IB Physics Quiz: Apply Forces And Momentum

20 questions · exam conditions

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Apply Forces And MomentumQuestion 1 of 20

A block of mass $m$ moving with speed $v$ makes a head-on perfectly inelastic collision with a stationary block of mass (3m). The two blocks stick together. What fraction of the initial kinetic energy is converted to other forms of energy?

1/4

1/3

2/3

3/4

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IB Physics Quiz

IB Physics Quiz: Apply Forces And Momentum

Practice Apply 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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This quiz focuses on Apply Forces And Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

1/4
1/3
2/3
3/4 (correct answer)

Explanation: Initial momentum is

p_i = mv

. Initial kinetic energy is

KE_i = \frac{1}{2}mv^2

. After the collision, the combined mass is (4m). By conservation of momentum,

mv = (4m)v_f

, so the final velocity is

v_f = v/4

. The final kinetic energy is

KE_f = \frac{1}{2}(4m)v_f^2 = \frac{1}{2}(4m)(v/4)^2 = \frac{1}{2}(4m)(v^2/16) = \frac{1}{8}mv^2

. The energy lost is

\Delta KE = KE_i - KE_f = \frac{1}{2}mv^2 - \frac{1}{8}mv^2 = \frac{3}{8}mv^2

. The fraction of energy lost is

\frac{\Delta KE}{KE_i} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4}

Question 2

A large truck of mass $M$ collides with a small car of mass $m$ (where $M > m$ ). During the collision, which of the following statements is always correct?

The force exerted by the truck on the car is greater than the force exerted by the car on the truck.
The change in momentum of the truck is equal in magnitude to the change in momentum of the car. (correct answer)
The acceleration of the truck is equal in magnitude to the acceleration of the car.
The change in kinetic energy of the truck is equal to the change in kinetic energy of the car.

Explanation: According to Newton's third law, the force the truck exerts on the car is equal in magnitude and opposite in direction to the force the car exerts on the truck. Since the forces are equal and act for the same time interval, the impulses (

F\Delta t

) on both vehicles are equal in magnitude. Since impulse is equal to the change in momentum, the magnitudes of the changes in momentum are equal.

Question 3

A block of mass 3.0 kg rests on a rough horizontal table (coefficient of kinetic friction 0.20). It is connected by a light string over a frictionless pulley to a hanging block of mass 1.0 kg. The system is released from rest. What is the acceleration of the system? (Use $g = 10 \text{ m s}^{-2}$ )

1.0 m s⁻² (correct answer)
1.3 m s⁻²
2.5 m s⁻²
3.3 m s⁻²

Explanation: The driving force is the weight of the hanging block:

F_{drive} = m_{hang}g = 1.0 \times 10 = 10 \text{ N}

. The opposing friction force is

F_f = \mu_k F_N = \mu_k m_{table}g = 0.20 \times 3.0 \times 10 = 6.0 \text{ N}

. The net force on the system is

F_{net} = F_{drive} - F_f = 10 - 6.0 = 4.0 \text{ N}

. This net force accelerates the total mass of the system,

M_{total} = m_{hang} + m_{table} = 1.0 + 3.0 = 4.0 \text{ kg}

. Using Newton's second law,

a = F_{net} / M_{total} = 4.0 / 4.0 = 1.0 \text{ m s}^{-2}

Question 4

A small sphere falls through a viscous fluid and reaches a terminal velocity $v_t$ . If the experiment is repeated with a sphere of the same size and material but in a fluid with four times the viscosity, what will be the new terminal velocity?

$v_t/4$ (correct answer)
$v_t/2$
$2v_t$
$4v_t$

Explanation: At terminal velocity, the net force on the sphere is zero. The downward gravitational force (weight) is balanced by the upward buoyant force and the viscous drag force. Thus,

F_g = F_b + F_d

. The drag force is given by Stokes' Law,

F_d = 6\pi\eta r v_t

. Since

F_g

and

F_b

are constant for the same sphere and fluid density, the drag force

F_d

must also be constant at terminal velocity. This means the product

\eta v_t

is constant. If the viscosity

\eta

is multiplied by 4, the terminal velocity

v_t

must be divided by 4 to keep the product constant.

Question 5

A person stands on a weighing scale in a lift. The scale shows a reading $R_1$ as the lift starts moving upwards from rest, $R_2$ as it moves upwards at a constant speed, and $R_3$ as it slows to a stop at a higher floor. Which statement correctly compares the magnitudes of the readings?

$R_1 > R_2 > R_3$ (correct answer)
$R_3 > R_2 > R_1$
$R_1 = R_3 > R_2$
$R_2 > R_1 > R_3$

Explanation: The scale reading equals the normal force

F_N

. Let

m

be the person's mass. When accelerating upwards (starting),

F_N - mg = ma

, so

R_1 = m(g+a) > mg

. When moving at constant speed,

a=0

, so

F_N - mg = 0

, and

R_2 = mg

. When decelerating upwards (slowing), the acceleration is downwards, so

F_N - mg = -ma

, and

R_3 = m(g-a) < mg

. Therefore, the order of the readings is

R_1 > R_2 > R_3

Question 6

An object of mass 5.0 kg is initially at rest. It is subjected to two perpendicular forces, $F_1 = 30$ N and $F_2 = 40$ N. What is the magnitude of the object's acceleration?

2.0 m s⁻²
10 m s⁻² (correct answer)
14 m s⁻²
50 m s⁻²

Explanation: The net force on the object is the vector sum of the two perpendicular forces. The magnitude of the net force can be found using the Pythagorean theorem:

F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ N}

. According to Newton's second law,

F_{net} = ma

. Therefore, the acceleration is

a = F_{net} / m = 50 \text{ N} / 5.0 \text{ kg} = 10 \text{ m s}^{-2}

Question 7

Three blocks with masses $m_1 = 1.0$ kg, $m_2 = 2.0$ kg, and $m_3 = 3.0$ kg are connected by light, inextensible strings. The blocks are pulled along a frictionless horizontal surface by a horizontal force $F = 12$ N applied to $m_3$ . What is the tension in the string connecting $m_1$ and $m_2$ ?

6.0 N
4.0 N
2.0 N (correct answer)
12 N

Explanation: First, find the acceleration of the entire system. The total mass is

M = m_1 + m_2 + m_3 = 1.0 + 2.0 + 3.0 = 6.0 \text{ kg}

. The net force on the system is

F = 12 \text{ N}

. The acceleration is

a = F/M = 12 \text{ N} / 6.0 \text{ kg} = 2.0 \text{ m s}^{-2}

. The tension in the string between

m_1

and

m_2

, let's call it

T_1

, is the force that accelerates

m_1

. Applying Newton's second law to

m_1

alone:

T_1 = m_1 a = 1.0 \text{ kg} \times 2.0 \text{ m s}^{-2} = 2.0 \text{ N}

Question 8

A block of mass $m$ rests on the floor of a lift that is accelerating upwards with acceleration $a$ . A horizontal force $F$ is applied to the block, but the block does not move. The coefficient of static friction between the block and the floor is $\mu_s$ . What is the maximum possible value of $F$ for which the block will not move?

$\mu_s m g$
$\mu_s m a$
$\mu_s m (g-a)$
$\mu_s m (g+a)$ (correct answer)

Explanation: The net vertical force on the block is

F_N - mg = ma

, where

F_N

is the normal force. Therefore,

F_N = m(g+a)

. The maximum static friction force is

F_{f,max} = \mu_s F_N

. For the block not to move, the applied horizontal force

F

must be less than or equal to the maximum static friction force. The maximum possible value of

F

is thus

F_{max} = \mu_s F_N = \mu_s m(g+a)

Question 9

A car travels at a constant speed over a circular hump-backed bridge of radius $r$ . What is the maximum speed the car can have at the highest point of the bridge without losing contact with the road?

$\sqrt{gr}$ (correct answer)
$\sqrt{2gr}$
$gr$
$\sqrt{g/r}$

Explanation: At the highest point of the bridge, the net force on the car provides the centripetal force required for circular motion. The forces acting on the car are gravity (

mg

, downwards) and the normal force from the road (

F_N

, upwards). The net force is

mg - F_N

, directed towards the center of the circle (downwards). So,

mg - F_N = mv^2/r

. The car loses contact with the road when the normal force

F_N

becomes zero. Setting

F_N = 0

, we get

mg = mv_{max}^2/r

. Solving for

v_{max}

gives

v_{max} = \sqrt{gr}

Question 10

A block of mass $m$ is on a horizontal surface with a coefficient of static friction $\mu_s$ . A pulling force $F$ is applied at an angle $\theta$ above the horizontal. What is the minimum magnitude of $F$ that will cause the block to start moving?

$\frac{\mu_s mg}{\cos\theta}$
$\frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$ (correct answer)
$\mu_s mg$
$\frac{\mu_s mg}{\cos\theta - \mu_s \sin\theta}$

Explanation: To start moving, the horizontal component of the applied force must overcome static friction:

F\cos\theta \ge F_f

. The normal force

F_N

is reduced by the upward component of

F

, so

F_N + F\sin\theta = mg

, which means

F_N = mg - F\sin\theta

. The maximum static friction is

F_{f,max} = \mu_s F_N = \mu_s(mg - F\sin\theta)

. Setting the forces equal for the minimum condition:

F\cos\theta = \mu_s(mg - F\sin\theta)

. Rearranging to solve for F:

F\cos\theta = \mu_s mg - \mu_s F\sin\theta

, so

F(\cos\theta + \mu_s \sin\theta) = \mu_s mg

. Thus,

F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}

Question 11

A 20 kg crate is pushed 5.0 m up a rough incline at a constant velocity by a force parallel to the incline. The incline is at 30° to the horizontal and the coefficient of kinetic friction is 0.20. What is the work done by the applied force? (Use $g = 10 \text{ m s}^{-2}$ )

500 J
670 J (correct answer)
700 J
1000 J

Explanation: Since velocity is constant, the net force is zero. The applied force

F_{app}

must balance the force of friction

F_f

and the component of gravity parallel to the incline,

mg\sin\theta

F_{app} = F_f + mg\sin\theta

. The normal force is

F_N = mg\cos\theta = 20 \times 10 \times \cos(30^{\circ}) = 173.2\text{ N}

. The friction force is

F_f = \mu_k F_N = 0.20 \times 173.2 = 34.64\text{ N}

. The gravity component is

mg\sin\theta = 20 \times 10 \times \sin(30^{\circ}) = 100\text{ N}

. So,

F_{app} = 34.64 + 100 = 134.64\text{ N}

. The work done is

W = F_{app} \times d = 134.64 \times 5.0 = 673.2\text{ J}

, which is approximately 670 J.

Question 12

A child and a skateboard have a combined mass of 49 kg. They are stationary on a frictionless horizontal surface. The child throws a 1.0 kg ball horizontally with a speed of 50 m s⁻¹ relative to the skateboard. What is the magnitude of the final velocity of the skateboard?

0.98 m s⁻¹
1.0 m s⁻¹ (correct answer)
1.02 m s⁻¹
1.04 m s⁻¹

Explanation: Let

M

be the mass of the child and skateboard (49 kg) and

m

be the mass of the ball (1.0 kg). Let

v_s

be the final velocity of the skateboard and

v_b

be the final velocity of the ball, both relative to the ground. The velocity of the ball relative to the skateboard is

v_{rel} = v_b - v_s = 50 \text{ m s}^{-1}

. So,

v_b = 50 + v_s

. Total momentum is conserved, and the initial momentum is zero.

Mv_s + mv_b = 0

. Substituting for

v_b

Mv_s + m(50 + v_s) = 0

(M+m)v_s + 50m = 0

v_s = -\frac{50m}{M+m} = -\frac{50 \times 1.0}{49+1.0} = -\frac{50}{50} = -1.0 \text{ m s}^{-1}

. The magnitude is 1.0 m s⁻¹.

Question 13

An object of mass 2.0 kg is initially at rest. A time-varying horizontal force acts on it. The force starts at 0 N, increases linearly to 10 N over 2.0 s, and then decreases linearly to 0 N over the next 2.0 s. What is the final speed of the object, assuming no friction?

5.0 m s⁻¹
10 m s⁻¹ (correct answer)
20 m s⁻¹
40 m s⁻¹

Explanation: The change in momentum of the object is equal to the impulse, which is the area under the force-time graph. The graph described is a triangle with a base of 4.0 s (2.0 s + 2.0 s) and a height of 10 N. The area is

\text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \text{ s} \times 10 \text{ N} = 20 \text{ Ns}

. Since impulse equals change in momentum (

J = \Delta p = m\Delta v

) and the object starts from rest,

20 \text{ Ns} = (2.0 \text{ kg}) \times v_f

. Solving for the final speed gives

v_f = 20 / 2.0 = 10 \text{ m s}^{-1}

Question 14

An object of mass $m$ is attached to a string and moves in a horizontal circle of radius $r$ as a conical pendulum. The string makes an angle $\theta$ with the vertical. What is the speed of the object?

$\sqrt{gr}$
$\sqrt{gr \tan\theta}$ (correct answer)
$\sqrt{gr \sin\theta}$
$\sqrt{gr / \tan\theta}$

Explanation: Let

T

be the tension in the string. Vertically, the forces are balanced:

T \cos\theta = mg

. Horizontally, the net force provides the centripetal force:

T \sin\theta = mv^2/r

. Dividing the second equation by the first gives

\frac{T \sin\theta}{T \cos\theta} = \frac{mv^2/r}{mg}

, which simplifies to

\tan\theta = v^2/(gr)

. Rearranging for the speed

v

gives

v = \sqrt{gr \tan\theta}

Question 15

A mass $m$ is attached to a vertical spring with spring constant $k$ . The mass is at rest at its equilibrium position. It is then displaced downwards by a distance $A$ and released. What is the magnitude of the net force on the mass when it is at a distance $A/2$ above the equilibrium position?

$kA/2$ (correct answer)
$mg - kA/2$
$mg + kA/2$
$kA$

Explanation: At the equilibrium position, the upward spring force balances the downward force of gravity,

kx_0 = mg

, where

x_0

is the equilibrium extension. When the mass is a distance

y = A/2

above equilibrium, the spring extension is

x_0 - y

. The upward spring force is

F_{spring} = k(x_0 - y) = kx_0 - ky = mg - ky

. The net force is

F_{net} = F_{spring} - mg = (mg - ky) - mg = -ky

. The magnitude of this restoring force is

ky

. Substituting

y=A/2

, the magnitude of the net force is

k(A/2)

Question 16

A rocket of total mass $M$ is at rest in deep space. It expels a small mass of gas $\Delta m$ with a high speed $u$ relative to the rocket. What is the approximate recoil speed, $\Delta v$ , of the rocket?

$u \frac{\Delta m}{M}$ (correct answer)
$u \frac{M}{\Delta m}$
$u \frac{\Delta m}{M - \Delta m}$
$u$

Explanation: The total momentum of the rocket-gas system is conserved and is initially zero. After the gas is expelled, the rocket has mass

M - \Delta m

and velocity

\Delta v

. The gas has mass

\Delta m

and its velocity relative to the rocket is

-u

. Its velocity relative to the initial rest frame is

v_{gas} = \Delta v - u

. By conservation of momentum:

0 = (M - \Delta m)\Delta v + \Delta m (\Delta v - u)

. Expanding gives

0 = M\Delta v - \Delta m \Delta v + \Delta m \Delta v - u\Delta m

, which simplifies to

M\Delta v = u\Delta m

. For a small expelled mass, we can approximate the rocket's mass as M, so

\Delta v \approx u \frac{\Delta m}{M}

Question 17

A stationary object of mass 3.0 kg explodes into three fragments of equal mass (1.0 kg each). Immediately after the explosion, one fragment moves east at 8.0 m s⁻¹ and a second fragment moves north at 6.0 m s⁻¹. What is the speed of the third fragment?

2.0 m s⁻¹
7.0 m s⁻¹
10 m s⁻¹ (correct answer)
14 m s⁻¹

Explanation: The total momentum of the system is conserved and is initially zero. Let the momentum vectors of the three fragments be

p_1

p_2

, and

p_3

. By conservation of momentum,

p_1 + p_2 + p_3 = 0

, so

p_3 = -(p_1 + p_2)

. Let east be the x-direction and north be the y-direction.

p_1 = (1.0 \text{ kg})(8.0 \text{ m s}^{-1}) \hat{i} = 8.0 \hat{i} \text{ Ns}

p_2 = (1.0 \text{ kg})(6.0 \text{ m s}^{-1}) \hat{j} = 6.0 \hat{j} \text{ Ns}

. So,

p_3 = -(8.0 \hat{i} + 6.0 \hat{j}) \text{ Ns}

. The magnitude of

p_3

|p_3| = \sqrt{(-8.0)^2 + (-6.0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ Ns}

. The speed of the third fragment is

v_3 = |p_3|/m_3 = 10 \text{ Ns} / 1.0 \text{ kg} = 10 \text{ m s}^{-1}

Question 18

An object with a volume of $5.0 \times 10^{-4} \text{ m}^3$ and a density of 800 kg m⁻³ is fully submerged in water (density 1000 kg m⁻³). It is held stationary by a vertical string attached to the bottom of the tank. What is the tension in the string? (Use $g = 10 \text{ m s}^{-2}$ )

5.0 N
4.0 N
1.0 N (correct answer)
9.0 N

Explanation: The forces acting on the object are its weight (downwards), the buoyant force (upwards), and the tension in the string (downwards). For equilibrium, the upward force equals the sum of the downward forces:

F_b = W + T

. The weight is

W = m_{obj}g = (\rho_{obj}V)g = (800 \times 5.0 \times 10^{-4}) \times 10 = 4.0 \text{ N}

. The buoyant force is

F_b = \rho_{water}Vg = (1000 \times 5.0 \times 10^{-4}) \times 10 = 5.0 \text{ N}

. The tension is

T = F_b - W = 5.0 \text{ N} - 4.0 \text{ N} = 1.0 \text{ N}

Question 19

A 50 g ball strikes a rigid wall horizontally at a speed of 20 m s⁻¹ and rebounds horizontally at a speed of 15 m s⁻¹. The ball is in contact with the wall for 10 ms. What is the magnitude of the average force exerted by the wall on the ball?

25 N
75 N
100 N
175 N (correct answer)

Explanation: The impulse is equal to the change in momentum,

J = \Delta p = F_{avg} \Delta t

. Momentum is a vector. Let the initial direction be positive. Initial momentum

p_i = 0.050 \text{ kg} \times 20 \text{ m s}^{-1} = 1.0 \text{ Ns}

. Final momentum

p_f = 0.050 \text{ kg} \times (-15 \text{ m s}^{-1}) = -0.75 \text{ Ns}

. Change in momentum

\Delta p = p_f - p_i = -0.75 - 1.0 = -1.75 \text{ Ns}

. The magnitude of the impulse is 1.75 Ns. The average force is

F_{avg} = |\Delta p| / \Delta t = 1.75 \text{ Ns} / (10 \times 10^{-3} \text{ s}) = 175 \text{ N}

Question 20

A rocket of initial mass 1500 kg burns fuel at a rate of 8.0 kg/s, ejecting exhaust at 2400 m/s relative to the rocket. The rocket starts from rest in space. What is the rocket's acceleration when its mass has decreased to 1200 kg?

12.8 m/s² because thrust equals mass flow rate times exhaust velocity
16.0 m/s² because acceleration increases as mass decreases linearly (correct answer)
14.4 m/s² because thrust depends on instantaneous mass and fuel consumption
18.2 m/s² because exhaust momentum creates proportional acceleration increase

Explanation: For a rocket, thrust

F = \frac{dm}{dt} v_e = 8.0 \times 2400 = 19200

N. When the rocket's mass is 1200 kg, acceleration

a = \frac{F}{m} = \frac{19200}{1200} = 16.0

m/s². The thrust remains constant as long as fuel burn rate and exhaust velocity are constant. Choice A incorrectly calculates thrust division by wrong mass. Choice C uses an incorrect formula mixing instantaneous mass with flow rate incorrectly. Choice D uses an incorrect proportionality assumption about momentum and acceleration.