Newton's Laws of Motion and Linear Momentum Practice Questions — JAMB UTME — Page 2

Question 21Question →

When a constant net external force is applied to a body for a given time interval, the magnitude of the change in the body's linear momentum depends only on the magnitude of the applied force and the time duration, and is completely independent of the mass of the body.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. By the impulse-momentum theorem (

\Delta p = F \Delta t

), the change in linear momentum is equal to the applied impulse, which depends only on force magnitude and time duration, regardless of the body's mass.

The statement is correct because impulse is defined as the product of net force and time interval (

J = F \Delta t

), and by the impulse-momentum theorem, impulse equals the change in momentum (

\Delta p = J

). Neither net force nor time interval depends on the mass of the body.

Step-by-Step Solution

1

Express Newton's Second Law of Motion in terms of linear momentum.

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

, where

F

is the net force and

\Delta p

is the change in linear momentum over time

\Delta t

.

Newton's second law states that the net force acting on an object equals the rate of change of its linear momentum.

2

Rearrange the equation to isolate the change in momentum (

\Delta p

).

\Delta p = F \Delta t

.

Multiplying both sides by

\Delta t

gives the impulse-momentum relationship.

3

Evaluate the dependence of

\Delta p

on the body's mass.

The expression

\Delta p = F \Delta t

contains only force and time, with no mass term (

m

).

Although a heavier body experiences smaller acceleration than a lighter body under the same force, both undergo the exact same total change in linear momentum over identical time intervals.

Key Concept

Impulse-Momentum Theorem and Newton's Second Law

Question 22Question →

A tennis ball of mass $0.2\text{ kg}$ moving horizontally at a speed of $15\text{ m s}^{-1}$ strikes a vertical wall and rebounds along its original path at a speed of $10\text{ m s}^{-1}$ . What is the magnitude of the impulse exerted by the wall on the ball?

5.0\text{ N s}

1.0\text{ N s}

3.0\text{ N s}

2.0\text{ N s}

Show answer & explanation

Answer: $5.0\text{ N s}$

Answer

5.0\text{ N s}

The correct answer accounts for the vector nature of velocity. Taking the initial direction as positive (

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

), the rebound velocity is negative (

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

). The change in velocity is

\Delta v = -10 - 15 = -25\text{ m s}^{-1}

. Multiplying by mass (

0.2\text{ kg}

) gives an impulse magnitude of

5.0\text{ N s}

.

Step-by-Step Solution

1

Assign direction signs to the initial and final velocity vectors

Initial velocity

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

, final velocity after rebound

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

Velocity is a vector quantity, so reversing direction requires a change in sign.

2

Apply the impulse-momentum theorem (

I = \Delta p = m(v - u)

)

I = 0.2 \times (-10 - 15) = 0.2 \times (-25) = -5.0\text{ N s}

Impulse is equal to the net change in linear momentum.

3

Determine the magnitude of the impulse

|I| = 5.0\text{ N s}

Magnitude is the absolute value of the vector quantity.

Key Concept

Impulse-Momentum Theorem in One-Dimensional Rebound Scenarios

Question 23Question →

Calculate the magnitude of the linear momentum, in $\text{kg m s}^{-1}$ , of a body of mass $6\text{ kg}$ moving at a constant velocity of $15\text{ m s}^{-1}$ .

Show answer & explanation

Answer: 90

Answer

The magnitude of the linear momentum is

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

.

Linear momentum (

p

) is the product of an object's mass (

m

) and its velocity (

v

). Substituting

m = 6\text{ kg}

and

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

yields

p = 6 \times 15 = 90\text{ kg m s}^{-1}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 6\text{ kg}

and velocity

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

These are the parameters provided in the problem statement.

2

Apply the linear momentum formula

p = mv = 6\text{ kg} \times 15\text{ m s}^{-1} = 90\text{ kg m s}^{-1}

Linear momentum is defined as the product of an object's mass and its velocity.

Key Concept

Linear Momentum Definition

Estimated Time:30s

Question 24Question →

A trolley P of mass $4.0\text{ kg}$ moving due east at $5.0\text{ m s}^{-1}$ collides head-on with a trolley Q of mass $1.0\text{ kg}$ moving due west at $10.0\text{ m s}^{-1}$ . If the two trolleys coalesce upon impact, what is their common velocity?

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

due east

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

due east

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

due east

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

due east

Show answer & explanation

Answer: $2.0\text{ m s}^{-1}$ due east

Answer

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

due east

Linear momentum is conserved in an isolated system. Taking east as positive, the initial momentum of trolley P is

+20.0\text{ kg m s}^{-1}

and trolley Q is

-10.0\text{ kg m s}^{-1}

, yielding a net initial momentum of

+10.0\text{ kg m s}^{-1}

. After impact, the total mass is

4.0\text{ kg} + 1.0\text{ kg} = 5.0\text{ kg}

. The common velocity is

\frac{10.0}{5.0} = +2.0\text{ m s}^{-1}

, where the positive sign denotes a direction due east.

Step-by-Step Solution

1

Assign a directional coordinate system

Let the eastward direction be positive (

+

) and the westward direction be negative (

-

)

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

2

Calculate total initial momentum (

p_i

)

p_i = m_P u_P + m_Q u_Q = (4.0 \times 5.0) + (1.0 \times (-10.0)) = 20.0 - 10.0 = +10.0\text{ kg m s}^{-1}

According to the principle of conservation of linear momentum, initial momentum equals final momentum.

3

Calculate final common velocity (

v

)

v = \frac{p_i}{m_P + m_Q} = \frac{+10.0}{4.0 + 1.0} = +2.0\text{ m s}^{-1}

Since the trolleys coalesce, they move together with a total combined mass of

5.0\text{ kg}

.

Key Concept

Conservation of Linear Momentum in 1D Inelastic Collisions

Question 25Question →

When a constant horizontal stream of water of fixed cross-sectional area strikes a flat vertical wall normally and comes to rest without rebounding, the magnitude of the force exerted on the wall is directly proportional to the square of the speed of the water stream.

True

False

Show answer & explanation

Answer: True

Answer

True. The force exerted on the wall is directly proportional to the square of the speed of the water stream (

F \propto v^2

).

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

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

). For a continuous fluid stream,

\frac{\Delta m}{\Delta t} = \rho A v

. Since each unit mass loses speed

v

upon impact,

F = \left(\frac{\Delta m}{\Delta t}\right) v = \rho A v^2

. Thus, force is directly proportional to the square of the speed.

Step-by-Step Solution

1

Determine the mass of water striking the wall in time interval

\Delta t

.

\Delta m = \rho A v \Delta t

, where

\rho

is fluid density,

A

is cross-sectional area, and

v

is speed.

The volume of water reaching the wall per unit time is given by

A v

.

2

Apply Newton's Second Law in terms of rate of change of linear momentum.

F = \frac{\Delta p}{\Delta t} = \frac{\Delta m \cdot v}{\Delta t} = \frac{(\rho A v \Delta t) v}{\Delta t} = \rho A v^2

.

Since the water comes to rest, its change in velocity per unit mass is

v

.

3

Analyze the dependence of force

F

on speed

v

.

Since density

\rho

and cross-sectional area

A

are constant,

F = (\text{constant}) \times v^2 \implies F \propto v^2

.

Both mass flow rate and momentum change per unit mass are proportional to

v

.

Key Concept

Newton's Second Law and continuous mass flow momentum change

Question 26Question →

A stationary object of mass $5.0\text{ kg}$ explodes into two fragments of masses $2.0\text{ kg}$ and $3.0\text{ kg}$ . If the $2.0\text{ kg}$ fragment moves due east at a velocity of $15\text{ m s}^{-1}$ , what is the velocity of the $3.0\text{ kg}$ fragment?

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

due west

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

due east

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

due west

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

due west

Show answer & explanation

Answer: $10\text{ m s}^{-1}$ due west

Answer

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

due west

Before the explosion, the system is stationary, giving an initial momentum of

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

. By the law of conservation of linear momentum, the total momentum after the explosion must also equal zero. Taking the eastward direction as positive, the

2.0\text{ kg}

fragment has a momentum of

+30\text{ kg m s}^{-1}

. To balance this, the

3.0\text{ kg}

fragment must possess a momentum of

-30\text{ kg m s}^{-1}

, yielding

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

, which means a speed of

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

directed due west.

Step-by-Step Solution

1

Determine the initial momentum of the system

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

The object is initially at rest.

2

Express final total linear momentum using vector direction (let east be positive)

p_f = m_1 v_1 + m_2 v_2 = (2.0 \times 15) + (3.0 \times v_2) = 30 + 3 v_2

Linear momentum is conserved in an isolated system.

3

Equate initial momentum to final momentum and solve for

v_2

0 = 30 + 3 v_2 \implies v_2 = -10\text{ m s}^{-1}

The negative sign indicates the direction is opposite to east, which is due west.

Key Concept

Conservation of Linear Momentum in Explosions

Estimated Time:1m 30s

Question 27Question →

A sand bag of mass $8.0\text{ kg}$ is suspended vertically by a light rope. A projectile of mass $0.50\text{ kg}$ moving horizontally at a speed of $170\text{ m s}^{-1}$ strikes the sand bag and becomes embedded in it. What is the common speed, in $\text{m s}^{-1}$ , of the sand bag and the embedded projectile immediately after collision?

Show answer & explanation

Answer: 10

Answer

The common speed of the sand bag and embedded projectile immediately after the collision is

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

.

According to the principle of conservation of linear momentum, total momentum before impact equals total momentum after impact. The initial momentum of the system is entirely from the projectile:

p_i = 0.50 \times 170 = 85\text{ kg m s}^{-1}

. After collision, both objects move together with a total mass of

0.50 + 8.0 = 8.5\text{ kg}

. Setting

8.5 v = 85

gives

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

.

Step-by-Step Solution

1

State the conservation of linear momentum equation for an inelastic collision

m_1 u_1 + m_2 u_2 = (m_1 + m_2) v

Since no net external horizontal force acts on the system during impact, total linear momentum is conserved.

2

Substitute the given values into the momentum balance equation

(0.50\text{ kg})(170\text{ m s}^{-1}) + (8.0\text{ kg})(0\text{ m s}^{-1}) = (0.50\text{ kg} + 8.0\text{ kg}) v

The projectile embeds into the sand bag, so they move together with a single combined mass.

3

Solve the linear equation for the common final velocity

v

85 = 8.5 v \implies v = 10\text{ m s}^{-1}

Dividing total initial momentum by total combined mass yields the final speed.

Key Concept

Conservation of Linear Momentum in Completely Inelastic Collisions

Estimated Time:1m 30s

Question 28Question →

A body of mass $4.0\text{ kg}$ is initially moving due east across a frictionless horizontal surface at a constant velocity of $15\text{ m s}^{-1}$ . A constant horizontal force of $20\text{ N}$ directed due west is then applied to the body for $5.0\text{ s}$ . What is the final velocity of the body?

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

due west

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

due east

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

due east

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

due west

Show answer & explanation

Answer: $10\text{ m s}^{-1}$ due west

Answer

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

due west

According to the impulse-momentum theorem (

F \Delta t = m v - m u

), setting East as the positive direction gives

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

,

F = -20\text{ N}

, and

t = 5.0\text{ s}

. Substituting these values yields

-100 = 4.0(v - 15)

, leading to

v - 15 = -25

, so

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

. The negative sign confirms the final velocity is

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

due west.

Step-by-Step Solution

1

Assign direction signs to vectors and determine initial momentum.

Taking East as positive (

+

) and West as negative (

-

), initial velocity

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

, force

F = -20\text{ N}

, and mass

m = 4.0\text{ kg}

. Initial momentum

p_i = m u = 4.0 \times (+15) = +60\text{ kg m s}^{-1}

.

Linear momentum is a vector quantity, so direction must be tracked consistently.

2

Calculate the impulse exerted by the retarding force.

\text{Impulse } I = F \Delta t = (-20) \times 5.0 = -100\text{ N s} \text{ (or kg m s}^{-1}\text{)}.

The impulse equals the change in linear momentum according to Newton's Second Law.

3

Determine the final momentum and final velocity.

Final momentum

p_f = p_i + I = +60 + (-100) = -40\text{ kg m s}^{-1}

. Final velocity

v = \frac{p_f}{m} = \frac{-40}{4.0} = -10\text{ m s}^{-1}

.

Dividing the final momentum by mass gives the final velocity, where the negative sign indicates motion due west.

Key Concept

Impulse-Momentum Theorem and Directional Vector Conventions

Alternative Method

Using Newton's Second Law directly: Acceleration

a = \frac{F}{m} = \frac{-20\text{ N}}{4.0\text{ kg}} = -5.0\text{ m s}^{-2}

. Using the kinematic formula

v = u + a t = 15 + (-5.0 \times 5.0) = 15 - 25 = -10\text{ m s}^{-1}

. Thus, the final velocity is

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

due west.

Estimated Time:1m 15s

Question 29Question →

A machine gun fires bullets, each of mass $0.020\text{ kg}$ , at a speed of $400\text{ m s}^{-1}$ . If the gun fires $5\text{ bullets per second}$ , what is the magnitude of the average recoil force exerted on the gun in newtons?

Show answer & explanation

Answer: 40

Answer

The magnitude of the average recoil force exerted on the gun is

40\text{ N}

.

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

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

). The total mass of ammunition leaving the gun per second is

5 \times 0.020\text{ kg} = 0.10\text{ kg s}^{-1}

. Multiplying this mass rate by the velocity (

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

) gives a rate of momentum change of

40\text{ N}

, which corresponds directly to the average recoil force.

Step-by-Step Solution

1

Calculate the mass of bullets fired per unit time (mass flow rate).

\frac{\Delta m}{\Delta t} = 5 \text{ bullets/s} \times 0.020 \text{ kg/bullet} = 0.10 \text{ kg s}^{-1}

.

Force is defined as the rate of change of linear momentum, which requires knowing the total mass delivered per second.

2

Apply Newton's second law in terms of momentum change per second.

F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) v = 0.10 \text{ kg s}^{-1} \times 400 \text{ m s}^{-1} = 40 \text{ N}

.

The rate of momentum change of the bullets equals the magnitude of the force exerted on them, which by Newton's third law equals the recoil force on the gun.

Key Concept

Newton's Second Law of Motion and Rate of Change of Linear Momentum

Question 30Question →

In an isolated system of colliding bodies where no net external force acts, total linear momentum is conserved only if the collision is perfectly elastic.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Total linear momentum is conserved in all isolated collisions regardless of whether the collision is elastic or inelastic.

Linear momentum conservation depends strictly on the absence of net external forces. Internal forces, such as those causing deformation or heat release in inelastic collisions, cancel out in equal and opposite pairs according to Newton's Third Law and therefore do not change total linear momentum.

Step-by-Step Solution

1

Analyze the condition for linear momentum conservation.

Linear momentum is conserved whenever the net external force acting on the system is zero (

∑ F_{\text{ext}} = 0

).

By Newton's Second Law written in terms of momentum (

F_{\text{net}} = \frac{\Delta p}{\Delta t}

), if

F_{\text{net}} = 0

, then

\Delta p = 0

, meaning initial total momentum equals final total momentum.

2

Differentiate between momentum conservation and kinetic energy conservation.

Total linear momentum is conserved in both elastic and inelastic collisions, whereas total kinetic energy is conserved only in elastic collisions.

In inelastic collisions, internal forces convert kinetic energy into heat, sound, or mechanical deformation, but internal forces cannot alter the net momentum of the system.

Key Concept

Conservation of Linear Momentum in Collisions

Newton's Laws of Motion and Linear Momentum

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes