Newton's Laws of Motion and Linear Momentum Practice Questions — JAMB UTME

Question 1Question →

When a stationary object in an isolated system explodes into two fragments of unequal mass, the fragment with the larger mass acquires a greater magnitude of linear momentum than the fragment with the smaller mass.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Both fragments acquire linear momentum of equal magnitude in opposite directions.

The statement is false because conservation of linear momentum requires the total initial momentum (zero) to equal the total final momentum. Therefore, the two fragments move in opposite directions with linear momenta of equal magnitude, regardless of their masses.

Step-by-Step Solution

1

Analyze internal forces acting during the explosion

By Newton's third law, the force exerted on the first fragment is equal in magnitude and opposite in direction to the force exerted on the second fragment (

F_1 = -F_2

).

Explosive forces are internal action-reaction pairs.

2

Apply the impulse-momentum theorem

Since both forces act for the exact same duration

\Delta t

, the impulse

J_1 = F_1 \Delta t

equals

-J_2 = -F_2 \Delta t

. Therefore, the change in linear momentum

\Delta p_1 = -\Delta p_2

.

Impulse delivered to an object equals its change in linear momentum.

3

Compare momentum magnitudes

|p_1| = |p_2|

, which means

m_1 v_1 = m_2 v_2

.

Initial momentum was zero (

p_{\text{initial}} = 0

), so the sum of final momentum vectors must be zero (

p_1 + p_2 = 0

).

Key Concept

Conservation of Linear Momentum and Newton's Third Law

Question 2Question →

A body $A$ of mass $4.0\text{ kg}$ moving due east at a velocity of $6.0\text{ m s}^{-1}$ collides head-on with a body $B$ of mass $2.0\text{ kg}$ moving due west at $3.0\text{ m s}^{-1}$ . If body $A$ continues to move due east after the collision with a speed of $1.0\text{ m s}^{-1}$ , what is the velocity of body $B$ after the collision?

7.0\text{ m s}^{-1}

due East

13.0\text{ m s}^{-1}

due East

7.0\text{ m s}^{-1}

due West

11.0\text{ m s}^{-1}

due East

Show answer & explanation

Answer: $7.0\text{ m s}^{-1}$ due East

Answer

7.0\text{ m s}^{-1}

due East

According to the principle of conservation of linear momentum, the total momentum before collision equals the total momentum after collision. Assigning positive to East and negative to West, total initial momentum is

4(6) + 2(-3) = 18\text{ kg m s}^{-1}

. Equating this to final momentum

4(1) + 2v_B

gives

2v_B = 14

, resulting in

v_B = +7.0\text{ m s}^{-1}

, which represents

7.0\text{ m s}^{-1}

due East.

Step-by-Step Solution

1

Define a reference direction for 1D momentum vectors

Let East be positive (+) and West be negative (-).

Linear momentum is a vector quantity, so opposite directions must have opposite algebraic signs.

2

Calculate the total initial momentum before collision (

p_i

)

p_i = m_A u_A + m_B u_B = (4.0\text{ kg})(+6.0\text{ m s}^{-1}) + (2.0\text{ kg})(-3.0\text{ m s}^{-1}) = 24.0 - 6.0 = 18.0\text{ kg m s}^{-1}

.

Body B moves west, so its initial velocity is

-3.0\text{ m s}^{-1}

.

3

Formulate the total final momentum after collision (

p_f

)

p_f = m_A v_A + m_B v_B = (4.0\text{ kg})(+1.0\text{ m s}^{-1}) + (2.0\text{ kg})v_B = 4.0 + 2.0 v_B

.

Body A moves east after collision, so its final velocity is

+1.0\text{ m s}^{-1}

.

4

Apply the Law of Conservation of Linear Momentum (

p_i = p_f

)

18.0 = 4.0 + 2.0 v_B \implies 2.0 v_B = 14.0 \implies v_B = +7.0\text{ m s}^{-1}

.

Since total initial momentum equals total final momentum in an isolated system.

5

Interpret the sign of the calculated velocity

Since

v_B

is positive (

+7.0\text{ m s}^{-1}

), body B moves at

7.0\text{ m s}^{-1}

due East.

Positive values correspond to the defined East direction.

Key Concept

Principle of Conservation of Linear Momentum in 1D head-on collisions

Question 3Question →

A constant net force of $20\text{ N}$ acts on an object of mass $4\text{ kg}$ that is initially at rest on a frictionless horizontal surface. What is the magnitude of the linear momentum of the object after $3\text{ s}$ ?

Show answer & explanation

Answer: 60

Answer

The magnitude of the linear momentum of the object after

3\text{ s}

is

60\text{ kg m s}^{-1}

.

According to Newton's Second Law in terms of momentum, force is the rate of change of linear momentum (

F = \frac{\Delta p}{\Delta t}

). Rearranging gives

\Delta p = F \Delta t

. Since the object is initially at rest (

p_i = 0\text{ kg m s}^{-1}

), the final momentum is

p_f = F \Delta t = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

.

Step-by-Step Solution

1

Apply the impulse-momentum theorem.

J = F \Delta t = \Delta p

The impulse of the net force acting on an object equals the change in its linear momentum.

2

Calculate the final linear momentum.

p_f = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

Since the object starts from rest, its initial momentum is zero (

p_i = 0

), so the final momentum is equal to the total impulse supplied.

Key Concept

Impulse-Momentum Theorem

Question 4Question →

A toy truck of mass $2.5\text{ kg}$ moving at a velocity of $6.0\text{ m s}^{-1}$ collides head-on with a toy car of mass $1.5\text{ kg}$ moving in the opposite direction at $2.0\text{ m s}^{-1}$ . If the two vehicles stick together upon impact, what is their combined velocity immediately after the collision?

3.0\text{ m s}^{-1}

4.5\text{ m s}^{-1}

4.8\text{ m s}^{-1}

2.0\text{ m s}^{-1}

Show answer & explanation

Answer: $3.0\text{ m s}^{-1}$

Answer

3.0\text{ m s}^{-1}

in the direction of the initial motion of the truck

According to the principle of conservation of linear momentum, total initial momentum equals total final momentum. Taking the direction of the toy truck as positive gives an initial momentum of

(2.5\text{ kg} \times 6.0\text{ m s}^{-1}) + (1.5\text{ kg} \times -2.0\text{ m s}^{-1}) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}

. Dividing this net momentum by the combined mass

(2.5\text{ kg} + 1.5\text{ kg} = 4.0\text{ kg})

yields a common final velocity of

3.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Assign direction signs to the velocity vectors

Let the initial direction of the truck be positive (

u_1 = +6.0\text{ m s}^{-1}

). The car moves in the opposite direction, so its velocity is negative (

u_2 = -2.0\text{ m s}^{-1}

).

Linear momentum is a vector quantity, so direction must be accounted for in one-dimensional motion.

2

Calculate the total initial momentum of the system

p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.5 \times 6.0) + (1.5 \times -2.0) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}

The total momentum before collision is the vector sum of individual momenta.

3

Apply the law of conservation of linear momentum to solve for common final velocity

v

p_{\text{final}} = (m_1 + m_2) v = (2.5 + 1.5) v = 4.0 v

. Setting

p_{\text{final}} = p_{\text{initial}} \Rightarrow 4.0 v = 12.0 \Rightarrow v = 3.0\text{ m s}^{-1}

.

Since the vehicles stick together, they move as a single combined mass.

Key Concept

Conservation of Linear Momentum in Inelastic Collisions

Estimated Time:50s

Question 5Question →

A shell of total mass $5.0\text{ kg}$ is moving horizontally with a velocity of $20\text{ m s}^{-1}$ when an internal explosion splits it into two fragments. One fragment of mass $2.0\text{ kg}$ is propelled backward along the original path at a speed of $10\text{ m s}^{-1}$ . What is the magnitude of the velocity, in $\text{m s}^{-1}$ , of the second fragment immediately after the explosion?

Show answer & explanation

Answer: 40

Answer

The magnitude of the velocity of the second fragment immediately after the explosion is

40\text{ m s}^{-1}

.

According to the Law of Conservation of Linear Momentum, the total momentum of a system remains constant when no net external horizontal force acts on it. Taking the original direction of motion as positive, the initial momentum is

100\text{ kg m s}^{-1}

. Since the

2.0\text{ kg}

fragment moves backward at

10\text{ m s}^{-1}

, its momentum is

-20\text{ kg m s}^{-1}

. For total momentum to remain

+100\text{ kg m s}^{-1}

, the remaining

3.0\text{ kg}

fragment must carry a momentum of

+120\text{ kg m s}^{-1}

, which corresponds to a velocity of

40\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the initial momentum of the shell prior to the explosion.

p_i = 5.0\text{ kg} \times 20\text{ m s}^{-1} = 100\text{ kg m s}^{-1}

in the initial forward direction.

Before the internal explosion, the system consists of a single mass moving with a constant velocity.

2

Apply the Law of Conservation of Linear Momentum taking vector direction into account.

p_i = m_1 v_1 + m_2 v_2 \implies 100 = 2.0(-10) + 3.0 v_2

In the absence of external forces, total momentum is conserved. The fragment propelled backward takes a negative sign relative to the initial forward motion.

3

Solve the algebraic equation for the unknown velocity

v_2

.

100 + 20 = 3.0 v_2 \implies 120 = 3.0 v_2 \implies v_2 = 40\text{ m s}^{-1}

Isolating

v_2

gives the forward velocity magnitude of the remaining

3.0\text{ kg}

piece.

Key Concept

Conservation of Linear Momentum in Explosions (1D Vector Sign Convention)

Question 6Question →

A horizontal jet of water issuing from a nozzle with a cross-sectional area of $2.0 \times 10^{-3}\text{ m}^2$ at a speed of $20\text{ m s}^{-1}$ strikes a vertical wall perpendicularly. If the water rebounds horizontally in the opposite direction at a speed of $5.0\text{ m s}^{-1}$ , what is the magnitude of the force exerted by the water stream on the wall? (Density of water = $1000\text{ kg m}^{-3}$ )

600\text{ N}

800\text{ N}

1000\text{ N}

50\text{ N}

Show answer & explanation

Answer: $1000\text{ N}$

Answer

The magnitude of the force exerted by the water jet on the wall is

1000\text{ N}

.

According to Newton's second law, force is the rate of change of linear momentum. The mass of water hitting the wall each second is

\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000 \times 2.0 \times 10^{-3} \times 20 = 40\text{ kg s}^{-1}

. Taking the initial direction as positive (

+20\text{ m s}^{-1}

), the rebounding velocity is opposite in direction (

-5.0\text{ m s}^{-1}

). The magnitude of the change in velocity is

|-5.0 - 20| = 25\text{ m s}^{-1}

. Multiplying mass flow rate by the change in velocity gives a force magnitude of

40 \times 25 = 1000\text{ N}

.

Step-by-Step Solution

1

Calculate the mass of water striking the wall per second (mass flow rate,

\frac{\Delta m}{\Delta t}

).

\frac{\Delta m}{\Delta t} = \rho A v_1 = 1000\text{ kg m}^{-3} \times (2.0 \times 10^{-3}\text{ m}^2) \times 20\text{ m s}^{-1} = 40\text{ kg s}^{-1}

.

The volume of water reaching the wall per second is given by the cross-sectional area multiplied by its initial speed.

2

Determine the change in velocity vector per unit mass of water

(\Delta v)

.

Taking the direction towards the wall as positive,

v_1 = +20\text{ m s}^{-1}

and

v_2 = -5.0\text{ m s}^{-1}

. Thus,

\Delta v = v_2 - v_1 = -5.0 - 20 = -25.0\text{ m s}^{-1}

.

Velocity is a vector quantity; rebounding in the opposite direction requires assigning opposite signs to the initial and final velocities.

3

Apply Newton's Second Law (

F = \frac{\Delta p}{\Delta t}

) to find the magnitude of force on the wall.

F = \frac{\Delta m}{\Delta t} |\Delta v| = 40\text{ kg s}^{-1} \times 25.0\text{ m s}^{-1} = 1000\text{ N}

.

By Newton's third law, the magnitude of force exerted on the wall equals the rate of change of momentum of the water stream.

Key Concept

Newton's Second Law and Linear Momentum Rate of Change

Question 7Question →

A ball of mass $0.20\text{ kg}$ moving horizontally towards a vertical wall at a speed of $15\text{ m s}^{-1}$ rebounds in the opposite direction at $10\text{ m s}^{-1}$ . If the impact with the wall lasts for $0.020\text{ s}$ , what is the magnitude of the average force exerted on the ball by the wall?

50\text{ N}

250\text{ N}

1250\text{ N}

0.10\text{ N}

Show answer & explanation

Answer: $250\text{ N}$

Answer

The magnitude of the average force exerted on the ball by the wall is

250\text{ N}

.

The average force is determined by Newton's second law (

F = \frac{\Delta p}{\Delta t}

). Because the ball rebounds in the opposite direction, velocity changes from

+15\text{ m s}^{-1}

to

-10\text{ m s}^{-1}

, yielding a total velocity change magnitude of

25\text{ m s}^{-1}

. Multiplying by mass (

0.20\text{ kg}

) gives an impulse magnitude of

5.0\text{ N s}

. Dividing impulse by the contact time (

0.020\text{ s}

) yields

250\text{ N}

.

Step-by-Step Solution

1

Establish vector direction and assign initial and final velocities

Initial velocity

u = +15\text{ m s}^{-1}

, final velocity

v = -10\text{ m s}^{-1}

Velocity is a vector quantity, so reversing direction requires a opposite sign convention.

2

Calculate the change in momentum (impulse)

\Delta p = m(v - u) = 0.20 \times (-10 - 15) = 0.20 \times (-25) = -5.0\text{ N s}

Impulse is equal to the change in linear momentum.

3

Calculate the magnitude of the average force

F = \frac{|\Delta p|}{\Delta t} = \frac{5.0\text{ N s}}{0.020\text{ s}} = 250\text{ N}

By Newton's second law, average force equals rate of change of momentum (F = \Delta p / \Delta t).

Key Concept

Impulse-Momentum Theorem and Vector Nature of Momentum

Estimated Time:1m 0s

Question 8Question →

A trolley of mass $0.40\text{ kg}$ moving to the right with a velocity of $5.0\text{ m s}^{-1}$ collides head-on with a second trolley of mass $0.60\text{ kg}$ moving to the left at $2.5\text{ m s}^{-1}$ . If the two trolleys stick together on impact, what is their combined velocity immediately after the collision?

0.50\text{ m s}^{-1}

to the right

3.50\text{ m s}^{-1}

to the right

2.50\text{ m s}^{-1}

to the right

1.25\text{ m s}^{-1}

to the right

Show answer & explanation

Answer: $0.50\text{ m s}^{-1}$ to the right

Answer

0.50\text{ m s}^{-1}

to the right

By adopting a standard vector sign convention where movement to the right is positive and movement to the left is negative, the initial momentum is

p_i = (0.40 \text{ kg} \times 5.0 \text{ m s}^{-1}) + (0.60 \text{ kg} \times -2.5 \text{ m s}^{-1}) = 2.0 - 1.5 = 0.50 \text{ kg m s}^{-1}

. Because the trolleys stick together, their total mass becomes

0.40 \text{ kg} + 0.60 \text{ kg} = 1.0 \text{ kg}

. Dividing the total momentum by the combined mass gives a final velocity of

+0.50 \text{ m s}^{-1}

, indicating movement to the right.

Step-by-Step Solution

1

Assign a directional sign convention for velocity vectors

Rightward velocity

u_1 = +5.0\text{ m s}^{-1}

, Leftward velocity

u_2 = -2.5\text{ m s}^{-1}

Linear momentum is a vector quantity, so direction must be accounted for using opposite algebraic signs.

2

Calculate total initial linear momentum of the system

p_i = m_1 u_1 + m_2 u_2 = (0.40 \times 5.0) + (0.60 \times (-2.5)) = 2.0 - 1.5 = +0.50\text{ kg m s}^{-1}

Sum the individual initial momenta of both trolleys.

3

Apply the principle of conservation of linear momentum to solve for final combined velocity

v = \frac{p_i}{m_1 + m_2} = \frac{+0.50}{0.40 + 0.60} = +0.50\text{ m s}^{-1}

Since no external net force acts on the system, total initial momentum equals total final momentum.

Key Concept

Conservation of Linear Momentum in Inelastic Collisions

Estimated Time:1m 30s

Question 9Question →

A constant force of $12\text{ N}$ acts for $4.0\text{ s}$ on a body of mass $3.0\text{ kg}$ that is initially moving at $5.0\text{ m s}^{-1}$ in the direction opposite to the force. What is the final velocity of the body in the direction of the applied force?

11\text{ m s}^{-1}

21\text{ m s}^{-1}

16\text{ m s}^{-1}

43\text{ m s}^{-1}

Show answer & explanation

Answer: $11\text{ m s}^{-1}$

Answer

The final velocity of the body in the direction of the force is

11\text{ m s}^{-1}

.

By the impulse-momentum theorem, the impulse

F \Delta t = 12 \times 4.0 = 48\text{ N s}

causes a velocity change

\Delta v = \frac{48}{3.0} = 16\text{ m s}^{-1}

in the direction of the force. Since the body initially moved in the opposite direction at

-5.0\text{ m s}^{-1}

, the final velocity is

-5.0 + 16 = 11\text{ m s}^{-1}

in the direction of the force.

Step-by-Step Solution

1

Assign direction signs to physical quantities based on a chosen coordinate system.

Let the direction of the applied force be positive (+). Then

F = +12\text{ N}

,

m = 3.0\text{ kg}

,

t = 4.0\text{ s}

, and initial velocity

u = -5.0\text{ m s}^{-1}

.

Velocity and force are vector quantities; motion opposite to the force must carry a negative sign.

2

Calculate the impulse delivered by the force.

\text{Impulse} = F \times \Delta t = 12\text{ N} \times 4.0\text{ s} = 48\text{ N s}

.

Impulse equals force multiplied by the time interval over which it acts.

3

Apply the impulse-momentum theorem

F \Delta t = m(v - u)

to solve for final velocity

v

.

48 = 3.0 \times (v - (-5.0)) \implies 16 = v + 5.0 \implies v = 11\text{ m s}^{-1}

.

The change in linear momentum of an object is equal to the net impulse applied to it.

Key Concept

Impulse-Momentum Theorem and Vector Sign Conventions

Question 10Question →

If two objects of unequal mass experience the same magnitude of net force for the same duration of time, the lighter object will undergo a greater change in linear momentum than the heavier object.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Both objects undergo the exact same change in linear momentum because impulse depends only on the force applied and the time duration, not on the mass of the object.

The statement is false because the change in momentum (

\Delta p

) is determined solely by the impulse applied (

F \Delta t

). Since the net force and time duration are identical for both objects, the change in momentum is the same for both, regardless of mass differences.

Step-by-Step Solution

1

State the relationship between force, time, and momentum change using Newton's Second Law.

Impulse

J = F \Delta t = \Delta p

, where

F

is the net force,

\Delta t

is the duration, and

\Delta p

is the change in linear momentum.

The impulse-momentum theorem directly links the net force acting over time to the resulting change in momentum.

2

Evaluate the given problem constraints for both the lighter mass (

m_1

) and heavier mass (

m_2

).

F_1 = F_2 = F

and

\Delta t_1 = \Delta t_2 = \Delta t

.

Both objects experience equal forces for equal durations.

3

Compare the resulting changes in linear momentum.

\Delta p_1 = F \Delta t

and

\Delta p_2 = F \Delta t

, so

\Delta p_1 = \Delta p_2

.

Because mass

m

does not alter the product

F \Delta t

, both objects experience identical momentum changes.

Key Concept

Impulse-Momentum Theorem and Newton's Second Law of Motion

Question 11Question →

A sledge of mass $8.0\text{ kg}$ sliding on ice with an initial velocity of $15\text{ m s}^{-1}$ enters a rough patch that exerts a constant retarding force of $24\text{ N}$ . Calculate the time, in seconds, required for the sledge to come to a complete stop.

Show answer & explanation

Answer: 5

Answer

The time required for the sledge to come to a complete stop is

5.0\text{ s}

.

By Newton's second law in terms of momentum, the rate of change of momentum is equal to the applied net force (

F = \frac{\Delta p}{\Delta t}

). Rearranging gives

\Delta t = \frac{m(v - u)}{F}

. Substituting

m = 8.0\text{ kg}

,

u = 15\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and retarding force

F = -24\text{ N}

yields

\Delta t = \frac{8.0 \times (0 - 15)}{-24} = 5.0\text{ s}

.

Step-by-Step Solution

1

Determine the change in linear momentum of the sledge.

The change in linear momentum is

\Delta p = m(v - u) = 8.0\text{ kg} \times (0\text{ m s}^{-1} - 15\text{ m s}^{-1}) = -120\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity.

2

Apply the impulse-momentum theorem to determine the time duration.

\Delta t = \frac{\Delta p}{F} = \frac{-120\text{ kg m s}^{-1}}{-24\text{ N}} = 5.0\text{ s}

.

Impulse delivered by a net force over a time interval equals the change in linear momentum.

Key Concept

Newton's Second Law and Impulse-Momentum Theorem

Question 12Question →

A rifle of mass $4.0\text{ kg}$ fires a bullet of mass $0.01\text{ kg}$ with a velocity of $400\text{ m s}^{-1}$ . What is the recoil velocity of the rifle?

-1.0\text{ m s}^{-1}

1.0\text{ m s}^{-1}

-100\text{ m s}^{-1}

4.0\text{ m s}^{-1}

Show answer & explanation

Answer: $-1.0\text{ m s}^{-1}$

Answer

The recoil velocity of the rifle is

-1.0\text{ m s}^{-1}

(or

1.0\text{ m s}^{-1}

in the direction opposite to the bullet).

The system starts at rest, so the initial total momentum is zero. By the law of conservation of linear momentum, the total final momentum must also be zero:

m_{rifle}v_{rifle} + m_{bullet}v_{bullet} = 0

. Substituting

m_{rifle} = 4.0\text{ kg}

,

m_{bullet} = 0.01\text{ kg}

, and

v_{bullet} = 400\text{ m s}^{-1}

gives

4.0 v_{rifle} + 4.0 = 0

, which yields

v_{rifle} = -1.0\text{ m s}^{-1}

. The negative sign indicates that the rifle moves in the direction opposite to the bullet.

Step-by-Step Solution

1

State the principle of conservation of linear momentum

Total Initial Momentum = Total Final Momentum = 0

Before firing, both the rifle and bullet are at rest.

2

Set up the linear momentum conservation equation

m_{rifle} v_{rifle} + m_{bullet} v_{bullet} = 0

The sum of the final momenta of the system components must equal zero.

3

Substitute the given numerical values into the equation and solve for recoil velocity

4.0 \cdot v_{rifle} + (0.01 \cdot 400) = 0 \implies 4.0 \cdot v_{rifle} + 4 = 0 \implies v_{rifle} = -1.0\text{ m s}^{-1}

Solving the linear algebraic equation yields the exact magnitude and direction of the recoil velocity.

Key Concept

Law of Conservation of Linear Momentum and Newton's Third Law of Motion

Estimated Time:45s

Question 13Question →

If a constant magnitude net force continuously acts on a moving body strictly perpendicular to its direction of motion, the magnitude of the body's linear momentum remains constant while its direction of motion continuously changes.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. A net force directed perpendicular to velocity changes only the direction of motion, keeping speed and the magnitude of linear momentum constant.

The statement is true because a force perpendicular to displacement does zero work, preserving kinetic energy and speed. Because speed is unchanged, the scalar magnitude of momentum remains constant while its vector direction curves continuously.

Step-by-Step Solution

1

Determine the work done by a perpendicular force.

The work done is

W = F \Delta s \cos(90^\circ) = 0\text{ J}

.

Force perpendicular to displacement performs no work on the object.

2

Relate work done to speed and magnitude of linear momentum.

Zero work means zero change in kinetic energy, so speed

v

is constant, making magnitude

p = mv

constant.

Linear momentum magnitude depends solely on mass and speed.

3

Evaluate the effect on momentum direction.

The force produces an acceleration vector perpendicular to velocity, changing the vector direction of linear momentum

\vec{p}

.

By Newton's Second Law (

\vec{F} = \frac{d\vec{p}}{dt}

), force dictates the rate of change of momentum vector.

Key Concept

Vector nature of linear momentum and perpendicular force action

Question 14Question →

A cart of mass $40\text{ kg}$ carrying a package of mass $10\text{ kg}$ is coasting along a straight horizontal track at a constant velocity of $6.0\text{ m s}^{-1}$ . The package is suddenly ejected horizontally backward (opposite to the direction of motion of the cart) at a speed of $15\text{ m s}^{-1}$ relative to the ground. What is the new velocity of the cart after the package is ejected?

3.75\text{ m s}^{-1}

9.75\text{ m s}^{-1}

11.25\text{ m s}^{-1}

2.25\text{ m s}^{-1}

Show answer & explanation

Answer: $11.25\text{ m s}^{-1}$

Answer

The new velocity of the cart is

11.25\text{ m s}^{-1}

in the forward direction.

By the law of conservation of linear momentum, the total initial momentum of the system (cart plus package,

50\text{ kg}

moving at

6.0\text{ m s}^{-1}

) equals

300\text{ kg m s}^{-1}

. When the

10\text{ kg}

package is ejected backward at

-15\text{ m s}^{-1}

, its momentum is

-150\text{ kg m s}^{-1}

. Setting

300 = -150 + 40 v_{\text{cart}}

yields

v_{\text{cart}} = 11.25\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the total initial momentum of the system before ejection.

Total mass

m_{\text{total}} = 40\text{ kg} + 10\text{ kg} = 50\text{ kg}

. Initial momentum

P_i = 50\text{ kg} \times 6.0\text{ m s}^{-1} = 300\text{ kg m s}^{-1}

.

The initial system consists of both the cart and the package moving together at

6.0\text{ m s}^{-1}

.

2

Set up the expression for final momentum considering direction.

Taking the forward direction as positive, the package's velocity is

v_{\text{pkg}} = -15\text{ m s}^{-1}

. Final momentum

P_f = (10 \times -15) + (40 \times v_{\text{cart}}) = -150 + 40 v_{\text{cart}}

.

Linear momentum is a vector quantity, so opposite motion must be assigned a negative sign.

3

Apply the law of conservation of linear momentum (

P_i = P_f

) to solve for the cart's final velocity.

300 = -150 + 40 v_{\text{cart}} \implies 450 = 40 v_{\text{cart}} \implies v_{\text{cart}} = 11.25\text{ m s}^{-1}

.

In the absence of external forces on the system, total linear momentum is conserved.

Key Concept

Conservation of Linear Momentum

Estimated Time:2m 0s

Question 15Question →

A wooden block of mass $4.0\text{ kg}$ is suspended vertically at rest. A bullet of mass $0.05\text{ kg}$ travelling horizontally at $400\text{ m s}^{-1}$ strikes the block, passes completely through it, and emerges on the opposite side with a reduced speed of $100\text{ m s}^{-1}$ . If a constant retarding force brings the moving block to rest in $0.25\text{ s}$ after the bullet emerges, calculate the magnitude of this retarding force in newtons.

Show answer & explanation

Answer: 60

Answer

The magnitude of the retarding force acting on the block is

60\text{ N}

.

During the impact, the bullet loses momentum equal to

\Delta p = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

. By the conservation of linear momentum, this exact amount of momentum is gained by the block. Applying Newton's second law (

F = \frac{\Delta p}{\Delta t}

), the magnitude of the constant retarding force needed to reduce the block's momentum to zero in

0.25\text{ s}

is

F = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

.

Step-by-Step Solution

1

Calculate the momentum lost by the bullet during penetration.

\Delta p_{\text{bullet}} = 0.05\text{ kg} \times (400\text{ m s}^{-1} - 100\text{ m s}^{-1}) = 15\text{ N s}

The momentum lost by the bullet equals its mass multiplied by the change in its horizontal velocity vector.

2

Determine the initial momentum imparted to the wooden block using the law of conservation of linear momentum.

p_{\text{block}} = \Delta p_{\text{bullet}} = 15\text{ N s}

Since no external horizontal force acts during the collision impact, the momentum lost by the bullet is fully transferred to the block.

3

Calculate the retarding force required to bring the block to rest using the impulse-momentum theorem.

F = \frac{\Delta p_{\text{block}}}{\Delta t} = \frac{15\text{ N s}}{0.25\text{ s}} = 60\text{ N}

According to Newton's second law of motion, the net force acting on a body equals the rate of change of momentum.

Key Concept

Conservation of Linear Momentum and Newton's Second Law

Question 16Question →

For an object of constant mass moving at constant speed along a complete circular path, the net vector impulse imparted to the object over one full revolution is zero, even though a continuous centripetal force acts on the object throughout the motion.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the impulse-momentum theorem states that net impulse equals the change in momentum (

\vec{J} = \Delta \vec{p}

). After one full revolution, the object's initial and final velocity vectors are identical, resulting in zero change in momentum.

The statement is correct because linear momentum is a vector quantity. Over a complete circular revolution, the initial and final velocity vectors are identical in both magnitude and direction, making the change in momentum—and therefore the net vector impulse—equal to zero.

Step-by-Step Solution

1

Recall the vector definition of impulse and the impulse-momentum theorem.

The net impulse

\vec{J}

acting on a body equals its change in linear momentum:

\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i

.

Impulse is a vector quantity dependent on the initial and final states of momentum over the time interval.

2

Evaluate the velocity vector of the object after one complete circular revolution at constant speed.

Since the speed is constant and the trajectory completes a closed loop, the final velocity vector

\vec{v}_f

has the exact same magnitude and direction as the initial velocity vector

\vec{v}_i

.

A complete revolution returns the object to its starting point with its velocity pointing in the initial direction.

3

Calculate the change in momentum

\Delta \vec{p}

.

\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(0) = \vec{0}

. Thus, net impulse

\vec{J} = \vec{0}

.

Subtracting identical vectors yields zero vector magnitude.

Key Concept

Impulse-Momentum Theorem and Vector Nature of Linear Momentum

Estimated Time:1m 30s

Question 17Question →

A body of mass $2.5\text{ kg}$ moving at a speed of $4.0\text{ m s}^{-1}$ along a straight horizontal path is brought to rest in $2.0\text{ s}$ by a constant retarding force. What is the magnitude of this retarding force in newtons?

Show answer & explanation

Answer: 5

Answer

The magnitude of the retarding force is

5.0\text{ N}

.

According to Newton's second law of motion, net force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}

). Substituting

m = 2.5\text{ kg}

,

u = 4.0\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and

t = 2.0\text{ s}

gives

F = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}

. The magnitude of this force is

5.0\text{ N}

.

Step-by-Step Solution

1

Determine the initial momentum and final momentum of the body.

Initial momentum

p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}

, and final momentum

p_f = 0\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity (

p = mv

).

2

Calculate the magnitude of the force applied using the impulse-momentum relationship

F = \frac{\Delta p}{\Delta t}

.

Magnitude of force

F = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}

.

Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Key Concept

Newton's Second Law and Impulse-Momentum Relationship

Estimated Time:45s

Question 18Question →

A body of mass $3.0\text{ kg}$ moving due east along a smooth horizontal track at $8.0\text{ m s}^{-1}$ collides head-on with a $2.0\text{ kg}$ body moving due west at $4.0\text{ m s}^{-1}$ . Immediately after the impact, the $2.0\text{ kg}$ body rebounds due east with a speed of $5.0\text{ m s}^{-1}$ . If the duration of the impact is $0.02\text{ s}$ , what is the magnitude of the average impact force exerted on the $3.0\text{ kg}$ body?

100\text{ N}

1200\text{ N}

900\text{ N}

1500\text{ N}

Show answer & explanation

Answer: $900\text{ N}$

Answer

The magnitude of the average impact force exerted on the

3.0\text{ kg}

body is

900\text{ N}

.

By designating East as positive, initial velocities are

u_1 = +8.0\text{ m s}^{-1}

and

u_2 = -4.0\text{ m s}^{-1}

. Conservation of linear momentum gives

(3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0)

, yielding

v_1 = +2.0\text{ m s}^{-1}

. The change in momentum of the

3.0\text{ kg}

body is

\Delta p = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Dividing the magnitude of this impulse by the impact time (

0.02\text{ s}

) yields an average force of

900\text{ N}

.

Step-by-Step Solution

1

Set up momentum conservation by assigning directional signs to velocities.

Taking East as positive (

+

):

u_1 = +8.0\text{ m s}^{-1}

,

u_2 = -4.0\text{ m s}^{-1}

,

v_2 = +5.0\text{ m s}^{-1}

.

Linear momentum is a vector quantity, so direction must be taken into account.

2

Calculate the initial total momentum and solve for the final velocity of the

3.0\text{ kg}

body (

v_1

).

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies (3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0) \implies 16.0 = 3.0 v_1 + 10.0 \implies v_1 = +2.0\text{ m s}^{-1}

.

Total momentum is conserved in the absence of external forces.

3

Calculate the magnitude of the impulse and average force acting on the

3.0\text{ kg}

body.

\Delta p_1 = m_1(v_1 - u_1) = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Force magnitude

F = \frac{|\Delta p_1|}{\Delta t} = \frac{18.0}{0.02} = 900\text{ N}

.

The average force equals the rate of change of linear momentum.

Key Concept

Law of Conservation of Linear Momentum and Impulse-Momentum Theorem

Estimated Time:2m 0s

Question 19Question →

In an isolated system where two colliding bodies of unequal mass undergo a perfectly elastic head-on collision, the body with the larger mass imparts a greater magnitude of impulse on the lighter body than the lighter body imparts on the heavier body.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. By Newton's Third Law, interaction forces are equal in magnitude and opposite in direction at all times, making the impulse imparted by each body on the other equal in magnitude.

The statement is false because Newton's Third Law dictates that the force exerted by object 1 on object 2 is equal in magnitude to the force exerted by object 2 on object 1 at every instant during contact. Integrating force over time yields equal magnitudes of impulse (

|J_1| = |J_2|

), independent of mass differences.

Step-by-Step Solution

1

Apply Newton's Third Law to interaction forces.

F_{12} = -F_{21}

, where

F_{12}

is the force exerted on body 2 by body 1, and

F_{21}

is the force exerted on body 1 by body 2.

Forces between interacting objects always occur in equal and opposite action-reaction pairs regardless of mass.

2

Integrate both forces over the duration of collision

\Delta t

.

J_{12} = \int F_{12} \, dt = -\int F_{21} \, dt = -J_{21}

.

Impulse is defined as the time-integral of force.

3

Compare impulse magnitudes.

|J_{12}| = |J_{21}|

.

Taking the magnitude of both sides shows that both bodies experience equal magnitudes of impulse regardless of mass ratio or collision elasticity.

Key Concept

Newton's Third Law and Equality of Mutual Impulse

Question 20Question →

The impulse imparted to an object by a net force is equal to the time rate of change of the object's linear momentum.

True

False

Show answer & explanation

Answer: False

Answer

False. Impulse is equal to the total change in linear momentum (

J = \Delta p

), while the time rate of change of linear momentum (

\frac{\Delta p}{\Delta t}

) represents the net force.

The statement is false. By definition, impulse is the product of net force and the time interval during which it acts (

J = F \Delta t

), which equals the total change in linear momentum (

\Delta p

). In contrast, the time rate of change of momentum (

\frac{\Delta p}{\Delta t}

) is equal to the net applied force.

Step-by-Step Solution

1

Define impulse according to the Impulse-Momentum Theorem

Impulse (

J

) is defined as

J = F \Delta t = \Delta p

, which represents the total change in momentum.

Establishing the mathematical definition of impulse.

2

Identify the physical quantity equal to the time rate of change of momentum

Newton's Second Law states that force

F = \frac{\Delta p}{\Delta t}

, which is the time rate of change of linear momentum.

Distinguishing between force and impulse.

3

Evaluate the statement

Because impulse equals total change in momentum (

\Delta p

) rather than rate of change (

\frac{\Delta p}{\Delta t}

), the statement is false.

Concluding the true/false verification.

Key Concept

Impulse-Momentum Theorem vs Newton's Second Law

Newton's Laws of Motion and Linear Momentum

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