Impulse
1
A ball is thrown with a force of 10 N for 5 seconds. What is the impulse delivered to the ball?
Practice
A
2 N·s
B
5 N·s
C
50 N·s
D
100 N·s
✅ Show Answer
✔ Correct Answer:
C
(50 N·s)
💡 Explanation
Given:
Force, Fₙₑₜ = 10 N
Time, Δt = 5 s
Law applied: Impulse-Momentum Theorem.
Impulse is given by:
J = Fₙₑₜ × Δt
Substituting the values:
J = 10 × 5
J = 50 N·s
Therefore:
Impulse = 50 N·s.
Force, Fₙₑₜ = 10 N
Time, Δt = 5 s
Law applied: Impulse-Momentum Theorem.
Impulse is given by:
J = Fₙₑₜ × Δt
Substituting the values:
J = 10 × 5
J = 50 N·s
Therefore:
Impulse = 50 N·s.
2
A ball is hit by a cricketer with a force of 10 N for 2 seconds. What is the impulse delivered to the ball?
Practice
A
5 N·s
B
10 N·s
C
20 N·s
D
40 N·s
✅ Show Answer
✔ Correct Answer:
C
(20 N·s)
💡 Explanation
Given:
Force, Fₙₑₜ = 10 N
Time, Δt = 2 s
Law applied: Impulse-Momentum Theorem.
Impulse is given by:
J = Fₙₑₜ × Δt
Substituting the values:
J = 10 × 2
J = 20 N·s
Therefore:
Impulse = 20 N·s.
Force, Fₙₑₜ = 10 N
Time, Δt = 2 s
Law applied: Impulse-Momentum Theorem.
Impulse is given by:
J = Fₙₑₜ × Δt
Substituting the values:
J = 10 × 2
J = 20 N·s
Therefore:
Impulse = 20 N·s.
3
A ball is moving toward a wall with a momentum of 10 kg·m/s. It hits the wall and rebounds with a momentum of 5 kg·m/s in the opposite direction. What is the impulse delivered to the ball?
Practice
A
5 N·s
B
10 N·s
C
15 N·s
D
20 N·s
✅ Show Answer
✔ Correct Answer:
C
(15 N·s)
💡 Explanation
Given:
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
The ball reverses direction after hitting the wall, so the final momentum is negative.
Law applied: Impulse-Momentum Theorem.
Impulse is equal to the change in momentum:
J = Δp
J = p_f - pᵢ
Substituting the values:
J = -5 - (+10)
J = -15 N·s
The negative sign indicates that the impulse is opposite to the ball's initial direction of motion.
Therefore:
Magnitude of impulse = 15 N·s.
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
The ball reverses direction after hitting the wall, so the final momentum is negative.
Law applied: Impulse-Momentum Theorem.
Impulse is equal to the change in momentum:
J = Δp
J = p_f - pᵢ
Substituting the values:
J = -5 - (+10)
J = -15 N·s
The negative sign indicates that the impulse is opposite to the ball's initial direction of motion.
Therefore:
Magnitude of impulse = 15 N·s.
4
A 2 kg ball is moving with a velocity of 10 m/s. It hits a wall and rebounds with a velocity of 5 m/s in the opposite direction. What is the impulse delivered to the ball?
Practice
A
10 N·s
B
20 N·s
C
30 N·s
D
40 N·s
✅ Show Answer
✔ Correct Answer:
C
(30 N·s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, v_f = -5 m/s
The ball rebounds, so its final velocity is in the opposite direction.
Step 1: Find the initial momentum.
pᵢ = mvᵢ
pᵢ = 2 × 10
pᵢ = 20 kg·m/s
Step 2: Find the final momentum.
p_f = mv_f
p_f = 2 × (-5)
p_f = -10 kg·m/s
Step 3: Find the impulse.
J = Δp
J = p_f - pᵢ
J = -10 - 20
J = -30 N·s
The negative sign indicates that the impulse acts opposite to the ball's initial direction of motion.
Therefore:
Magnitude of impulse = 30 N·s.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, v_f = -5 m/s
The ball rebounds, so its final velocity is in the opposite direction.
Step 1: Find the initial momentum.
pᵢ = mvᵢ
pᵢ = 2 × 10
pᵢ = 20 kg·m/s
Step 2: Find the final momentum.
p_f = mv_f
p_f = 2 × (-5)
p_f = -10 kg·m/s
Step 3: Find the impulse.
J = Δp
J = p_f - pᵢ
J = -10 - 20
J = -30 N·s
The negative sign indicates that the impulse acts opposite to the ball's initial direction of motion.
Therefore:
Magnitude of impulse = 30 N·s.
5
A 2 kg ball hits a wall and remains in contact with the wall for 2 seconds. During the contact, the ball experiences an acceleration of 2 m/s² away from the wall. What is the impulse delivered by the wall to the ball?
Practice
A
4 N·s
B
8 N·s
C
10 N·s
D
16 N·s
✅ Show Answer
✔ Correct Answer:
B
(8 N·s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Acceleration during contact, a = 2 m/s²
Contact time, Δt = 2 s
Law applied: Impulse is the product of net force and time.
First, find the net force using Newton's Second Law:
Fₙₑₜ = ma
Fₙₑₜ = 2 × 2
Fₙₑₜ = 4 N
Now calculate the impulse:
J = Fₙₑₜ × Δt
J = 4 × 2
J = 8 N·s
Therefore:
Impulse = 8 N·s.
The impulse acts away from the wall, in the direction of the acceleration.
Mass of the ball, m = 2 kg
Acceleration during contact, a = 2 m/s²
Contact time, Δt = 2 s
Law applied: Impulse is the product of net force and time.
First, find the net force using Newton's Second Law:
Fₙₑₜ = ma
Fₙₑₜ = 2 × 2
Fₙₑₜ = 4 N
Now calculate the impulse:
J = Fₙₑₜ × Δt
J = 4 × 2
J = 8 N·s
Therefore:
Impulse = 8 N·s.
The impulse acts away from the wall, in the direction of the acceleration.
6
A 2 kg ball is thrown with an acceleration of 5 m/s² for 2 seconds. What is the impulse acting on the ball?
Practice
A
5 N·s
B
10 N·s
C
20 N·s
D
40 N·s
✅ Show Answer
✔ Correct Answer:
C
(20 N·s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Acceleration, a = 5 m/s²
Time, Δt = 2 s
Law applied: Impulse-Momentum Theorem.
First, find the net force using Newton's Second Law:
Fₙₑₜ = ma
Fₙₑₜ = 2 × 5
Fₙₑₜ = 10 N
Now calculate the impulse:
J = Fₙₑₜ × Δt
J = 10 × 2
J = 20 N·s
Therefore:
Impulse = 20 N·s.
The impulse acts in the direction of the acceleration.
Mass of the ball, m = 2 kg
Acceleration, a = 5 m/s²
Time, Δt = 2 s
Law applied: Impulse-Momentum Theorem.
First, find the net force using Newton's Second Law:
Fₙₑₜ = ma
Fₙₑₜ = 2 × 5
Fₙₑₜ = 10 N
Now calculate the impulse:
J = Fₙₑₜ × Δt
J = 10 × 2
J = 20 N·s
Therefore:
Impulse = 20 N·s.
The impulse acts in the direction of the acceleration.
7
An impulse of 10 N·s acts on a ball with a net force of 20 N. What is the time for which the force acts on the ball?
Practice
A
0.25 s
B
0.5 s
C
1 s
D
2 s
✅ Show Answer
✔ Correct Answer:
B
(0.5 s)
💡 Explanation
Given:
Impulse, J = 10 N·s
Net force, Fₙₑₜ = 20 N
Law applied: Impulse is the product of net force and time.
J = Fₙₑₜ × Δt
Rearranging for time:
Δt = J / Fₙₑₜ
Δt = 10 / 20
Δt = 0.5 s
Therefore:
Time taken = 0.5 seconds.
Impulse, J = 10 N·s
Net force, Fₙₑₜ = 20 N
Law applied: Impulse is the product of net force and time.
J = Fₙₑₜ × Δt
Rearranging for time:
Δt = J / Fₙₑₜ
Δt = 10 / 20
Δt = 0.5 s
Therefore:
Time taken = 0.5 seconds.
8
A 2 kg ball hits a wall with a velocity of 10 m/s and rebounds with a velocity of 5 m/s in the opposite direction. If the wall exerts a force of 10 N on the ball, what is the contact time between the ball and the wall?
Practice
A
1 s
B
2 s
C
3 s
D
5 s
✅ Show Answer
✔ Correct Answer:
C
(3 s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, v_f = -5 m/s
Force exerted by the wall, Fₙₑₜ = 10 N
The ball rebounds, so the final velocity is in the opposite direction.
Step 1: Find the initial momentum.
pᵢ = mvᵢ
pᵢ = 2 × 10
pᵢ = 20 kg·m/s
Step 2: Find the final momentum.
p_f = mv_f
p_f = 2 × (-5)
p_f = -10 kg·m/s
Step 3: Find the impulse.
J = Δp
J = p_f - pᵢ
J = -10 - 20
J = -30 N·s
The magnitude of the impulse is 30 N·s.
Step 4: Use the impulse formula to find contact time.
J = Fₙₑₜ × Δt
Δt = J / Fₙₑₜ
Δt = 30 / 10
Δt = 3 s
Therefore:
Contact time = 3 seconds.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, v_f = -5 m/s
Force exerted by the wall, Fₙₑₜ = 10 N
The ball rebounds, so the final velocity is in the opposite direction.
Step 1: Find the initial momentum.
pᵢ = mvᵢ
pᵢ = 2 × 10
pᵢ = 20 kg·m/s
Step 2: Find the final momentum.
p_f = mv_f
p_f = 2 × (-5)
p_f = -10 kg·m/s
Step 3: Find the impulse.
J = Δp
J = p_f - pᵢ
J = -10 - 20
J = -30 N·s
The magnitude of the impulse is 30 N·s.
Step 4: Use the impulse formula to find contact time.
J = Fₙₑₜ × Δt
Δt = J / Fₙₑₜ
Δt = 30 / 10
Δt = 3 s
Therefore:
Contact time = 3 seconds.