IB Physics Quiz: Understand Forces And Momentum

20 questions · exam conditions

0:00

Understand Forces And MomentumQuestion 1 of 20

A 2.0 kg cart moving at 3.0 m s⁻¹ on a frictionless track collides with a stationary 1.0 kg cart. The two carts stick together after the collision. What is the loss in kinetic energy during this collision?

0 J

3.0 J

6.0 J

9.0 J

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IB Physics Quiz

IB Physics Quiz: Understand Forces And Momentum

Practice Understand Forces And Momentum in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

0 J
3.0 J (correct answer)
6.0 J
9.0 J

Explanation: This is a perfectly inelastic collision, so momentum is conserved but kinetic energy is not. Initial momentum

p_i = m_1v_1 = (2.0 \text{ kg})(3.0 \text{ m s⁻¹}) = 6.0 \text{ kg m s⁻¹}

. After the collision, the total mass is

M = 2.0 + 1.0 = 3.0 \text{ kg}

. Final momentum

p_f = Mv_f = 3.0v_f

. By conservation of momentum,

p_i=p_f

, so

6.0 = 3.0v_f

, which gives the final velocity

v_f = 2.0 \text{ m s⁻¹}

. The initial kinetic energy is

E_{ki} = \frac{1}{2}m_1v_1^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}

. The final kinetic energy is

E_{kf} = \frac{1}{2}Mv_f^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}

. The loss in kinetic energy is

E_{ki} - E_{kf} = 9.0 \text{ J} - 6.0 \text{ J} = 3.0 \text{ J}

Question 2

A person of mass 60 kg stands on a weighing scale in an elevator. The elevator is moving downwards and slowing down with a deceleration of 2.0 m s⁻². What is the reading on the weighing scale? (Use $g = 10$ m s⁻²).

120 N
480 N
600 N
720 N (correct answer)

Explanation: The weighing scale reads the normal force

F_N

. Decelerating while moving downwards is equivalent to having an upward acceleration of

a = +2.0 \text{ m s⁻²}

. According to Newton's second law, the net force is

F_{net} = ma

. The forces acting on the person are the normal force

F_N

(upwards) and gravity

mg

(downwards). So,

F_{net} = F_N - mg

. Setting this equal to

ma

gives

F_N - mg = ma

. Rearranging for

F_N

gives

F_N = mg + ma = m(g+a)

. Plugging in the values:

F_N = 60 \text{ kg} \times (10 \text{ m s⁻²} + 2.0 \text{ m s⁻²}) = 60 \times 12 = 720 \text{ N}

Question 3

A stationary object of mass (3M) explodes into two pieces. One piece of mass $M$ moves to the right with speed $v$ . What is the velocity of the second piece, which has mass (2M)?

Speed $v/2$ to the right
Speed $v/2$ to the left (correct answer)
Speed $2v$ to the left
Speed $v$ to the left

Explanation: The total momentum of the system must be conserved. The initial momentum is zero since the object is stationary. Let the velocity of the second piece be

v_2

. The total final momentum is the sum of the momenta of the two pieces:

p_{final} = (M)(+v) + (2M)(v_2)

. By conservation of momentum,

p_{initial} = p_{final}

, so

0 = Mv + 2Mv_2

. Solving for

v_2

gives

2Mv_2 = -Mv

, so

v_2 = -Mv / (2M) = -v/2

. The negative sign indicates the direction is to the left.

Question 4

A block of mass $m_1 = 3.0$ kg rests on a frictionless horizontal table. It is connected by a light string that passes over a frictionless pulley to a second, hanging block of mass $m_2 = 1.0$ kg. The system is released from rest. What is the magnitude of the acceleration of the system? (Use $g = 10$ m s⁻²).

10 m s⁻²
5.0 m s⁻²
3.3 m s⁻²
2.5 m s⁻² (correct answer)

Explanation: The net force driving the system's motion is the weight of the hanging mass,

F_{net} = m_2g = (1.0 \text{ kg})(10 \text{ m s⁻²}) = 10 \text{ N}

. This force must accelerate the total mass of the system, which is

m_{total} = m_1 + m_2 = 3.0 \text{ kg} + 1.0 \text{ kg} = 4.0 \text{ kg}

. Using Newton's second law for the entire system,

F_{net} = m_{total}a

, we can solve for the acceleration:

a = F_{net} / m_{total} = 10 \text{ N} / 4.0 \text{ kg} = 2.5 \text{ m s⁻²}

Question 5

An object rests on a horizontal surface. The coefficient of static friction is $\mu_s$ and the coefficient of kinetic friction is $\mu_d$ . It is known that $\mu_s > \mu_d$ . A horizontal force $F$ is applied to the object, and its magnitude is gradually increased from zero. The object starts to move when $F$ reaches the value $F_{start}$ . To keep the object moving at a constant velocity, the force must be adjusted to a value $F_{keep}$ . Which statement is correct?

$F_{start} > F_{keep}$ (correct answer)
$F_{start} < F_{keep}$
$F_{start} = F_{keep}$
The relationship depends on the mass of the object.

Explanation: The force required to start the motion,

F_{start}

, must overcome the maximum static friction, which is

F_{s,max} = \mu_s F_N

. The force required to keep the object moving at a constant velocity,

F_{keep}

, must be equal in magnitude to the kinetic friction force,

F_d = \mu_d F_N

. Since the problem states

\mu_s > \mu_d

(which is physically typical), it follows that

\mu_s F_N > \mu_d F_N

, and therefore

F_{start} > F_{keep}

Question 6

A car is successfully negotiating a flat, unbanked circular turn at a constant speed. Which force provides the necessary centripetal force for the car's circular motion?

The normal force from the road.
The kinetic friction force between the tires and the road.
The static friction force between the tires and the road. (correct answer)
The gravitational force acting on the car.

Explanation: Centripetal force is a net force that acts towards the center of a circular path. For a car on a flat track, this inward force is provided by friction between the tires and the road. Since the tires are rolling without slipping, the point of contact with the road is momentarily at rest relative to the road surface. Therefore, it is the force of static friction that acts towards the center of the turn, providing the centripetal force.

Question 7

A billiard ball of mass $m$ moving with speed $v$ collides head-on elastically with an identical billiard ball that is initially at rest. What are the velocities of the two balls immediately after the collision?

The first ball stops, and the second ball moves with speed $v$ . (correct answer)
Both balls move together with a combined speed of $v/2$ .
The first ball rebounds with speed $v$ , and the second ball remains at rest.
The first ball moves with speed $v/2$ , and the second ball moves with speed $v/2$ .

Explanation: In a one-dimensional elastic collision between two objects of equal mass, where one is initially at rest, the objects exchange velocities. The moving object comes to a stop, and the stationary object moves off with the initial velocity of the first object. This can be proven by applying the principles of conservation of momentum (

mv = mv_{1f} + mv_{2f}

) and conservation of kinetic energy (

\frac{1}{2}mv^2 = \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2

). Solving these two equations simultaneously yields

v_{1f} = 0

and

v_{2f} = v

Question 8

Two small objects, P and Q, are on a horizontal turntable rotating at a constant angular velocity $\omega$ . Object P is at a distance $R$ from the center, and object Q is at a distance $2R$ from the center. What is the ratio of the centripetal acceleration of P to the centripetal acceleration of Q, $a_P / a_Q$ ?

2
1
1/2 (correct answer)
1/4

Explanation: The formula for centripetal acceleration

a

in terms of angular velocity

\omega

and radius

r

a = \omega^2 r

. Since both objects are on the same turntable, they have the same angular velocity

\omega

. Therefore, the centripetal acceleration is directly proportional to the radius

r

. The acceleration of P is

a_P = \omega^2 R

, and the acceleration of Q is

a_Q = \omega^2 (2R)

. The ratio is

a_P / a_Q = (\omega^2 R) / (\omega^2 2R) = 1/2

Question 9

Two identical balls are thrown horizontally at a vertical wall with the same speed, $v$ . Ball 1 hits the wall and comes to a complete stop. Ball 2 hits the wall and rebounds horizontally with speed $v$ . What is the ratio of the magnitude of the impulse delivered by the wall to Ball 2 compared to Ball 1?

1:1
$\sqrt{2}:1$
2:1 (correct answer)
1:2

Explanation: Impulse is the change in momentum (

J = \Delta p

). Let the initial momentum be

p = mv

. For Ball 1, the final momentum is 0, so the change in momentum is

\Delta p_1 = 0 - mv = -mv

. The magnitude is

|J_1| = mv

. For Ball 2, the final momentum is in the opposite direction, so it is

-mv

. The change in momentum is

\Delta p_2 = (-mv) - mv = -2mv

. The magnitude is

|J_2| = 2mv

. The ratio of the magnitudes

|J_2| / |J_1|

(2mv) / (mv) = 2

, which is a ratio of 2:1.

Question 10

A block of mass 0.50 kg is attached to a horizontal spring with a spring constant of 200 N m⁻¹. The block is pulled 0.10 m from its equilibrium position on a frictionless surface and then released from rest. What is the magnitude of the initial acceleration of the block?

10 m s⁻²
20 m s⁻²
40 m s⁻² (correct answer)
400 m s⁻²

Explanation: First, calculate the restoring force from the spring using Hooke's Law,

F = kx

. The magnitude of the force is

F = (200 \text{ N m⁻¹})(0.10 \text{ m}) = 20 \text{ N}

. This is the net force acting on the block at the moment of release. Next, use Newton's second law,

F = ma

, to find the acceleration. Rearranging gives

a = F/m = 20 \text{ N} / 0.50 \text{ kg} = 40 \text{ m s⁻²}

Question 11

A ball of mass 0.20 kg is initially at rest. A force acts on it for 0.050 s. The magnitude of the force is not constant, but its average value over the time interval is 40 N. What is the final speed of the ball?

2.0 m s⁻¹
8.0 m s⁻¹
10 m s⁻¹ (correct answer)
200 m s⁻¹

Explanation: The impulse-momentum theorem states that the impulse delivered to an object is equal to its change in momentum (

J = \Delta p

). The impulse can be calculated using the average force:

J = F_{avg} \Delta t

. So,

F_{avg} \Delta t = p_{final} - p_{initial}

. Since the ball starts from rest,

p_{initial} = 0

. Therefore,

F_{avg} \Delta t = mv_{final}

. Plugging in the values:

(40 \text{ N})(0.050 \text{ s}) = (0.20 \text{ kg})v_{final}

. This gives

2.0 \text{ Ns} = 0.20v_{final}

. Solving for the final speed,

v_{final} = 2.0 / 0.20 = 10 \text{ m s⁻¹}

Question 12

A 5.0 kg block is pulled by a constant horizontal force of 20 N across a rough horizontal surface, causing it to move at a constant velocity of 2.0 m s⁻¹. What is the coefficient of kinetic friction, $\mu_d$ , between the block and the surface? (Assume the acceleration of free fall $g$ is 10 m s⁻²).

0.20
0.40 (correct answer)
0.80
4.0

Explanation: Since the block moves at a constant velocity, its acceleration is zero. According to Newton's second law, the net force on the block must be zero. The horizontal forces are the applied force (20 N) and the kinetic friction force (

F_f

). For the net force to be zero,

F_f = 20

N. The normal force

F_N

is equal to the weight of the block,

F_N = mg = 5.0 \text{ kg} \times 10 \text{ m s⁻²} = 50

N. The coefficient of kinetic friction is given by

\mu_d = F_f / F_N = 20 \text{ N} / 50 \text{ N} = 0.40

Question 13

An object of mass 2.0 kg has an initial velocity of 3.0 m s⁻¹ in the positive x-direction. A constant force of 8.0 N is then applied to the object for 1.0 s in the positive y-direction. What is the magnitude of the final momentum of the object?

8.0 kg m s⁻¹
10 kg m s⁻¹ (correct answer)
14 kg m s⁻¹
2.0 kg m s⁻¹

Explanation: Momentum is a vector quantity. The initial momentum is entirely in the x-direction:

p_x = mv_x = 2.0 \text{ kg} \times 3.0 \text{ m s⁻¹} = 6.0 \text{ kg m s⁻¹}

. The force creates an impulse

J_y = F_y \Delta t = 8.0 \text{ N} \times 1.0 \text{ s} = 8.0 \text{ Ns}

in the y-direction. This impulse is equal to the change in momentum in the y-direction. Since the initial y-momentum was zero, the final y-momentum is

p_y = 8.0 \text{ kg m s⁻¹}

. The final momentum is the vector sum of the components. The magnitude is found using the Pythagorean theorem:

p_{final} = \sqrt{p_x^2 + p_y^2} = \sqrt{(6.0)^2 + (8.0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ kg m s⁻¹}

Question 14

An object is held in static equilibrium by two ropes attached to a horizontal ceiling. One rope makes an angle of 30° with the ceiling, and the other makes an angle of 60° with the ceiling. The tension in the 30° rope is $T_1$ and the tension in the 60° rope is $T_2$ . What is the ratio $T_1 / T_2$ ?

$\sqrt{3}$
1
2
$1/\sqrt{3}$ (correct answer)

Explanation: For the object to be in static equilibrium, the net force in both the horizontal and vertical directions must be zero. Considering the horizontal forces: the horizontal component of

T_1

T_1 \cos(30°)

and it pulls in one direction. The horizontal component of

T_2

T_2 \cos(60°)

and it pulls in the opposite direction. For equilibrium, these must be equal:

T_1 \cos(30°) = T_2 \cos(60°)

. We want the ratio

T_1 / T_2

, so we rearrange the equation:

T_1 / T_2 = \cos(60°) / \cos(30°)

. Using the values

\cos(60°) = 1/2

and

\cos(30°) = \sqrt{3}/2

, the ratio is

(1/2) / (\sqrt{3}/2) = 1/\sqrt{3}

Question 15

A car of mass $m$ travels at a constant speed $v$ over the crest of a hill. The crest of the hill can be modeled as a circular arc of radius $r$ . What is the magnitude of the normal force exerted by the road on the car at the highest point of the hill?

$mg + \frac{mv^2}{r}$
$\frac{mv^2}{r}$
$mg - \frac{mv^2}{r}$ (correct answer)
$mg$

Explanation: At the crest of the hill, the car is undergoing circular motion. The net force towards the center of the circle (downwards) provides the centripetal force. The forces acting on the car are its weight

mg

(downwards) and the normal force

F_N

from the road (upwards). The net force is

F_{net} = mg - F_N

. This net force must equal the centripetal force,

F_c = \frac{mv^2}{r}

. Therefore,

mg - F_N = \frac{mv^2}{r}

. Rearranging for the normal force gives

F_N = mg - \frac{mv^2}{r}

Question 16

A large truck collides head-on with a small car. During the collision, the magnitude of the force exerted by the truck on the car is $F_{tc}$ and the magnitude of the force exerted by the car on the truck is $F_{ct}$ . Which statement is correct about these forces?

$F_{tc} = F_{ct}$ at all moments during the collision. (correct answer)
$F_{tc} > F_{ct}$ because the truck has a larger mass.
$F_{tc} < F_{ct}$ because the car experiences a greater change in velocity.
The relationship between $F_{tc}$ and $F_{ct}$ depends on the initial speeds of the vehicles.

Explanation: According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The force the truck exerts on the car and the force the car exerts on the truck form an action-reaction pair. Therefore, their magnitudes must be equal at all times during the collision. The difference in mass results in different accelerations (

a = F/m

), but the forces are identical in magnitude.

Question 17

A 0.50 kg ball is thrown vertically upward with an initial speed of 10 m/s. At the moment when the ball is moving downward at 6.0 m/s, what is the magnitude of the change in momentum from the initial throw?

2.0 kg⋅m/s
8.0 kg⋅m/s (correct answer)
10 kg⋅m/s
16 kg⋅m/s

Explanation: Taking upward as positive, initial momentum is

p_i = mv_i = 0.50(10) = 5.0\text{ kg⋅m/s}

. When moving downward at 6.0 m/s, the momentum is

p_f = mv_f = 0.50(-6.0) = -3.0\text{ kg⋅m/s}

. The change in momentum is

\Delta p = p_f - p_i = -3.0 - 5.0 = -8.0\text{ kg⋅m/s}

. The magnitude is 8.0 kg⋅m/s. Choice A represents the difference in speeds only. Choice C uses only initial momentum. Choice D incorrectly adds the momentum magnitudes.

Question 18

Two identical 2.0 kg blocks are connected by a light string over a massless pulley. One block hangs vertically while the other slides on a horizontal surface with coefficient of kinetic friction μ = 0.30. What is the tension in the string when the system is accelerating?

8.8 N
11.2 N
12.7 N (correct answer)
19.6 N

Explanation: For the hanging block:

mg - T = ma

. For the sliding block:

T - \mu mg = ma

. Adding these equations:

mg - \mu mg = 2ma

, so

a = \frac{g(1-\mu)}{2} = \frac{9.8(1-0.30)}{2} = \frac{9.8(0.70)}{2} = 3.43\text{ m/s}^2

. Substituting back:

T = mg - ma = 2.0(9.8) - 2.0(3.43) = 19.6 - 6.86 = 12.7\text{ N}

. Choice A uses

T = \mu mg

. Choice B uses incorrect acceleration. Choice D ignores friction entirely.

Question 19

A 1200 kg car traveling at 20 m/s applies brakes and skids to a stop over a distance of 80 m. If the same car were traveling at 30 m/s and applied the same braking force, what would be the stopping distance?

120 m
150 m
180 m (correct answer)
240 m

Explanation: First, find the braking force from the first scenario. Using

v^2 = v_0^2 + 2as

0 = (20)^2 + 2a(80)

, so

a = -\frac{400}{160} = -2.5\text{ m/s}^2

. The braking force is

F = ma = 1200(-2.5) = -3000\text{ N}

. For the second scenario with the same force,

a = -2.5\text{ m/s}^2

. Using

v^2 = v_0^2 + 2as

0 = (30)^2 + 2(-2.5)s

, so

s = \frac{900}{5} = 180\text{ m}

. Choice A assumes linear scaling. Choice B uses incorrect proportional reasoning. Choice D assumes quadratic scaling without considering the physics.

Question 20

The kinetic energy of an object of mass $m$ is increased by a factor of 16. By what factor does the magnitude of its momentum change?

It increases by a factor of 4. (correct answer)
It increases by a factor of 16.
It increases by a factor of 256.
It increases by a factor of 8.

Explanation: Kinetic energy (

E_k

) and momentum (

p

) are related by the equation

E_k = \frac{p^2}{2m}

. Rearranging for momentum gives

p = \sqrt{2mE_k}

. This shows that momentum is proportional to the square root of the kinetic energy (

p \propto \sqrt{E_k}

). If the kinetic energy is multiplied by 16, the new momentum

p'

will be proportional to

\sqrt{16E_k} = 4\sqrt{E_k}

. Therefore, the momentum increases by a factor of 4.