
The impulse delivered to the object is 20 N·s.
🎯 Conclusion
For a constant force, impulse is the product of force and time.
27
A 2 kg cart moving at 5 m/s to the right collides with a stationary 3 kg cart. They stick together after the collision. What is their common velocity?
Practice
A
1 m/s to the right
B
2 m/s to the right
C
3 m/s to the right
D
5 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(2 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 2 kg
Initial velocity of the first cart, v₁ᵢ = 5 m/s
Mass of the second cart, m₂ = 3 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Since the carts stick together, their final velocity is vₓ.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Therefore:
(2)(5) + (3)(0) = (2 + 3)vₓ
10 = 5vₓ
vₓ = 2 m/s
Therefore:
The common velocity is 2 m/s to the right.
Mass of the first cart, m₁ = 2 kg
Initial velocity of the first cart, v₁ᵢ = 5 m/s
Mass of the second cart, m₂ = 3 kg
Initial velocity of the second cart, v₂ᵢ = 0 m/s
Since the carts stick together, their final velocity is vₓ.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Therefore:
(2)(5) + (3)(0) = (2 + 3)vₓ
10 = 5vₓ
vₓ = 2 m/s
Therefore:
The common velocity is 2 m/s to the right.
🎯 Conclusion
When two objects stick together, their common velocity is found by conserving the total linear momentum.
28
A 4 kg cart moving at 6 m/s to the right collides with a 2 kg cart moving at 3 m/s to the left. If they stick together, what is their common velocity?
Practice
A
1 m/s to the right
B
3 m/s to the right
C
3 m/s to the left
D
5 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(3 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = +6 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = -3 m/s
Take the right direction as positive.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(4)(6) + (2)(-3) = (4 + 2)vₓ
24 − 6 = 6vₓ
18 = 6vₓ
vₓ = 3 m/s
Therefore:
The common velocity is 3 m/s to the right.
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = +6 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = -3 m/s
Take the right direction as positive.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(4)(6) + (2)(-3) = (4 + 2)vₓ
24 − 6 = 6vₓ
18 = 6vₓ
vₓ = 3 m/s
Therefore:
The common velocity is 3 m/s to the right.
🎯 Conclusion
In a collision, the final direction of a combined system is determined by the net initial momentum.
29
A 10 kg object initially at rest explodes into two pieces. A 2 kg piece moves at 20 m/s to the right. What is the velocity of the 8 kg piece?
Practice
A
5 m/s to the left
B
10 m/s to the left
C
20 m/s to the left
D
40 m/s to the right
✅ Show Answer
✔ Correct Answer:
A
(5 m/s to the left)
💡 Explanation
Given:
Total initial mass = 10 kg
Initial velocity = 0 m/s
Mass of piece 1, m₁ = 2 kg
Velocity of piece 1, v₁ₓ = +20 m/s
Mass of piece 2, m₂ = 8 kg
Velocity of piece 2, v₂ₓ = ?
Since the object was initially at rest:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(2)(20) + (8)v₂ₓ = 0
40 + 8v₂ₓ = 0
8v₂ₓ = -40
v₂ₓ = -5 m/s
The negative sign indicates motion to the left.
Therefore:
The 8 kg piece moves at 5 m/s to the left.
Total initial mass = 10 kg
Initial velocity = 0 m/s
Mass of piece 1, m₁ = 2 kg
Velocity of piece 1, v₁ₓ = +20 m/s
Mass of piece 2, m₂ = 8 kg
Velocity of piece 2, v₂ₓ = ?
Since the object was initially at rest:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(2)(20) + (8)v₂ₓ = 0
40 + 8v₂ₓ = 0
8v₂ₓ = -40
v₂ₓ = -5 m/s
The negative sign indicates motion to the left.
Therefore:
The 8 kg piece moves at 5 m/s to the left.
🎯 Conclusion
When a stationary system explodes, the momenta of the resulting pieces are equal in magnitude and opposite in direction.
30
A 5 kg cannon is initially at rest and fires a 0.5 kg projectile at 20 m/s to the right. What is the recoil velocity of the cannon?
Practice
A
1 m/s to the left
B
2 m/s to the left
C
4 m/s to the left
D
10 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(2 m/s to the left)
💡 Explanation
Given:
Mass of cannon, m₁ = 5 kg
Mass of projectile, m₂ = 0.5 kg
Initial velocity of the system = 0 m/s
Final velocity of projectile, v₂ₓ = +20 m/s
Recoil velocity of cannon, v₁ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.5)(20) = 0
5v₁ₓ + 10 = 0
5v₁ₓ = -10
v₁ₓ = -2 m/s
The negative sign indicates that the cannon recoils to the left.
Therefore:
The recoil velocity of the cannon is 2 m/s to the left.
Mass of cannon, m₁ = 5 kg
Mass of projectile, m₂ = 0.5 kg
Initial velocity of the system = 0 m/s
Final velocity of projectile, v₂ₓ = +20 m/s
Recoil velocity of cannon, v₁ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.5)(20) = 0
5v₁ₓ + 10 = 0
5v₁ₓ = -10
v₁ₓ = -2 m/s
The negative sign indicates that the cannon recoils to the left.
Therefore:
The recoil velocity of the cannon is 2 m/s to the left.
🎯 Conclusion
In recoil, the backward momentum of the heavier object balances the forward momentum of the lighter object.
31
A 3 kg cart moving at 4 m/s to the right collides with a 2 kg cart moving at 1 m/s to the right. After the collision, the 3 kg cart moves at 2 m/s to the right. What is the velocity of the 2 kg cart after the collision?
Practice
A
1 m/s to the right
B
2 m/s to the right
C
4 m/s to the right
D
5 m/s to the right
✅ Show Answer
✔ Correct Answer:
C
(4 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 3 kg
Initial velocity of the first cart, v₁ᵢ = 4 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 1 m/s
Final velocity of the first cart, v₁ₓ = 2 m/s
Final velocity of the second cart, v₂ₓ = ?
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
(3)(4) + (2)(1) = (3)(2) + (2)v₂ₓ
12 + 2 = 6 + 2v₂ₓ
14 = 6 + 2v₂ₓ
8 = 2v₂ₓ
v₂ₓ = 4 m/s
Therefore:
The velocity of the second cart after the collision is 4 m/s to the right.
Mass of the first cart, m₁ = 3 kg
Initial velocity of the first cart, v₁ᵢ = 4 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 1 m/s
Final velocity of the first cart, v₁ₓ = 2 m/s
Final velocity of the second cart, v₂ₓ = ?
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
(3)(4) + (2)(1) = (3)(2) + (2)v₂ₓ
12 + 2 = 6 + 2v₂ₓ
14 = 6 + 2v₂ₓ
8 = 2v₂ₓ
v₂ₓ = 4 m/s
Therefore:
The velocity of the second cart after the collision is 4 m/s to the right.
🎯 Conclusion
For an isolated system, an unknown final velocity can be found by equating total initial momentum to total final momentum.
32
Two objects have the same mass of 4 kg. Object A moves at 3 m/s and Object B moves at 7 m/s in the same direction. Which object has greater linear momentum?
Practice
A
Object A has greater momentum.
B
Object B has greater momentum.
C
Both have the same momentum.
D
Neither object has momentum.
✅ Show Answer
✔ Correct Answer:
B
(Object B has greater momentum.)
💡 Explanation
Given:
Mass of Object A, m₁ = 4 kg
Velocity of Object A, v₁ = 3 m/s
Mass of Object B, m₂ = 4 kg
Velocity of Object B, v₂ = 7 m/s
Momentum of Object A:
p₁ = m₁v₁
p₁ = 4 × 3
p₁ = 12 kg·m/s
Momentum of Object B:
p₂ = m₂v₂
p₂ = 4 × 7
p₂ = 28 kg·m/s
Since 28 kg·m/s > 12 kg·m/s:
Object B has greater momentum.
Mass of Object A, m₁ = 4 kg
Velocity of Object A, v₁ = 3 m/s
Mass of Object B, m₂ = 4 kg
Velocity of Object B, v₂ = 7 m/s
Momentum of Object A:
p₁ = m₁v₁
p₁ = 4 × 3
p₁ = 12 kg·m/s
Momentum of Object B:
p₂ = m₂v₂
p₂ = 4 × 7
p₂ = 28 kg·m/s
Since 28 kg·m/s > 12 kg·m/s:
Object B has greater momentum.
🎯 Conclusion
For the same mass, linear momentum is directly proportional to velocity.
33
Two objects move with the same velocity of 5 m/s in the same direction. Object A has a mass of 2 kg and Object B has a mass of 6 kg. Which object has greater linear momentum?
Practice
A
Object A has greater momentum.
B
Object B has greater momentum.
C
Both have the same momentum.
D
Both have zero momentum.
✅ Show Answer
✔ Correct Answer:
B
(Object B has greater momentum.)
💡 Explanation
Given:
Mass of Object A, m₁ = 2 kg
Mass of Object B, m₂ = 6 kg
Velocity of both objects, v = 5 m/s
Momentum of Object A:
p₁ = m₁v
p₁ = 2 × 5
p₁ = 10 kg·m/s
Momentum of Object B:
p₂ = m₂v
p₂ = 6 × 5
p₂ = 30 kg·m/s
Therefore:
Object B has greater momentum.
Mass of Object A, m₁ = 2 kg
Mass of Object B, m₂ = 6 kg
Velocity of both objects, v = 5 m/s
Momentum of Object A:
p₁ = m₁v
p₁ = 2 × 5
p₁ = 10 kg·m/s
Momentum of Object B:
p₂ = m₂v
p₂ = 6 × 5
p₂ = 30 kg·m/s
Therefore:
Object B has greater momentum.
🎯 Conclusion
For the same velocity, linear momentum is directly proportional to mass.
34
Object A has a mass of 2 kg and moves at 10 m/s. Object B has a mass of 5 kg and moves at 4 m/s. Both move in the same direction. Which statement is correct about their linear momenta?
Practice
A
Object A has greater momentum.
B
Object B has greater momentum.
C
Both have the same momentum.
D
Object A has twice the momentum of Object B.
✅ Show Answer
✔ Correct Answer:
C
(Both have the same momentum.)
💡 Explanation
Given:
For Object A:
m₁ = 2 kg
v₁ = 10 m/s
For Object B:
m₂ = 5 kg
v₂ = 4 m/s
Momentum of Object A:
p₁ = m₁v₁
p₁ = 2 × 10
p₁ = 20 kg·m/s
Momentum of Object B:
p₂ = m₂v₂
p₂ = 5 × 4
p₂ = 20 kg·m/s
Therefore:
p₁ = p₂
Both objects have the same linear momentum.
For Object A:
m₁ = 2 kg
v₁ = 10 m/s
For Object B:
m₂ = 5 kg
v₂ = 4 m/s
Momentum of Object A:
p₁ = m₁v₁
p₁ = 2 × 10
p₁ = 20 kg·m/s
Momentum of Object B:
p₂ = m₂v₂
p₂ = 5 × 4
p₂ = 20 kg·m/s
Therefore:
p₁ = p₂
Both objects have the same linear momentum.
🎯 Conclusion
When mass and velocity change together, two objects can have equal momentum if their products mv are equal.
35
A 3 kg object increases its speed from 4 m/s to 9 m/s in the same direction. What is the change in its linear momentum?
Practice
A
15 kg·m/s
B
18 kg·m/s
C
27 kg·m/s
D
39 kg·m/s
✅ Show Answer
✔ Correct Answer:
A
(15 kg·m/s)
💡 Explanation
Given:
Mass, m = 3 kg
Initial velocity, vᵢ = 4 m/s
Final velocity, vₓ = 9 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 4
pᵢ = 12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 9
pₓ = 27 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 27 − 12
Δp = 15 kg·m/s
Therefore:
The change in momentum is 15 kg·m/s.
Mass, m = 3 kg
Initial velocity, vᵢ = 4 m/s
Final velocity, vₓ = 9 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 3 × 4
pᵢ = 12 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 3 × 9
pₓ = 27 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 27 − 12
Δp = 15 kg·m/s
Therefore:
The change in momentum is 15 kg·m/s.
🎯 Conclusion
When an object speeds up in the same direction, its linear momentum increases in that direction.
36
A 4 kg object moving at 10 m/s to the right slows down to 6 m/s in the same direction. What is the change in its linear momentum?
Practice
A
16 kg·m/s to the right
B
16 kg·m/s to the left
C
24 kg·m/s to the left
D
40 kg·m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(16 kg·m/s to the left)
💡 Explanation
Given:
Mass, m = 4 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, vₓ = +6 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 10
pᵢ = +40 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 6
pₓ = +24 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 24 − 40
Δp = -16 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
Mass, m = 4 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, vₓ = +6 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 4 × 10
pᵢ = +40 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 4 × 6
pₓ = +24 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = 24 − 40
Δp = -16 kg·m/s
The negative sign indicates that the change in momentum is toward the left.
🎯 Conclusion
When an object slows down, its change in momentum is opposite to its direction of motion.
37
A 2 kg ball is moving at 5 m/s to the right. It reverses its direction and moves at 5 m/s to the left. What is the magnitude of the change in its linear momentum?
Practice
A
0 kg·m/s
B
5 kg·m/s
C
10 kg·m/s
D
20 kg·m/s
✅ Show Answer
✔ Correct Answer:
D
(20 kg·m/s)
💡 Explanation
Given:
Mass, m = 2 kg
Initial velocity, vᵢ = +5 m/s
Final velocity, vₓ = -5 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-5)
pₓ = -10 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -10 − 10
Δp = -20 kg·m/s
Magnitude:
|Δp| = 20 kg·m/s
Therefore:
The magnitude of the change in momentum is 20 kg·m/s.
Mass, m = 2 kg
Initial velocity, vᵢ = +5 m/s
Final velocity, vₓ = -5 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 2 × 5
pᵢ = +10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 2 × (-5)
pₓ = -10 kg·m/s
Change in momentum:
Δp = pₓ − pᵢ
Δp = -10 − 10
Δp = -20 kg·m/s
Magnitude:
|Δp| = 20 kg·m/s
Therefore:
The magnitude of the change in momentum is 20 kg·m/s.
🎯 Conclusion
When an object reverses direction, the change in momentum accounts for both the initial and final momentum directions.
38
A 5 kg object is moving at 8 m/s to the right. It is brought to rest. What is the magnitude of the change in its linear momentum?
Practice
A
8 kg·m/s
B
13 kg·m/s
C
40 kg·m/s
D
45 kg·m/s
✅ Show Answer
✔ Correct Answer:
C
(40 kg·m/s)
💡 Explanation
Given:
Mass, m = 5 kg
Initial velocity, vᵢ = 8 m/s
Final velocity, vₓ = 0 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 5 × 8
pᵢ = 40 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 5 × 0
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 40
Δp = -40 kg·m/s
Magnitude:
|Δp| = 40 kg·m/s
Therefore:
The magnitude of the change in momentum is 40 kg·m/s.
Mass, m = 5 kg
Initial velocity, vᵢ = 8 m/s
Final velocity, vₓ = 0 m/s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 5 × 8
pᵢ = 40 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 5 × 0
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 40
Δp = -40 kg·m/s
Magnitude:
|Δp| = 40 kg·m/s
Therefore:
The magnitude of the change in momentum is 40 kg·m/s.
🎯 Conclusion
When an object is brought to rest, the magnitude of its change in momentum equals its initial momentum.
39
A constant force of 12 N acts on an object for 3 s. What is the impulse delivered to the object?
Practice
A
4 N·s
B
15 N·s
C
36 N·s
D
48 N·s
✅ Show Answer
✔ Correct Answer:
C
(36 N·s)
💡 Explanation
Given:
Force, F = 12 N
Time, t = 3 s
Impulse is given by:
J = F × t
Therefore:
J = 12 × 3
J = 36 N·s
Therefore:
The impulse delivered to the object is 36 N·s.
Force, F = 12 N
Time, t = 3 s
Impulse is given by:
J = F × t
Therefore:
J = 12 × 3
J = 36 N·s
Therefore:
The impulse delivered to the object is 36 N·s.
🎯 Conclusion
Impulse is equal to the product of force and the time interval for which the force acts.
40
A 2 kg object initially at rest receives an impulse of 10 N·s to the right. What is its final velocity?
Practice
A
2 m/s to the right
B
5 m/s to the right
C
10 m/s to the right
D
20 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(5 m/s to the right)
💡 Explanation
Given:
Mass, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Impulse, J = 10 N·s to the right
Using the impulse-momentum theorem:
J = Δp
Therefore:
J = pₓ − pᵢ
Since the object is initially at rest:
pᵢ = 0
Therefore:
J = pₓ
J = mvₓ
Substituting the values:
10 = 2vₓ
vₓ = 5 m/s
Therefore:
The final velocity is 5 m/s to the right.
Mass, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Impulse, J = 10 N·s to the right
Using the impulse-momentum theorem:
J = Δp
Therefore:
J = pₓ − pᵢ
Since the object is initially at rest:
pᵢ = 0
Therefore:
J = pₓ
J = mvₓ
Substituting the values:
10 = 2vₓ
vₓ = 5 m/s
Therefore:
The final velocity is 5 m/s to the right.
🎯 Conclusion
An impulse changes an object's momentum; for an object initially at rest, impulse determines its final momentum.
41
A force of 18 N acts on an object for 0.4 s. What is the impulse delivered to the object?
Practice
A
4.5 N·s
B
7.2 N·s
C
18.4 N·s
D
45 N·s
✅ Show Answer
✔ Correct Answer:
B
(7.2 N·s)
💡 Explanation
Given:
Force, F = 18 N
Time, t = 0.4 s
Impulse is given by:
J = F × t
Therefore:
J = 18 × 0.4
J = 7.2 N·s
Therefore:
The impulse delivered to the object is 7.2 N·s.
Force, F = 18 N
Time, t = 0.4 s
Impulse is given by:
J = F × t
Therefore:
J = 18 × 0.4
J = 7.2 N·s
Therefore:
The impulse delivered to the object is 7.2 N·s.
🎯 Conclusion
Impulse is equal to the product of force and the time for which the force acts.
42
An impulse of 24 N·s changes the momentum of an object in 0.6 s. What is the magnitude of the average force acting on the object?
Practice
A
14.4 N
B
24 N
C
40 N
D
60 N
✅ Show Answer
✔ Correct Answer:
C
(40 N)
💡 Explanation
Given:
Impulse, J = 24 N·s
Time, t = 0.6 s
Using the impulse-momentum relation:
J = Fₐᵥₑ × t
Therefore:
Fₐᵥₑ = J / t
Fₐᵥₑ = 24 / 0.6
Fₐᵥₑ = 40 N
Therefore:
The magnitude of the average force is 40 N.
Impulse, J = 24 N·s
Time, t = 0.6 s
Using the impulse-momentum relation:
J = Fₐᵥₑ × t
Therefore:
Fₐᵥₑ = J / t
Fₐᵥₑ = 24 / 0.6
Fₐᵥₑ = 40 N
Therefore:
The magnitude of the average force is 40 N.
🎯 Conclusion
For a given impulse, the average force is inversely proportional to the time interval.
43
A constant force of 15 N produces an impulse of 45 N·s. For how much time does the force act?
Practice
A
0.33 s
B
2 s
C
3 s
D
4.5 s
✅ Show Answer
✔ Correct Answer:
C
(3 s)
💡 Explanation
Given:
Force, F = 15 N
Impulse, J = 45 N·s
Impulse is given by:
J = F × t
Therefore:
t = J / F
t = 45 / 15
t = 3 s
Therefore:
The force acts for 3 s.
Force, F = 15 N
Impulse, J = 45 N·s
Impulse is given by:
J = F × t
Therefore:
t = J / F
t = 45 / 15
t = 3 s
Therefore:
The force acts for 3 s.
🎯 Conclusion
For a constant force, the time of action is directly proportional to impulse and inversely proportional to force.
44
A 2 kg object is initially moving at 3 m/s to the right. An impulse of 10 N·s acts on it in the same direction. What is its change in velocity?
Practice
A
2 m/s
B
5 m/s
C
7 m/s
D
10 m/s
✅ Show Answer
✔ Correct Answer:
B
(5 m/s)
💡 Explanation
Given:
Mass, m = 2 kg
Initial velocity, vᵢ = 3 m/s
Impulse, J = 10 N·s
Using the impulse-momentum theorem:
J = Δp
Since mass is constant:
Δp = mΔv
Therefore:
J = mΔv
So:
Δv = J / m
Δv = 10 / 2
Δv = 5 m/s
Therefore:
The change in velocity is 5 m/s to the right.
Mass, m = 2 kg
Initial velocity, vᵢ = 3 m/s
Impulse, J = 10 N·s
Using the impulse-momentum theorem:
J = Δp
Since mass is constant:
Δp = mΔv
Therefore:
J = mΔv
So:
Δv = J / m
Δv = 10 / 2
Δv = 5 m/s
Therefore:
The change in velocity is 5 m/s to the right.
🎯 Conclusion
For a given mass, the change in velocity is directly proportional to the impulse.
45
A constant force of 20 N acts on an object for 2.5 s. What is the impulse delivered to the object?
Practice
A
8 N·s
B
22.5 N·s
C
50 N·s
D
80 N·s
✅ Show Answer
✔ Correct Answer:
C
(50 N·s)
💡 Explanation
Given:
Force, F = 20 N
Time, t = 2.5 s
For a constant force:
J = F × t
Therefore:
J = 20 × 2.5
J = 50 N·s
Therefore:
The impulse delivered to the object is 50 N·s.
Force, F = 20 N
Time, t = 2.5 s
For a constant force:
J = F × t
Therefore:
J = 20 × 2.5
J = 50 N·s
Therefore:
The impulse delivered to the object is 50 N·s.
🎯 Conclusion
For a constant force, impulse increases directly with the duration of the force.
46
A force of 10 N acts on an object for 2 s, followed immediately by a force of 20 N acting for 3 s in the same direction. What is the total impulse?
Practice
A
40 N·s
B
60 N·s
C
80 N·s
D
100 N·s
✅ Show Answer
✔ Correct Answer:
C
(80 N·s)
💡 Explanation
Given:
First force, F₁ = 10 N
First time, t₁ = 2 s
Second force, F₂ = 20 N
Second time, t₂ = 3 s
Impulse due to the first force:
J₁ = F₁t₁
J₁ = 10 × 2
J₁ = 20 N·s
Impulse due to the second force:
J₂ = F₂t₂
J₂ = 20 × 3
J₂ = 60 N·s
Since both forces act in the same direction:
J = J₁ + J₂
J = 20 + 60
J = 80 N·s
Therefore:
The total impulse is 80 N·s.
First force, F₁ = 10 N
First time, t₁ = 2 s
Second force, F₂ = 20 N
Second time, t₂ = 3 s
Impulse due to the first force:
J₁ = F₁t₁
J₁ = 10 × 2
J₁ = 20 N·s
Impulse due to the second force:
J₂ = F₂t₂
J₂ = 20 × 3
J₂ = 60 N·s
Since both forces act in the same direction:
J = J₁ + J₂
J = 20 + 60
J = 80 N·s
Therefore:
The total impulse is 80 N·s.
🎯 Conclusion
When forces act successively in the same direction, their individual impulses add to give the total impulse.
47
Force A of 30 N acts for 0.4 s, while Force B of 12 N acts for 1 s. Which statement is correct about the impulses produced by the two forces?
Practice
A
Force A produces a greater impulse.
B
Force B produces a greater impulse.
C
Both forces produce the same impulse.
D
The impulses cannot be compared.
✅ Show Answer
✔ Correct Answer:
A
(Force A produces a greater impulse.)
💡 Explanation
Given:
For Force A:
F₁ = 30 N
t₁ = 0.4 s
For Force B:
F₂ = 12 N
t₂ = 1 s
Impulse due to Force A:
J₁ = F₁t₁
J₁ = 30 × 0.4
J₁ = 12 N·s
Impulse due to Force B:
J₂ = F₂t₂
J₂ = 12 × 1
J₂ = 12 N·s
Therefore:
J₁ = J₂ = 12 N·s
Both forces produce the same impulse.
For Force A:
F₁ = 30 N
t₁ = 0.4 s
For Force B:
F₂ = 12 N
t₂ = 1 s
Impulse due to Force A:
J₁ = F₁t₁
J₁ = 30 × 0.4
J₁ = 12 N·s
Impulse due to Force B:
J₂ = F₂t₂
J₂ = 12 × 1
J₂ = 12 N·s
Therefore:
J₁ = J₂ = 12 N·s
Both forces produce the same impulse.
🎯 Conclusion
Different force-time combinations can produce the same impulse when their products Ft are equal.
48
A 4 kg cart moving at 3 m/s to the right collides with a 2 kg cart moving at 1 m/s to the right. If they stick together, what is their common velocity after the collision?
Practice
A
1.67 m/s to the right
B
2.33 m/s to the right
C
2.67 m/s to the right
D
3 m/s to the right
✅ Show Answer
✔ Correct Answer:
C
(2.67 m/s to the right)
💡 Explanation
Given:
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = 3 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 1 m/s
Since the carts stick together, they have a common final velocity, vₓ.
By conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting:
(4)(3) + (2)(1) = (4 + 2)vₓ
12 + 2 = 6vₓ
14 = 6vₓ
vₓ = 14/6
vₓ = 2.33 m/s
Therefore:
The common velocity is 2.33 m/s to the right.
Mass of the first cart, m₁ = 4 kg
Initial velocity of the first cart, v₁ᵢ = 3 m/s
Mass of the second cart, m₂ = 2 kg
Initial velocity of the second cart, v₂ᵢ = 1 m/s
Since the carts stick together, they have a common final velocity, vₓ.
By conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
Substituting:
(4)(3) + (2)(1) = (4 + 2)vₓ
12 + 2 = 6vₓ
14 = 6vₓ
vₓ = 14/6
vₓ = 2.33 m/s
Therefore:
The common velocity is 2.33 m/s to the right.
🎯 Conclusion
When objects move in the same direction and stick together, their common velocity is the total initial momentum divided by the total mass.
49
Two objects move toward each other along a straight line. A 4 kg object moves at 5 m/s to the right, while a 2 kg object moves at 3 m/s to the left. What is the total linear momentum of the system?
Practice
A
14 kg·m/s to the right
B
14 kg·m/s to the left
C
26 kg·m/s to the right
D
26 kg·m/s to the left
✅ Show Answer
✔ Correct Answer:
A
(14 kg·m/s to the right)
💡 Explanation
Take the right direction as positive.
For object 1:
m₁ = 4 kg
v₁ᵢ = +5 m/s
For object 2:
m₂ = 2 kg
v₂ᵢ = -3 m/s
Momentum of object 1:
p₁ᵢ = m₁v₁ᵢ
p₁ᵢ = 4 × 5
p₁ᵢ = +20 kg·m/s
Momentum of object 2:
p₂ᵢ = m₂v₂ᵢ
p₂ᵢ = 2 × (-3)
p₂ᵢ = -6 kg·m/s
Total momentum:
pₜ = p₁ᵢ + p₂ᵢ
pₜ = 20 − 6
pₜ = 14 kg·m/s
Therefore:
The total linear momentum is 14 kg·m/s to the right.
For object 1:
m₁ = 4 kg
v₁ᵢ = +5 m/s
For object 2:
m₂ = 2 kg
v₂ᵢ = -3 m/s
Momentum of object 1:
p₁ᵢ = m₁v₁ᵢ
p₁ᵢ = 4 × 5
p₁ᵢ = +20 kg·m/s
Momentum of object 2:
p₂ᵢ = m₂v₂ᵢ
p₂ᵢ = 2 × (-3)
p₂ᵢ = -6 kg·m/s
Total momentum:
pₜ = p₁ᵢ + p₂ᵢ
pₜ = 20 − 6
pₜ = 14 kg·m/s
Therefore:
The total linear momentum is 14 kg·m/s to the right.
🎯 Conclusion
When objects move in opposite directions, their momenta have opposite signs and must be added algebraically.
50
A 6 kg cart moving at 4 m/s to the right collides with a stationary 2 kg cart. If they stick together, what is their common velocity after the collision?
Practice
A
2 m/s to the right
B
3 m/s to the right
C
4 m/s to the right
D
6 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(3 m/s to the right)
💡 Explanation
Given:
m₁ = 6 kg
v₁ᵢ = 4 m/s to the right
m₂ = 2 kg
v₂ᵢ = 0 m/s
Since the carts stick together, their final velocity is vₓ.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(6)(4) + (2)(0) = (6 + 2)vₓ
24 = 8vₓ
vₓ = 3 m/s
Therefore:
The common velocity is 3 m/s to the right.
m₁ = 6 kg
v₁ᵢ = 4 m/s to the right
m₂ = 2 kg
v₂ᵢ = 0 m/s
Since the carts stick together, their final velocity is vₓ.
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(6)(4) + (2)(0) = (6 + 2)vₓ
24 = 8vₓ
vₓ = 3 m/s
Therefore:
The common velocity is 3 m/s to the right.
🎯 Conclusion
When one object is initially at rest, the initial momentum of the moving object is shared by the combined mass after they stick together.
51
A 3 kg cart moving at 6 m/s to the right collides with a 2 kg cart moving at 1 m/s to the left. If they stick together, what is their common velocity after the collision?
Practice
A
2.2 m/s to the right
B
3.2 m/s to the right
C
3.2 m/s to the left
D
4.4 m/s to the right
✅ Show Answer
✔ Correct Answer:
B
(3.2 m/s to the right)
💡 Explanation
Take the right direction as positive.
m₁ = 3 kg
v₁ᵢ = +6 m/s
m₂ = 2 kg
v₂ᵢ = -1 m/s
Since the carts stick together:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(3)(6) + (2)(-1) = (3 + 2)vₓ
18 − 2 = 5vₓ
16 = 5vₓ
vₓ = 3.2 m/s
Therefore:
The common velocity is 3.2 m/s to the right.
m₁ = 3 kg
v₁ᵢ = +6 m/s
m₂ = 2 kg
v₂ᵢ = -1 m/s
Since the carts stick together:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(3)(6) + (2)(-1) = (3 + 2)vₓ
18 − 2 = 5vₓ
16 = 5vₓ
vₓ = 3.2 m/s
Therefore:
The common velocity is 3.2 m/s to the right.
🎯 Conclusion
In a perfectly inelastic collision, total momentum is conserved even though the objects move together afterward.
52
A 4 kg cart moving at 5 m/s to the right collides with a 2 kg cart moving at 1 m/s to the right. After the collision, the 4 kg cart moves at 2 m/s to the right. What is the final velocity of the 2 kg cart?
Practice
A
3 m/s to the right
B
5 m/s to the right
C
7 m/s to the right
D
9 m/s to the right
✅ Show Answer
✔ Correct Answer:
C
(7 m/s to the right)
💡 Explanation
Given:
m₁ = 4 kg
v₁ᵢ = 5 m/s
m₂ = 2 kg
v₂ᵢ = 1 m/s
v₁ₓ = 2 m/s
v₂ₓ = ?
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
(4)(5) + (2)(1) = (4)(2) + (2)v₂ₓ
20 + 2 = 8 + 2v₂ₓ
22 = 8 + 2v₂ₓ
14 = 2v₂ₓ
v₂ₓ = 7 m/s
Therefore:
The final velocity of the 2 kg cart is 7 m/s to the right.
m₁ = 4 kg
v₁ᵢ = 5 m/s
m₂ = 2 kg
v₂ᵢ = 1 m/s
v₁ₓ = 2 m/s
v₂ₓ = ?
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁ₓ + m₂v₂ₓ
(4)(5) + (2)(1) = (4)(2) + (2)v₂ₓ
20 + 2 = 8 + 2v₂ₓ
22 = 8 + 2v₂ₓ
14 = 2v₂ₓ
v₂ₓ = 7 m/s
Therefore:
The final velocity of the 2 kg cart is 7 m/s to the right.
🎯 Conclusion
In a collision, conservation of momentum can be used to find an unknown final velocity.
53
A 3 kg cart moving at 6 m/s to the right collides with a stationary cart. They stick together and move at 2 m/s to the right. What is the mass of the stationary cart?
Practice
A
3 kg
B
4 kg
C
6 kg
D
9 kg
✅ Show Answer
✔ Correct Answer:
C
(6 kg)
💡 Explanation
Given:
Mass of moving cart, m₁ = 3 kg
Initial velocity, v₁ᵢ = 6 m/s
Mass of stationary cart, m₂ = ?
Initial velocity of stationary cart, v₂ᵢ = 0 m/s
Common final velocity, vₓ = 2 m/s
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(3)(6) + m₂(0) = (3 + m₂)(2)
18 = 6 + 2m₂
12 = 2m₂
m₂ = 6 kg
Therefore:
The mass of the stationary cart is 6 kg.
Mass of moving cart, m₁ = 3 kg
Initial velocity, v₁ᵢ = 6 m/s
Mass of stationary cart, m₂ = ?
Initial velocity of stationary cart, v₂ᵢ = 0 m/s
Common final velocity, vₓ = 2 m/s
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(3)(6) + m₂(0) = (3 + m₂)(2)
18 = 6 + 2m₂
12 = 2m₂
m₂ = 6 kg
Therefore:
The mass of the stationary cart is 6 kg.
🎯 Conclusion
In a collision, conservation of momentum can be rearranged to determine an unknown mass.
54
A 2 kg cart moving at an unknown velocity to the right collides with a 4 kg cart moving at 2 m/s to the left. They stick together and move at 1 m/s to the right. What was the initial velocity of the 2 kg cart?
Practice
A
3 m/s
B
4 m/s
C
5 m/s
D
6 m/s
✅ Show Answer
✔ Correct Answer:
D
(6 m/s)
💡 Explanation
Take the right direction as positive.
m₁ = 2 kg
v₁ᵢ = ?
m₂ = 4 kg
v₂ᵢ = -2 m/s
vₓ = +1 m/s
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(2)v₁ᵢ + (4)(-2) = (2 + 4)(1)
2v₁ᵢ − 8 = 6
2v₁ᵢ = 14
v₁ᵢ = 7 m/s
Therefore:
The initial velocity of the 2 kg cart was 7 m/s to the right.
m₁ = 2 kg
v₁ᵢ = ?
m₂ = 4 kg
v₂ᵢ = -2 m/s
vₓ = +1 m/s
Using conservation of linear momentum:
m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vₓ
(2)v₁ᵢ + (4)(-2) = (2 + 4)(1)
2v₁ᵢ − 8 = 6
2v₁ᵢ = 14
v₁ᵢ = 7 m/s
Therefore:
The initial velocity of the 2 kg cart was 7 m/s to the right.
🎯 Conclusion
Conservation of momentum can be used to determine an unknown initial velocity when the other masses and velocities are known.
55
A 10 kg object initially at rest explodes into two pieces. One piece has a mass of 4 kg and moves at 6 m/s to the right. What is the velocity of the 6 kg piece?
Practice
A
2 m/s to the left
B
4 m/s to the left
C
6 m/s to the left
D
9 m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(4 m/s to the left)
💡 Explanation
Given:
Initial velocity of the system = 0 m/s
m₁ = 4 kg
v₁ₓ = +6 m/s
m₂ = 6 kg
v₂ₓ = ?
Since the object was initially at rest:
pᵢ = 0
By conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(4)(6) + (6)v₂ₓ = 0
24 + 6v₂ₓ = 0
6v₂ₓ = -24
v₂ₓ = -4 m/s
The negative sign indicates motion to the left.
Therefore:
The 6 kg piece moves at 4 m/s to the left.
Initial velocity of the system = 0 m/s
m₁ = 4 kg
v₁ₓ = +6 m/s
m₂ = 6 kg
v₂ₓ = ?
Since the object was initially at rest:
pᵢ = 0
By conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(4)(6) + (6)v₂ₓ = 0
24 + 6v₂ₓ = 0
6v₂ₓ = -24
v₂ₓ = -4 m/s
The negative sign indicates motion to the left.
Therefore:
The 6 kg piece moves at 4 m/s to the left.
🎯 Conclusion
When an object initially at rest explodes, the total momentum remains zero, so the fragments move with equal and opposite total momenta.
56
A 5 kg cannon is initially at rest. It fires a 0.5 kg bullet at 30 m/s to the right. What is the recoil velocity of the cannon?
Practice
A
2 m/s to the left
B
3 m/s to the left
C
5 m/s to the left
D
6 m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(3 m/s to the left)
💡 Explanation
Given:
Mass of cannon, m₁ = 5 kg
Mass of bullet, m₂ = 0.5 kg
Initial velocity of the system = 0 m/s
Final velocity of bullet, v₂ₓ = +30 m/s
Recoil velocity of cannon, v₁ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.5)(30) = 0
5v₁ₓ + 15 = 0
5v₁ₓ = -15
v₁ₓ = -3 m/s
The negative sign indicates that the cannon recoils to the left.
Therefore:
The recoil velocity of the cannon is 3 m/s to the left.
Mass of cannon, m₁ = 5 kg
Mass of bullet, m₂ = 0.5 kg
Initial velocity of the system = 0 m/s
Final velocity of bullet, v₂ₓ = +30 m/s
Recoil velocity of cannon, v₁ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.5)(30) = 0
5v₁ₓ + 15 = 0
5v₁ₓ = -15
v₁ₓ = -3 m/s
The negative sign indicates that the cannon recoils to the left.
Therefore:
The recoil velocity of the cannon is 3 m/s to the left.
🎯 Conclusion
When a cannon fires a bullet, the cannon recoils in the opposite direction so that total momentum remains conserved.
57
A 5 kg gun fires a 0.05 kg bullet at 200 m/s to the right. If the gun was initially at rest, what is its recoil velocity?
Practice
A
1 m/s to the left
B
2 m/s to the left
C
5 m/s to the left
D
10 m/s to the left
✅ Show Answer
✔ Correct Answer:
B
(2 m/s to the left)
💡 Explanation
Given:
Mass of gun, m₁ = 5 kg
Mass of bullet, m₂ = 0.05 kg
Initial velocity of system = 0 m/s
Bullet velocity, v₂ₓ = +200 m/s
Recoil velocity of gun, v₁ₓ = ?
Initial momentum:
pᵢ = 0
By conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.05)(200) = 0
5v₁ₓ + 10 = 0
v₁ₓ = -2 m/s
The negative sign indicates motion to the left.
Therefore:
The recoil velocity of the gun is 2 m/s to the left.
Mass of gun, m₁ = 5 kg
Mass of bullet, m₂ = 0.05 kg
Initial velocity of system = 0 m/s
Bullet velocity, v₂ₓ = +200 m/s
Recoil velocity of gun, v₁ₓ = ?
Initial momentum:
pᵢ = 0
By conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(5)v₁ₓ + (0.05)(200) = 0
5v₁ₓ + 10 = 0
v₁ₓ = -2 m/s
The negative sign indicates motion to the left.
Therefore:
The recoil velocity of the gun is 2 m/s to the left.
🎯 Conclusion
Gun recoil occurs because the backward momentum of the gun balances the forward momentum of the bullet.
58
A person of mass 60 kg jumps from a stationary boat of mass 40 kg with a velocity of 4 m/s to the right relative to the water. What is the velocity of the boat immediately after the jump?
Practice
A
2 m/s to the left
B
4 m/s to the left
C
6 m/s to the left
D
8 m/s to the left
✅ Show Answer
✔ Correct Answer:
C
(6 m/s to the left)
💡 Explanation
Given:
Mass of person, m₁ = 60 kg
Mass of boat, m₂ = 40 kg
Initial velocity of system = 0 m/s
Final velocity of person, v₁ₓ = +4 m/s
Final velocity of boat, v₂ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(60)(4) + (40)v₂ₓ = 0
240 + 40v₂ₓ = 0
40v₂ₓ = -240
v₂ₓ = -6 m/s
The negative sign indicates motion to the left.
Therefore:
The boat moves at 6 m/s to the left.
Mass of person, m₁ = 60 kg
Mass of boat, m₂ = 40 kg
Initial velocity of system = 0 m/s
Final velocity of person, v₁ₓ = +4 m/s
Final velocity of boat, v₂ₓ = ?
Initial total momentum:
pᵢ = 0
Using conservation of linear momentum:
m₁v₁ₓ + m₂v₂ₓ = 0
(60)(4) + (40)v₂ₓ = 0
240 + 40v₂ₓ = 0
40v₂ₓ = -240
v₂ₓ = -6 m/s
The negative sign indicates motion to the left.
Therefore:
The boat moves at 6 m/s to the left.
🎯 Conclusion
When a person jumps from a stationary boat, the boat moves in the opposite direction to conserve total momentum.
59
A 4 kg object moves at 5 m/s to the left. If the positive direction is chosen to be to the right, what is its linear momentum?
Practice
A
20 kg·m/s to the right
B
20 kg·m/s to the left
C
-20 kg·m/s
D
0 kg·m/s
✅ Show Answer
✔ Correct Answer:
C
(-20 kg·m/s)
💡 Explanation
Given:
Mass, m = 4 kg
Positive direction = right
Velocity, v = -5 m/s
The velocity is negative because the object moves to the left.
Linear momentum:
p = mv
p = 4 × (-5)
p = -20 kg·m/s
Therefore:
The linear momentum is -20 kg·m/s.
Mass, m = 4 kg
Positive direction = right
Velocity, v = -5 m/s
The velocity is negative because the object moves to the left.
Linear momentum:
p = mv
p = 4 × (-5)
p = -20 kg·m/s
Therefore:
The linear momentum is -20 kg·m/s.
🎯 Conclusion
In one-dimensional problems, the sign of momentum is determined by the chosen positive direction.
60
A 3 kg object has a velocity of 4 m/s to the right, while a 2 kg object has a velocity of 3 m/s to the left. Taking right as positive, what is the total momentum?
Practice
A
6 kg·m/s to the right
B
6 kg·m/s to the left
C
18 kg·m/s to the right
D
18 kg·m/s to the left
✅ Show Answer
✔ Correct Answer:
A
(6 kg·m/s to the right)
💡 Explanation
Take the right direction as positive.
For object 1:
m₁ = 3 kg
v₁ = +4 m/s
p₁ = 3 × 4 = +12 kg·m/s
For object 2:
m₂ = 2 kg
v₂ = -3 m/s
p₂ = 2 × (-3) = -6 kg·m/s
Total momentum:
pₜ = p₁ + p₂
pₜ = 12 − 6
pₜ = +6 kg·m/s
Therefore:
The resultant momentum is 6 kg·m/s to the right.
For object 1:
m₁ = 3 kg
v₁ = +4 m/s
p₁ = 3 × 4 = +12 kg·m/s
For object 2:
m₂ = 2 kg
v₂ = -3 m/s
p₂ = 2 × (-3) = -6 kg·m/s
Total momentum:
pₜ = p₁ + p₂
pₜ = 12 − 6
pₜ = +6 kg·m/s
Therefore:
The resultant momentum is 6 kg·m/s to the right.
🎯 Conclusion
The resultant momentum in one dimension is the algebraic sum of all individual momenta.
61
A 2 kg object moves with a momentum of 6 kg·m/s to the right, while another 3 kg object moves with a momentum of 8 kg·m/s upward. What is the magnitude of the resultant momentum?
Practice
A
2 kg·m/s
B
10 kg·m/s
C
14 kg·m/s
D
48 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(10 kg·m/s)
💡 Explanation
The two momentum vectors are perpendicular.
Horizontal momentum:
pₓ = 6 kg·m/s
Vertical momentum:
pᵧ = 8 kg·m/s
For perpendicular momentum components:
p = √(pₓ² + pᵧ²)
Therefore:
p = √(6² + 8²)
p = √(36 + 64)
p = √100
p = 10 kg·m/s
Therefore:
The magnitude of the resultant momentum is 10 kg·m/s.
Horizontal momentum:
pₓ = 6 kg·m/s
Vertical momentum:
pᵧ = 8 kg·m/s
For perpendicular momentum components:
p = √(pₓ² + pᵧ²)
Therefore:
p = √(6² + 8²)
p = √(36 + 64)
p = √100
p = 10 kg·m/s
Therefore:
The magnitude of the resultant momentum is 10 kg·m/s.
🎯 Conclusion
For perpendicular momentum components, the magnitude of resultant momentum is found using the Pythagorean theorem.
62
A ball has a momentum of 12 kg·m/s to the east and 5 kg·m/s to the north. What is the magnitude of its resultant momentum?
Practice
A
7 kg·m/s
B
13 kg·m/s
C
17 kg·m/s
D
60 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(13 kg·m/s)
💡 Explanation
Given:
Eastward momentum, pₓ = 12 kg·m/s
Northward momentum, pᵧ = 5 kg·m/s
The two components are perpendicular.
Resultant momentum:
p = √(pₓ² + pᵧ²)
Therefore:
p = √(12² + 5²)
p = √(144 + 25)
p = √169
p = 13 kg·m/s
Therefore:
The magnitude of the resultant momentum is 13 kg·m/s.
Eastward momentum, pₓ = 12 kg·m/s
Northward momentum, pᵧ = 5 kg·m/s
The two components are perpendicular.
Resultant momentum:
p = √(pₓ² + pᵧ²)
Therefore:
p = √(12² + 5²)
p = √(144 + 25)
p = √169
p = 13 kg·m/s
Therefore:
The magnitude of the resultant momentum is 13 kg·m/s.
🎯 Conclusion
The resultant momentum of perpendicular components is greater than either individual component and is found from their vector sum.
63
A 2 kg ball moving at 6 m/s to the east collides with a stationary 2 kg ball. After the collision, the second ball moves at 6 m/s north. Assuming momentum is conserved, what is the magnitude of the momentum of the first ball after the collision?
Practice
A
6 kg·m/s
B
8.49 kg·m/s
C
12 kg·m/s
D
24 kg·m/s
✅ Show Answer
✔ Correct Answer:
B
(8.49 kg·m/s)
💡 Explanation
Initial momentum of the system:
The first ball has:
m₁ = 2 kg
v₁ᵢ = 6 m/s east
Therefore:
pᵢ = m₁v₁ᵢ
pᵢ = 2 × 6
pᵢ = 12 kg·m/s east
After the collision, the second ball has:
p₂ₓ = 2 × 6
p₂ₓ = 12 kg·m/s north
Momentum conservation requires:
p₁ₓ + p₂ₓ = pᵢ
Thus the first ball's final momentum vector must be:
p₁ₓ = (12 kg·m/s east) − (12 kg·m/s north)
Its magnitude is:
|p₁ₓ| = √(12² + 12²)
|p₁ₓ| = √288
|p₁ₓ| ≈ 16.97 kg·m/s
Therefore, the correct magnitude is approximately 16.97 kg·m/s.
The first ball has:
m₁ = 2 kg
v₁ᵢ = 6 m/s east
Therefore:
pᵢ = m₁v₁ᵢ
pᵢ = 2 × 6
pᵢ = 12 kg·m/s east
After the collision, the second ball has:
p₂ₓ = 2 × 6
p₂ₓ = 12 kg·m/s north
Momentum conservation requires:
p₁ₓ + p₂ₓ = pᵢ
Thus the first ball's final momentum vector must be:
p₁ₓ = (12 kg·m/s east) − (12 kg·m/s north)
Its magnitude is:
|p₁ₓ| = √(12² + 12²)
|p₁ₓ| = √288
|p₁ₓ| ≈ 16.97 kg·m/s
Therefore, the correct magnitude is approximately 16.97 kg·m/s.
🎯 Conclusion
In a two-dimensional collision, momentum must be conserved separately as a vector, so the final momentum is determined from its perpendicular components.
64
Two objects have the same linear momentum. Object A has a mass of 2 kg and Object B has a mass of 8 kg. Which object has greater kinetic energy?
Practice
A
Object A has greater kinetic energy.
B
Object B has greater kinetic energy.
C
Both have the same kinetic energy.
D
Both have zero kinetic energy.
✅ Show Answer
✔ Correct Answer:
A
(Object A has greater kinetic energy.)
💡 Explanation
Given:
Both objects have the same momentum, p.
Mass of Object A, m₁ = 2 kg
Mass of Object B, m₂ = 8 kg
Kinetic energy can be written in terms of momentum as:
K = p² / 2m
Since both objects have the same momentum, p² is the same for both.
Therefore, kinetic energy is inversely proportional to mass.
Since Object A has the smaller mass, it has greater kinetic energy.
Therefore:
Object A has greater kinetic energy.
Both objects have the same momentum, p.
Mass of Object A, m₁ = 2 kg
Mass of Object B, m₂ = 8 kg
Kinetic energy can be written in terms of momentum as:
K = p² / 2m
Since both objects have the same momentum, p² is the same for both.
Therefore, kinetic energy is inversely proportional to mass.
Since Object A has the smaller mass, it has greater kinetic energy.
Therefore:
Object A has greater kinetic energy.
🎯 Conclusion
For the same linear momentum, kinetic energy is inversely proportional to mass.
65
Two carts collide on a smooth horizontal track. An external force acts on the system during the collision. Can the total momentum of the two-cart system be assumed to remain constant?
Practice
A
Yes, always.
B
No, because an external force changes the total momentum.
C
Yes, because kinetic energy is conserved.
D
No, because momentum can never be conserved.
✅ Show Answer
✔ Correct Answer:
B
(No, because an external force changes the total momentum.)
💡 Explanation
The conservation of linear momentum applies when the net external impulse on the system is zero or negligible.
Here, an external force acts on the two-cart system during the collision.
Therefore, the external force produces an external impulse:
Jₑₓₜ = FₑₓₜΔt
Hence, the total momentum of the system changes according to:
Δp = Jₑₓₜ
Therefore, the total momentum cannot automatically be assumed to remain constant.
Here, an external force acts on the two-cart system during the collision.
Therefore, the external force produces an external impulse:
Jₑₓₜ = FₑₓₜΔt
Hence, the total momentum of the system changes according to:
Δp = Jₑₓₜ
Therefore, the total momentum cannot automatically be assumed to remain constant.
🎯 Conclusion
Total linear momentum is conserved only when the net external impulse on the chosen system is zero or negligible.
66
A 0.5 kg ball moving at 20 m/s is brought to rest in 0.1 s during a collision. What is the magnitude of the average force acting on the ball?
Practice
A
50 N
B
100 N
C
150 N
D
200 N
✅ Show Answer
✔ Correct Answer:
D
(200 N)
💡 Explanation
Given:
Mass, m = 0.5 kg
Initial velocity, vᵢ = 20 m/s
Final velocity, vₓ = 0 m/s
Collision time, Δt = 0.1 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 0.5 × 20
pᵢ = 10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 10
Δp = -10 kg·m/s
Magnitude of change in momentum:
|Δp| = 10 kg·m/s
Using the impulse-momentum theorem:
FₐᵥₑΔt = |Δp|
Therefore:
Fₐᵥₑ = |Δp| / Δt
Fₐᵥₑ = 10 / 0.1
Fₐᵥₑ = 100 N
Therefore:
The magnitude of the average force is 100 N.
Mass, m = 0.5 kg
Initial velocity, vᵢ = 20 m/s
Final velocity, vₓ = 0 m/s
Collision time, Δt = 0.1 s
Initial momentum:
pᵢ = mvᵢ
pᵢ = 0.5 × 20
pᵢ = 10 kg·m/s
Final momentum:
pₓ = mvₓ
pₓ = 0
Change in momentum:
Δp = pₓ − pᵢ
Δp = 0 − 10
Δp = -10 kg·m/s
Magnitude of change in momentum:
|Δp| = 10 kg·m/s
Using the impulse-momentum theorem:
FₐᵥₑΔt = |Δp|
Therefore:
Fₐᵥₑ = |Δp| / Δt
Fₐᵥₑ = 10 / 0.1
Fₐᵥₑ = 100 N
Therefore:
The magnitude of the average force is 100 N.
🎯 Conclusion
The average force during a collision equals the change in momentum divided by the collision time.
67
A passenger's momentum must change by the same amount during two collisions. In Collision A, the passenger is stopped in 0.05 s. In Collision B, the passenger is stopped in 0.5 s. Which collision produces the smaller average force?
Practice
A
Collision A
B
Collision B
C
Both produce the same average force.
D
The force cannot be compared.
✅ Show Answer
✔ Correct Answer:
B
(Collision B)
💡 Explanation
The change in momentum is the same in both collisions.
Average force is given by:
Fₐᵥₑ = Δp / Δt
Since Δp is the same:
Fₐᵥₑ is inversely proportional to Δt.
Collision A:
Δt = 0.05 s
Collision B:
Δt = 0.5 s
Collision B has a much longer collision time.
Therefore, it produces a smaller average force.
Average force is given by:
Fₐᵥₑ = Δp / Δt
Since Δp is the same:
Fₐᵥₑ is inversely proportional to Δt.
Collision A:
Δt = 0.05 s
Collision B:
Δt = 0.5 s
Collision B has a much longer collision time.
Therefore, it produces a smaller average force.
🎯 Conclusion
For the same change in momentum, increasing the collision time reduces the average force.
68
Why does a person bend their knees while landing after jumping from a height?
Practice
A
To increase the change in momentum.
B
To increase the person's mass.
C
To increase the time over which the momentum changes and reduce the average force.
D
To make the person's momentum zero before landing.
✅ Show Answer
✔ Correct Answer:
C
(To increase the time over which the momentum changes and reduce the average force.)
💡 Explanation
When a person lands, their downward momentum must change to zero.
The change in momentum is approximately fixed for a given landing condition.
Average force is given by:
Fₐᵥₑ = Δp / Δt
By bending the knees, the person increases the time taken to come to rest.
Therefore, the same change in momentum occurs over a longer time interval, which reduces the average force on the body.
This makes the landing safer.
The change in momentum is approximately fixed for a given landing condition.
Average force is given by:
Fₐᵥₑ = Δp / Δt
By bending the knees, the person increases the time taken to come to rest.
Therefore, the same change in momentum occurs over a longer time interval, which reduces the average force on the body.
This makes the landing safer.
🎯 Conclusion
Increasing the time taken to change momentum reduces the average force, which is why bending the knees helps during landing.