Mechanics Practice Questions — JAMB UTME — Page 12

Question 221Question →

A projectile is launched from level ground with a horizontal velocity component of $18\text{ m/s}$ and an initial vertical velocity component of $24\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at the apex of its trajectory?

0\text{ m/s}

18\text{ m/s}

24\text{ m/s}

30\text{ m/s}

Show answer & explanation

Answer: $18\text{ m/s}$

Answer

18\text{ m/s}

At the apex of a projectile's flight, vertical motion momentarily halts (

v_y = 0\text{ m/s}

), but horizontal motion continues unchanged (

v_x = 18\text{ m/s}

). Thus, the magnitude of the velocity at the apex equals the horizontal velocity component,

18\text{ m/s}

.

Step-by-Step Solution

1

Analyze the velocity components at the trajectory apex (highest point)

Vertical velocity component

v_y = 0\text{ m/s}

and horizontal velocity component

v_x = u_x = 18\text{ m/s}

Gravity acts downward, reducing the vertical speed component to zero at maximum height. In the absence of air resistance, no horizontal forces act on the projectile, so

v_x

remains constant throughout the flight.

2

Calculate the magnitude of the total velocity at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{18^2 + 0^2} = 18\text{ m/s}

The total velocity magnitude is obtained by vector addition of its orthogonal components.

Key Concept

Independence of horizontal and vertical motion components in projectile motion.

Estimated Time:1m 0s

Question 222Question →

An object is launched vertically upward from the ground with an initial velocity of $40\text{ m/s}$ . At the exact same instant, a second object is dropped from rest from a height of $100\text{ m}$ directly above the first object. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , at what time (in seconds) after launch will the two objects meet?

Show answer & explanation

Answer: 2.5

Answer

The two objects meet after 2.5 seconds.

Because both objects experience identical downward gravitational acceleration (

g = 10\text{ m/s}^2

), their relative acceleration is zero. The relative velocity between them remains constant at

40\text{ m/s}

. The time to cover the initial separation distance of

100\text{ m}

is calculated directly as

t = \frac{\text{distance}}{\text{relative velocity}} = \frac{100}{40} = 2.5\text{ s}

.

Step-by-Step Solution

1

Set up the position-time equations for both objects taking ground level as

y = 0\text{ m}

.

For the upward-launched object:

y_1(t) = u t - \frac{1}{2}gt^2 = 40t - 5t^2

. For the dropped object:

y_2(t) = h_0 - \frac{1}{2}gt^2 = 100 - 5t^2

.

Kinematic equations of motion under uniform gravitational acceleration apply to both bodies.

2

Equate the two vertical position equations to solve for the meeting time

t

.

40t - 5t^2 = 100 - 5t^2 \implies 40t = 100

When the objects meet, they share the exact same vertical position

y_1(t) = y_2(t)

at time

t

.

3

Calculate the value of

t

.

t = \frac{100}{40} = 2.5\text{ s}

Direct algebraic division yields the time elapsed before collision/meeting.

Key Concept

Relative vertical motion under uniform gravity

Estimated Time:1m 30s

Question 223Question →

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 224Question →

A particle traveling along a straight line accelerates uniformly from an initial speed $u$ to a final speed of $25\text{ m/s}$ over a distance of $150\text{ m}$ in a time duration of $8\text{ s}$ . What is the initial speed $u$ in $\text{m/s}$ ?

Show answer & explanation

Answer: 12.5

Answer

The initial speed of the particle is

12.5\text{ m/s}

.

Under uniform acceleration, displacement is given by the product of average velocity and time:

s = \frac{u + v}{2}t

. Substituting

s = 150\text{ m}

,

v = 25\text{ m/s}

, and

t = 8\text{ s}

gives

150 = 4(u + 25)

, which yields

u = 12.5\text{ m/s}

.

Step-by-Step Solution

1

Identify the given parameters and select the appropriate kinematic equation.

Givens:

s = 150\text{ m}

,

v = 25\text{ m/s}

,

t = 8\text{ s}

. Formula:

s = \frac{u + v}{2}t

.

This formula relates displacement directly to average velocity and time under uniform acceleration without needing the acceleration variable.

2

Substitute the given values into the formula.

150 = \left(\frac{u + 25}{2}\right) \times 8 = 4(u + 25)

.

Simplifying

\frac{8}{2}

gives a factor of

4

multiplying

(u + 25)

.

3

Isolate and solve for the unknown initial speed

u

.

u + 25 = \frac{150}{4} = 37.5 \implies u = 12.5\text{ m/s}

.

Subtracting

25

from

37.5

yields the value of

u

.

Key Concept

Kinematics with uniform linear acceleration

Question 225Question →

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 226Question →

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 227Question →

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

Mechanics

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