Newton's second law of motion
1
A 1 kg ball is moving with an acceleration of 5 m/s². What is the net force, F_net, acting on the ball, and in which direction does it act?
Practice
A
F_net = 5 N, in the direction of the ball's velocity
B
F_net = 5 N, opposite to the ball's velocity
C
F_net = 5 N, in the direction of the acceleration
D
F_net = 0 N, because the ball is moving
✅ Show Answer
✔ Correct Answer:
C
(F_net = 5 N, in the direction of the acceleration)
💡 Explanation
Given:
Mass of the ball, m = 1 kg
Acceleration, a = 5 m/s²
Law applied: Newton's Second Law of Motion.
Using Newton's Second Law:
F_net = ma
F_net = 1 × 5
F_net = 5 N
Direction:
The net force always acts in the same direction as the acceleration.
Therefore, the net force is 5 N in the direction of the acceleration.
Important:
The direction of motion (velocity) does not necessarily determine the direction of the net force.
The acceleration determines the direction of the net force.
Therefore:
Net force = 5 N.
Direction = same as the direction of acceleration.
Mass of the ball, m = 1 kg
Acceleration, a = 5 m/s²
Law applied: Newton's Second Law of Motion.
Using Newton's Second Law:
F_net = ma
F_net = 1 × 5
F_net = 5 N
Direction:
The net force always acts in the same direction as the acceleration.
Therefore, the net force is 5 N in the direction of the acceleration.
Important:
The direction of motion (velocity) does not necessarily determine the direction of the net force.
The acceleration determines the direction of the net force.
Therefore:
Net force = 5 N.
Direction = same as the direction of acceleration.
2
A 1 kg ball is initially at rest. A net force of 10 N is applied to the ball. What is the acceleration of the ball?
Practice
A
0 m/s²
B
5 m/s²
C
10 m/s²
D
20 m/s²
✅ Show Answer
✔ Correct Answer:
C
(10 m/s²)
💡 Explanation
Given:
Mass of the ball, m = 1 kg
Net force, Fₙₑₜ = 10 N
Law applied: Newton's Second Law of Motion.
Newton's Second Law:
Fₙₑₜ = ma
Rearranging for acceleration:
a = Fₙₑₜ / m
a = 10 / 1
a = 10 m/s²
Therefore:
Acceleration = 10 m/s².
The acceleration is in the same direction as Fₙₑₜ.
Important:
The fact that the ball is initially at rest does not mean its acceleration is zero.
A net force of 10 N causes the ball to accelerate at 10 m/s².
Mass of the ball, m = 1 kg
Net force, Fₙₑₜ = 10 N
Law applied: Newton's Second Law of Motion.
Newton's Second Law:
Fₙₑₜ = ma
Rearranging for acceleration:
a = Fₙₑₜ / m
a = 10 / 1
a = 10 m/s²
Therefore:
Acceleration = 10 m/s².
The acceleration is in the same direction as Fₙₑₜ.
Important:
The fact that the ball is initially at rest does not mean its acceleration is zero.
A net force of 10 N causes the ball to accelerate at 10 m/s².
3
A ball is initially at rest. A net force of 10 N is applied to it, producing an acceleration of 5 m/s². What is the mass of the ball?
Practice
A
1 kg
B
2 kg
C
5 kg
D
10 kg
✅ Show Answer
✔ Correct Answer:
B
(2 kg)
💡 Explanation
Given:
Net force, Fₙₑₜ = 10 N
Acceleration, a = 5 m/s²
Law applied: Newton's Second Law of Motion.
Newton's Second Law:
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 10 / 5
m = 2 kg
Therefore:
Mass of the ball = 2 kg.
Net force, Fₙₑₜ = 10 N
Acceleration, a = 5 m/s²
Law applied: Newton's Second Law of Motion.
Newton's Second Law:
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 10 / 5
m = 2 kg
Therefore:
Mass of the ball = 2 kg.
4
A 2 kg ball is initially at rest. After a force is applied, the ball moves with a velocity of 20 m/s in 10 seconds. What is the net force, Fₙₑₜ, acting on the ball?
Practice
A
2 N
B
4 N
C
10 N
D
40 N
✅ Show Answer
✔ Correct Answer:
B
(4 N)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Time, t = 10 s
Step 1: Find acceleration.
a = (v - vᵢ) / t
a = (20 - 0) / 10
a = 2 m/s²
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Fₙₑₜ = 2 × 2
Fₙₑₜ = 4 N
Therefore:
Net force, Fₙₑₜ = 4 N.
The net force acts in the direction of the acceleration.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Time, t = 10 s
Step 1: Find acceleration.
a = (v - vᵢ) / t
a = (20 - 0) / 10
a = 2 m/s²
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Fₙₑₜ = 2 × 2
Fₙₑₜ = 4 N
Therefore:
Net force, Fₙₑₜ = 4 N.
The net force acts in the direction of the acceleration.
5
A 2 kg ball is initially at rest. A net force of 4 N is applied to it, and the ball reaches a velocity of 20 m/s. How much time does the ball take to reach this velocity?
Practice
A
5 s
B
10 s
C
20 s
D
40 s
✅ Show Answer
✔ Correct Answer:
B
(10 s)
💡 Explanation
Given:
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Net force, Fₙₑₜ = 4 N
Step 1: Find acceleration using Newton's Second Law of Motion.
Fₙₑₜ = ma
a = Fₙₑₜ / m
a = 4 / 2
a = 2 m/s²
Step 2: Find the time using the equation of motion.
v = vᵢ + at
20 = 0 + (2 × t)
t = 20 / 2
t = 10 s
Therefore:
Time taken = 10 seconds.
Mass of the ball, m = 2 kg
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Net force, Fₙₑₜ = 4 N
Step 1: Find acceleration using Newton's Second Law of Motion.
Fₙₑₜ = ma
a = Fₙₑₜ / m
a = 4 / 2
a = 2 m/s²
Step 2: Find the time using the equation of motion.
v = vᵢ + at
20 = 0 + (2 × t)
t = 20 / 2
t = 10 s
Therefore:
Time taken = 10 seconds.
6
A net force of 20 N is applied to a ball of mass m, which is initially at rest. The ball moves with a velocity of 20 m/s in 10 seconds. What is the mass of the ball?
Practice
A
2 kg
B
5 kg
C
10 kg
D
20 kg
✅ Show Answer
✔ Correct Answer:
C
(10 kg)
💡 Explanation
Given:
Net force, Fₙₑₜ = 20 N
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Time, t = 10 s
Step 1: Find acceleration.
a = (v - vᵢ) / t
a = (20 - 0) / 10
a = 2 m/s²
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 20 / 2
m = 10 kg
Therefore:
Mass of the ball = 10 kg.
Net force, Fₙₑₜ = 20 N
Initial velocity, vᵢ = 0 m/s
Final velocity, v = 20 m/s
Time, t = 10 s
Step 1: Find acceleration.
a = (v - vᵢ) / t
a = (20 - 0) / 10
a = 2 m/s²
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 20 / 2
m = 10 kg
Therefore:
Mass of the ball = 10 kg.
7
A ball is initially at rest. Its final momentum is 10 kg·m/s after 10 seconds. What is the net force, Fₙₑₜ, acting on the ball?
Practice
A
0.5 N
B
1 N
C
2 N
D
10 N
✅ Show Answer
✔ Correct Answer:
B
(1 N)
💡 Explanation
Given:
Initial momentum, pᵢ = 0 kg·m/s
Final momentum, p_f = 10 kg·m/s
Time, t = 10 s
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = 10 - 0
Δp = 10 kg·m/s
Therefore:
Fₙₑₜ = 10 / 10
Fₙₑₜ = 1 N
The net force acts in the direction of the change in momentum.
Initial momentum, pᵢ = 0 kg·m/s
Final momentum, p_f = 10 kg·m/s
Time, t = 10 s
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = 10 - 0
Δp = 10 kg·m/s
Therefore:
Fₙₑₜ = 10 / 10
Fₙₑₜ = 1 N
The net force acts in the direction of the change in momentum.
8
A ball hits a wall with a momentum of 10 kg·m/s and rebounds with a momentum of 5 kg·m/s in the opposite direction. If the collision takes 5 seconds, what is the force exerted by the wall on the ball?
Practice
A
1 N in the direction of the ball's original motion
B
3 N opposite to the ball's original direction of motion
C
5 N opposite to the ball's original direction of motion
D
15 N in the direction of the ball's original motion
✅ Show Answer
✔ Correct Answer:
B
(3 N opposite to the ball's original direction of motion)
💡 Explanation
Given:
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
Important:
The ball reverses direction after hitting the wall.
Therefore, the final momentum is negative when the initial direction is taken as positive.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign indicates that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N opposite to the ball's original direction of motion.
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
Important:
The ball reverses direction after hitting the wall.
Therefore, the final momentum is negative when the initial direction is taken as positive.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign indicates that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N opposite to the ball's original direction of motion.
9
A ball hits a vertical wall with a momentum of 10 kg·m/s at 180° to the horizontal plane. It rebounds with a momentum of 5 kg·m/s in the opposite direction. If the collision takes 5 seconds, what is the force exerted by the wall on the ball?
Practice
A
1 N toward the wall
B
2 N away from the wall
C
3 N away from the wall
D
15 N toward the wall
✅ Show Answer
✔ Correct Answer:
C
(3 N away from the wall)
💡 Explanation
Given:
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
The ball reverses direction after hitting the vertical wall.
Therefore, if the direction toward the wall is taken as positive, the final momentum is negative.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign means that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N away from the wall.
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
The ball reverses direction after hitting the vertical wall.
Therefore, if the direction toward the wall is taken as positive, the final momentum is negative.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign means that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N away from the wall.
10
A ball hits a vertical wall with a momentum of 10 kg·m/s at 90° to the horizontal plane. It rebounds with a momentum of 5 kg·m/s in the opposite direction. If the collision takes 5 seconds, what is the force exerted by the wall on the ball?
Practice
A
1 N toward the wall
B
2 N away from the wall
C
3 N away from the wall
D
15 N toward the wall
✅ Show Answer
✔ Correct Answer:
C
(3 N away from the wall)
💡 Explanation
Given:
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
The ball hits the vertical wall at 90° to the horizontal plane and rebounds in the opposite direction.
Take the direction toward the wall as positive. Therefore, the final momentum is negative.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign indicates that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N away from the wall.
Initial momentum, pᵢ = +10 kg·m/s
Final momentum, p_f = -5 kg·m/s
Time, t = 5 s
The ball hits the vertical wall at 90° to the horizontal plane and rebounds in the opposite direction.
Take the direction toward the wall as positive. Therefore, the final momentum is negative.
Law applied: Newton's Second Law of Motion in terms of momentum.
Fₙₑₜ = Δp / Δt
Change in momentum:
Δp = p_f - pᵢ
Δp = -5 - (+10)
Δp = -15 kg·m/s
Therefore:
Fₙₑₜ = -15 / 5
Fₙₑₜ = -3 N
The negative sign indicates that the force acts opposite to the ball's original direction of motion.
Therefore:
Force exerted by the wall = 3 N away from the wall.
11
A ball hits a vertical wall with a momentum of 10 kg·m/s at 45° to the horizontal plane. It rebounds with a momentum of 5 kg·m/s in the opposite horizontal direction. If the collision takes 5 seconds, what is the force exerted by the wall on the ball?
Practice
A
1.06 N away from the wall
B
2.12 N away from the wall
C
3.00 N away from the wall
D
4.24 N away from the wall
✅ Show Answer
✔ Correct Answer:
B
(2.12 N away from the wall)
💡 Explanation
Given:
Initial momentum, pᵢ = 10 kg·m/s
Final momentum, p_f = 5 kg·m/s
Angle, θ = 45°
Time, t = 5 s
The wall is vertical, so it changes the horizontal component of momentum.
The horizontal component of the initial momentum is:
pᵢₓ = 10 cos 45°
The horizontal component of the final momentum is opposite in direction:
p_fₓ = -5 cos 45°
Change in horizontal momentum:
Δpₓ = p_fₓ - pᵢₓ
Δpₓ = -5 cos 45° - 10 cos 45°
Δpₓ = -15 cos 45°
Δpₓ ≈ -10.61 kg·m/s
Using Newton's Second Law in terms of momentum:
Fₙₑₜ = Δpₓ / Δt
Fₙₑₜ = -10.61 / 5
Fₙₑₜ ≈ -2.12 N
The negative sign indicates that the force acts opposite to the ball's incoming horizontal direction.
Therefore:
Force exerted by the wall = 2.12 N away from the wall.
Initial momentum, pᵢ = 10 kg·m/s
Final momentum, p_f = 5 kg·m/s
Angle, θ = 45°
Time, t = 5 s
The wall is vertical, so it changes the horizontal component of momentum.
The horizontal component of the initial momentum is:
pᵢₓ = 10 cos 45°
The horizontal component of the final momentum is opposite in direction:
p_fₓ = -5 cos 45°
Change in horizontal momentum:
Δpₓ = p_fₓ - pᵢₓ
Δpₓ = -5 cos 45° - 10 cos 45°
Δpₓ = -15 cos 45°
Δpₓ ≈ -10.61 kg·m/s
Using Newton's Second Law in terms of momentum:
Fₙₑₜ = Δpₓ / Δt
Fₙₑₜ = -10.61 / 5
Fₙₑₜ ≈ -2.12 N
The negative sign indicates that the force acts opposite to the ball's incoming horizontal direction.
Therefore:
Force exerted by the wall = 2.12 N away from the wall.
12
A ball hits a vertical wall with a momentum of 10 kg·m/s at 30° to the horizontal plane. It rebounds with a momentum of 5 kg·m/s in the opposite horizontal direction. If the collision takes 5 seconds, what is the force exerted by the wall on the ball?
Practice
A
1.30 N away from the wall
B
2.60 N away from the wall
C
3.00 N away from the wall
D
5.20 N away from the wall
✅ Show Answer
✔ Correct Answer:
B
(2.60 N away from the wall)
💡 Explanation
Given:
Initial momentum, pᵢ = 10 kg·m/s
Final momentum, p_f = 5 kg·m/s
Angle, θ = 30°
Time, t = 5 s
The wall is vertical, so it changes the horizontal component of momentum.
Initial horizontal momentum:
pᵢₓ = 10 cos 30°
Final horizontal momentum is in the opposite direction:
p_fₓ = -5 cos 30°
Change in horizontal momentum:
Δpₓ = p_fₓ - pᵢₓ
Δpₓ = -5 cos 30° - 10 cos 30°
Δpₓ = -15 cos 30°
Δpₓ ≈ -12.99 kg·m/s
Using Newton's Second Law in terms of momentum:
Fₙₑₜ = Δpₓ / Δt
Fₙₑₜ = -12.99 / 5
Fₙₑₜ ≈ -2.60 N
The negative sign indicates that the force acts opposite to the ball's incoming horizontal direction.
Therefore:
Force exerted by the wall = 2.60 N away from the wall.
Initial momentum, pᵢ = 10 kg·m/s
Final momentum, p_f = 5 kg·m/s
Angle, θ = 30°
Time, t = 5 s
The wall is vertical, so it changes the horizontal component of momentum.
Initial horizontal momentum:
pᵢₓ = 10 cos 30°
Final horizontal momentum is in the opposite direction:
p_fₓ = -5 cos 30°
Change in horizontal momentum:
Δpₓ = p_fₓ - pᵢₓ
Δpₓ = -5 cos 30° - 10 cos 30°
Δpₓ = -15 cos 30°
Δpₓ ≈ -12.99 kg·m/s
Using Newton's Second Law in terms of momentum:
Fₙₑₜ = Δpₓ / Δt
Fₙₑₜ = -12.99 / 5
Fₙₑₜ ≈ -2.60 N
The negative sign indicates that the force acts opposite to the ball's incoming horizontal direction.
Therefore:
Force exerted by the wall = 2.60 N away from the wall.
13
A force F = 3i + 4j N is applied on a ball. What is the magnitude of the force and its direction with respect to the positive horizontal axis?
Practice
A
Magnitude = 3 N, direction = 53.13°
B
Magnitude = 4 N, direction = 36.87°
C
Magnitude = 5 N, direction = 53.13°
D
Magnitude = 5 N, direction = 36.87°
✅ Show Answer
✔ Correct Answer:
C
(Magnitude = 5 N, direction = 53.13°)
💡 Explanation
Given:
Force vector, F = 3i + 4j N
Therefore, Fₓ = 3 N and Fᵧ = 4 N
Step 1: Find the magnitude of the force.
|F| = √(Fₓ² + Fᵧ²)
|F| = √(3² + 4²)
|F| = √(9 + 16)
|F| = √25
|F| = 5 N
Step 2: Find the direction.
tan θ = Fᵧ / Fₓ
tan θ = 4 / 3
θ = tan⁻¹(4 / 3)
θ ≈ 53.13°
Therefore:
Magnitude of force = 5 N.
Direction = 53.13° above the positive horizontal axis.
Force vector, F = 3i + 4j N
Therefore, Fₓ = 3 N and Fᵧ = 4 N
Step 1: Find the magnitude of the force.
|F| = √(Fₓ² + Fᵧ²)
|F| = √(3² + 4²)
|F| = √(9 + 16)
|F| = √25
|F| = 5 N
Step 2: Find the direction.
tan θ = Fᵧ / Fₓ
tan θ = 4 / 3
θ = tan⁻¹(4 / 3)
θ ≈ 53.13°
Therefore:
Magnitude of force = 5 N.
Direction = 53.13° above the positive horizontal axis.
14
A force F = 3i + 4j N is applied to a 2 kg ball. What is the acceleration of the ball and its direction with respect to the positive horizontal axis?
Practice
A
Acceleration = 2.5 m/s², direction = 53.13°
B
Acceleration = 2 m/s², direction = 36.87°
C
Acceleration = 2.5 m/s², direction = 36.87°
D
Acceleration = 5 m/s², direction = 53.13°
✅ Show Answer
✔ Correct Answer:
A
(Acceleration = 2.5 m/s², direction = 53.13°)
💡 Explanation
Given:
Force, F = 3i + 4j N
Mass, m = 2 kg
Using Newton's Second Law of Motion:
Fₙₑₜ = ma
Step 1: Find the acceleration components.
a = Fₙₑₜ / m
a = (3i + 4j) / 2
a = 1.5i + 2j m/s²
Step 2: Find the magnitude of acceleration.
|a| = √(1.5² + 2²)
|a| = √(2.25 + 4)
|a| = √6.25
|a| = 2.5 m/s²
Step 3: Find the direction.
tan θ = aᵧ / aₓ
tan θ = 2 / 1.5
θ = tan⁻¹(2 / 1.5)
θ ≈ 53.13°
Therefore:
Acceleration = 2.5 m/s².
Direction = 53.13° above the positive horizontal axis.
Force, F = 3i + 4j N
Mass, m = 2 kg
Using Newton's Second Law of Motion:
Fₙₑₜ = ma
Step 1: Find the acceleration components.
a = Fₙₑₜ / m
a = (3i + 4j) / 2
a = 1.5i + 2j m/s²
Step 2: Find the magnitude of acceleration.
|a| = √(1.5² + 2²)
|a| = √(2.25 + 4)
|a| = √6.25
|a| = 2.5 m/s²
Step 3: Find the direction.
tan θ = aᵧ / aₓ
tan θ = 2 / 1.5
θ = tan⁻¹(2 / 1.5)
θ ≈ 53.13°
Therefore:
Acceleration = 2.5 m/s².
Direction = 53.13° above the positive horizontal axis.
15
A force F = 3i + 4j N is applied to a ball. The ball has an acceleration of 2.5 m/s². What is the mass of the ball?
Practice
A
1 kg
B
2 kg
C
2.5 kg
D
5 kg
✅ Show Answer
✔ Correct Answer:
B
(2 kg)
💡 Explanation
Given:
Force, F = 3i + 4j N
Acceleration, a = 2.5 m/s²
Step 1: Find the magnitude of the force.
|F| = √(3² + 4²)
|F| = √25
|F| = 5 N
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 5 / 2.5
m = 2 kg
Therefore:
Mass of the ball = 2 kg.
Force, F = 3i + 4j N
Acceleration, a = 2.5 m/s²
Step 1: Find the magnitude of the force.
|F| = √(3² + 4²)
|F| = √25
|F| = 5 N
Step 2: Apply Newton's Second Law of Motion.
Fₙₑₜ = ma
Rearranging for mass:
m = Fₙₑₜ / a
m = 5 / 2.5
m = 2 kg
Therefore:
Mass of the ball = 2 kg.