Mechanics Practice Questions — JAMB UTME — Page 4

Question 61Question →

A boat moves due north across a river at a speed of $4.0\text{ m s}^{-1}$ relative to the water. If the river current flows due east at a speed of $3.0\text{ m s}^{-1}$ , what is the magnitude of the resultant velocity of the boat relative to the riverbank?

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

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

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

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

Show answer & explanation

Answer: $5.0\text{ m s}^{-1}$

Answer

The magnitude of the resultant velocity of the boat relative to the riverbank is

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

.

The resultant velocity magnitude is found by applying vector addition for perpendicular components:

v_{resultant} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the given velocity components and their directions

Northward velocity component

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

, Eastward velocity component

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

.

The motion of the boat and the movement of the river current act in perpendicular directions (

90^\circ

to each other).

2

Calculate the resultant magnitude using vector addition (Pythagorean theorem)

v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ m s}^{-1}

.

For two orthogonal vectors, the magnitude of the vector sum equals the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector Addition of Perpendicular Quantities

Estimated Time:1m 0s

Question 62Question →

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 63Question →

A solid block of mass $0.60\text{ kg}$ and density $600\text{ kg/m}^3$ is held fully submerged in water of density $1000\text{ kg/m}^3$ by a light vertical string attached to the bottom of a container. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

10.0\text{ N}

4.0\text{ N}

6.0\text{ N}

16.0\text{ N}

Show answer & explanation

Answer: $4.0\text{ N}$

Answer

The tension in the string is

4.0\text{ N}

.

First, the volume of the block is computed as

V = \frac{m}{\rho} = \frac{0.60}{600} = 1.0 \times 10^{-3}\text{ m}^3

. According to Archimedes' principle, the upthrust exerted by the displaced water is

U = \rho_{\text{water}} V g = 1000 \times 1.0 \times 10^{-3} \times 10 = 10.0\text{ N}

. The weight of the block is

W = mg = 0.60 \times 10 = 6.0\text{ N}

. For the block to remain completely submerged in equilibrium, the upward upthrust must balance the downward forces (the weight of the block and the tension

T

pulling downward). Thus,

T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

.

Step-by-Step Solution

1

Calculate the volume of the block using its mass and density.

V = \frac{m}{\rho_{\text{block}}} = \frac{0.60\text{ kg}}{600\text{ kg/m}^3} = 1.0 \times 10^{-3}\text{ m}^3

The volume of fluid displaced equals the total volume of the fully submerged block.

2

Calculate the upward upthrust force exerted by the water.

U = \rho_{\text{water}} \cdot V \cdot g = 1000\text{ kg/m}^3 \times 1.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 10.0\text{ N}

Archimedes' principle states upthrust equals the weight of the displaced fluid.

3

Calculate the downward gravitational weight of the block.

W = m \cdot g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force exerted on the mass of the block by gravity.

4

Apply vertical force equilibrium to solve for string tension.

U = W + T \Rightarrow T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

The string is tied to the bottom, so tension acts downward to hold the buoyant block in equilibrium.

Key Concept

Archimedes' Principle and Static Equilibrium of Submerged Bodies

Question 64Question →

An inclined plane of length $5\text{ m}$ is used to lift a load of $400\text{ N}$ through a vertical height of $1\text{ m}$ . If an effort of $100\text{ N}$ is applied parallel to the incline, what is the efficiency of the machine?

80\%

125\%

25\%

20\%

Show answer & explanation

Answer: $80\%$

Answer

The efficiency of the machine is

80\%

.

The correct answer of

80\%

is obtained by finding the velocity ratio (distance moved by effort divided by distance moved by load,

5 / 1 = 5

) and the mechanical advantage (load divided by effort,

400 / 100 = 4

), then calculating efficiency as

(\text{MA} / \text{VR}) \times 100\% = (4 / 5) \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the Velocity Ratio (VR) of the inclined plane

\text{VR} = \frac{\text{Length of incline}}{\text{Height}} = \frac{5\text{ m}}{1\text{ m}} = 5

Velocity Ratio is the distance moved by the effort divided by the distance moved by the load.

2

Calculate the Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical Advantage measures how many times a machine multiplies the applied force.

3

Calculate the Efficiency

\text{Efficiency} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4}{5}\right) \times 100\% = 80\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Key Concept

Efficiency of Simple Machines (Inclined Plane)

Estimated Time:45s

Question 65Question →

A variable force $F$ acts on a body moving along a straight horizontal path. The force increases linearly from $0\text{ N}$ at position $x = 0\text{ m}$ to $20\text{ N}$ at $x = 4\text{ m}$ , remains constant at $20\text{ N}$ from $x = 4\text{ m}$ to $x = 7\text{ m}$ , and then decreases linearly back to $0\text{ N}$ at $x = 10\text{ m}$ . What is the total work done by the force in Joules over the $10\text{ m}$ displacement?

Show answer & explanation

Answer: 130

Answer

130 J

The work done by a variable force is equal to the total area under the force-displacement (

F

-

x

) graph. The region under the graph forms a trapezoid bounded by parallel sides of length

10\text{ m}

(total displacement) and

3\text{ m}

(constant force interval), with a height of

20\text{ N}

. Calculating the area yields

W = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}

.

Step-by-Step Solution

1

Relate work done to the force-displacement graph

Work done

W

equals the total area under the

F

-

x

graph between

x = 0\text{ m}

and

x = 10\text{ m}

.

By definition,

W = \int F \, dx

, which corresponds to the geometric area under the force-displacement curve.

2

Calculate the geometric area under each section of the graph

First section (

0

to

4\text{ m}

): Triangle area =

\frac{1}{2} \times 4 \times 20 = 40\text{ J}

. Second section (

4

to

7\text{ m}

): Rectangle area =

(7 - 4) \times 20 = 60\text{ J}

. Third section (

7

to

10\text{ m}

): Triangle area =

\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}

.

Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.

3

Sum the areas of all sections

Total Work

W = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}

.

The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Key Concept

Work Done by a Variable Force (Area under Force-Displacement Graph)

Question 66Question →

A crate on a smooth surface is simultaneously pulled by two horizontal forces. One force of magnitude $9.0\text{ N}$ acts towards the East, and another force of magnitude $12.0\text{ N}$ acts towards the North. What is the magnitude of the resultant force acting on the crate in newtons?

Show answer & explanation

Answer: 15

Answer

The magnitude of the resultant force acting on the crate is

15.0\text{ N}

.

Because the two pulling forces are perpendicular, their resultant magnitude is calculated using vector addition via the Pythagorean theorem:

R = \sqrt{9.0^2 + 12.0^2} = 15.0\text{ N}

.

Step-by-Step Solution

1

Determine the angle between the two given vectors

The forces are perpendicular (

90^\circ

).

East and North cardinal directions are orthogonal to each other.

2

Compute the resultant magnitude using vector synthesis

R = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ N}

For perpendicular force vectors, the resultant magnitude is the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector addition of perpendicular forces using the Pythagorean theorem

Estimated Time:1m 15s

Question 67Question →

A solid sphere of mass $0.40\text{ kg}$ and relative density $2.5$ is held fully submerged in a liquid of density $800\text{ kg/m}^3$ by a light string attached to a fixed support. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

2.72\text{ N}

3.60\text{ N}

1.28\text{ N}

2.40\text{ N}

Show answer & explanation

Answer: $2.72\text{ N}$

Answer

The tension in the string is

2.72\text{ N}

.

The sphere has a weight of

4.0\text{ N}

acting downward. When fully submerged, it displaces a volume of liquid equal to its own volume (

1.6 \times 10^{-4}\text{ m}^3

). The displaced liquid of density

800\text{ kg/m}^3

exerts an upward buoyant force (upthrust) of

1.28\text{ N}

. Tension balances the remaining downward force:

T = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

.

Step-by-Step Solution

1

Calculate the weight of the sphere

W = mg = 0.40\text{ kg} \times 10\text{ m/s}^2 = 4.0\text{ N}

The force of gravity acting downward on the mass.

2

Determine the volume of the sphere using its relative density

Density of sphere

\rho_s = 2.5 \times 1000\text{ kg/m}^3 = 2500\text{ kg/m}^3

. Volume

V = \frac{m}{\rho_s} = \frac{0.40}{2500} = 1.6 \times 10^{-4}\text{ m}^3

.

Relative density is the ratio of the substance's density to the density of water (

1000\text{ kg/m}^3

).

3

Calculate the upthrust exerted by the liquid

U = V \cdot \rho_{\text{liquid}} \cdot g = (1.6 \times 10^{-4}) \times 800 \times 10 = 1.28\text{ N}

By Archimedes' principle, upthrust equals the weight of the fluid displaced by the submerged volume.

4

Calculate tension in the string

T = W - U = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

For vertical equilibrium of the submerged body, weight acts downward while upthrust and tension act upward.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:1m 30s

Question 68Question →

A box of mass $2\text{ kg}$ slides down a rough inclined plane from a height of $5\text{ m}$ . If it reaches the bottom of the incline with a speed of $6\text{ m s}^{-1}$ , what is the work done against friction during the descent? (Take $g = 10\text{ m s}^{-2}$ )

64\text{ J}

36\text{ J}

100\text{ J}

136\text{ J}

Show answer & explanation

Answer: $64\text{ J}$

Answer

64\text{ J}

According to the principle of conservation of energy, the work done against friction equals the loss in total mechanical energy. The initial potential energy is

mgh = 2 \times 10 \times 5 = 100\text{ J}

, and the final kinetic energy is

\frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 6^2 = 36\text{ J}

. Subtracting final kinetic energy from initial potential energy yields

100\text{ J} - 36\text{ J} = 64\text{ J}

.

Step-by-Step Solution

1

Calculate the initial potential energy (

E_p

) at height

h = 5\text{ m}

.

E_p = mgh = 2\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 100\text{ J}

At the top of the incline, all mechanical energy is stored as gravitational potential energy.

2

Calculate the final kinetic energy (

E_k

) at the bottom where

v = 6\text{ m s}^{-1}

.

E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2\text{ kg} \times (6\text{ m s}^{-1})^2 = 36\text{ J}

At the bottom of the incline, the remaining mechanical energy is kinetic energy.

3

Apply the work-energy theorem to find the work done against friction (

W_f

).

W_f = E_p - E_k = 100\text{ J} - 36\text{ J} = 64\text{ J}

The non-conservative friction force reduces the total mechanical energy by doing work against motion.

Key Concept

Work-Energy Theorem and Conservation of Energy with Friction

Question 69Question →

A uniform metre rule of mass $120\text{ g}$ is balanced horizontally on a pivot placed at the $40\text{ cm}$ mark when an unknown mass $m$ is suspended at the $10\text{ cm}$ mark. What is the value of the mass $m$ in grams?

Show answer & explanation

Answer: 40

Answer

The mass

m

required to balance the metre rule horizontally is

40\text{ g}

.

According to the Principle of Moments, a system is in rotational equilibrium when the sum of anticlockwise moments equals the sum of clockwise moments about the pivot. The weight of the

120\text{ g}

uniform metre rule acts at its center of gravity (

50\text{ cm}

mark), which is

10\text{ cm}

to the right of the pivot at

40\text{ cm}

. This creates a clockwise moment of

120\text{ g} \times 10\text{ cm} = 1200\text{ g}\cdot\text{cm}

. The mass

m

is suspended at the

10\text{ cm}

mark, which is

30\text{ cm}

to the left of the pivot, creating an anticlockwise moment of

m \times 30\text{ cm}

. Setting

30m = 1200

yields

m = 40\text{ g}

.

Step-by-Step Solution

1

Determine the position of the center of gravity of the metre rule.

Center of gravity is at the

50\text{ cm}

mark.

A uniform metre rule has its weight concentrated at its geometric midpoint.

2

Calculate perpendicular distances from the pivot at

40\text{ cm}

to all acting forces.

Distance to rule weight =

50\text{ cm} - 40\text{ cm} = 10\text{ cm}

; Distance to mass

m

=

40\text{ cm} - 10\text{ cm} = 30\text{ cm}

.

Moments are calculated by multiplying force (or mass) by perpendicular distance from the turning point.

3

Apply the Principle of Moments for equilibrium.

Sum of anticlockwise moments = Sum of clockwise moments

\implies m \times 30 = 120 \times 10

.

For rotational equilibrium, the total clockwise moment must equal the total anticlockwise moment about the pivot.

4

Solve for the unknown mass

m

.

m = 40\text{ g}

.

Dividing

1200\text{ g}\cdot\text{cm}

by

30\text{ cm}

gives

40\text{ g}

.

Key Concept

Principle of Moments and Center of Gravity of a Uniform Body

Estimated Time:50s

Question 70Question →

A rigid rod $AB$ of length $2.0\text{ m}$ is hinged at end $A$ . A force of $50\text{ N}$ is applied at end $B$ at an angle of $30^\circ$ to the rod. What is the moment of this force about the hinge $A$ ?

25\text{ N m}

50\text{ N m}

86.6\text{ N m}

100\text{ N m}

Show answer & explanation

Answer: $50\text{ N m}$

Answer

The moment of the force about hinge

A

is

50\text{ N m}

.

The moment of a force about a pivot is calculated by multiplying the force magnitude by the perpendicular distance to the pivot line of action (

F \times L \sin\theta

). Substituting

F = 50\text{ N}

,

L = 2.0\text{ m}

, and

\sin(30^\circ) = 0.5

gives

50 \times 2.0 \times 0.5 = 50\text{ N m}

.

Step-by-Step Solution

1

Identify the formula for the moment of a force applied at an angle.

\text{Moment} = F \times d_\perp = F \times L \sin\theta

Moment is defined as the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the pivot point.

2

Substitute the given values into the formula.

\text{Moment} = 50\text{ N} \times 2.0\text{ m} \times \sin(30^\circ)

The force

F = 50\text{ N}

, distance

L = 2.0\text{ m}

, and angle

\theta = 30^\circ

.

3

Calculate the final moment value.

\text{Moment} = 50 \times 2.0 \times 0.5 = 50\text{ N m}

Since

\sin(30^\circ) = 0.5

, the perpendicular distance is

1.0\text{ m}

, giving a moment of

50\text{ N m}

.

Key Concept

Moment of a Force at an Angle

Estimated Time:45s

Question 71Question →

A body of mass $2\text{ kg}$ is projected vertically upward from ground level with an initial speed of $30\text{ m s}^{-1}$ . During its entire flight, it experiences a constant resistive force due to air resistance of $5\text{ N}$ . Taking $g = 10\text{ m s}^{-2}$ , calculate the kinetic energy of the body in Joules when it returns to the ground level.

Show answer & explanation

Answer: 540

Answer

The kinetic energy of the body when it returns to ground level is

540\text{ J}

.

At launch, the body possesses an initial kinetic energy of

E_i = \frac{1}{2}(2)(30)^2 = 900\text{ J}

. During ascent, both gravity (

20\text{ N}

) and air resistance (

5\text{ N}

) retard the motion, giving a net downward force of

25\text{ N}

and a deceleration of

12.5\text{ m s}^{-2}

. The maximum height reached is

h = \frac{30^2}{2(12.5)} = 36\text{ m}

. Since the body travels up and down, total distance covered is

72\text{ m}

. Non-conservative work done against air resistance is

W = 5\text{ N} \times 72\text{ m} = 360\text{ J}

. Therefore, the remaining kinetic energy upon returning to the ground is

900\text{ J} - 360\text{ J} = 540\text{ J}

.

Step-by-Step Solution

1

Calculate initial kinetic energy of launch

E_i = 900\text{ J}

Kinetic energy is given by

\frac{1}{2}m u^2 = \frac{1}{2}(2\text{ kg})(30\text{ m s}^{-1})^2 = 900\text{ J}

.

2

Determine maximum height reached during ascent

h = 36\text{ m}

Net upward retarding force

F = mg + F_{\text{air}} = 20 + 5 = 25\text{ N}

, yielding a deceleration

a = 12.5\text{ m s}^{-2}

. Using

0 = u^2 - 2ah

,

h = \frac{900}{25} = 36\text{ m}

.

3

Calculate energy lost to air resistance over total trajectory

W_{\text{air}} = 360\text{ J}

Air resistance acts continuously over both ascent and descent (total distance

2h = 72\text{ m}

). Work dissipated

= F_{\text{air}} \times 2h = 5 \times 72 = 360\text{ J}

.

4

Subtract non-conservative work loss from initial mechanical energy

E_f = 540\text{ J}

By mechanical energy balance, final kinetic energy

E_f = E_i - W_{\text{air}} = 900 - 360 = 540\text{ J}

.

Key Concept

Work-Energy Theorem and Mechanical Energy Dissipation by Non-Conservative Forces

Question 72Question →

A car initially traveling at a constant speed of $15\text{ m/s}$ accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$ . It then maintains the acquired maximum speed for $10\text{ s}$ before coming to rest under uniform retardation in $4\text{ s}$ . What is the total distance covered by the car during the entire motion?

Show answer & explanation

Answer: 400

Answer

The total distance covered by the car during the entire motion is

400\text{ m}

.

The total distance is calculated by summing the distances covered in the three distinct phases of motion: acceleration (

100\text{ m}

), uniform velocity (

250\text{ m}

), and uniform retardation (

50\text{ m}

), yielding a total distance of

400\text{ m}

.

Step-by-Step Solution

1

Calculate final speed and distance for the acceleration phase.

Final speed

v = 25\text{ m/s}

and distance

s_1 = 100\text{ m}

.

Using kinematic equations

v = u + a t_1 = 15 + (2)(5) = 25\text{ m/s}

and

s_1 = u t_1 + \frac{1}{2}a t_1^2 = 15(5) + \frac{1}{2}(2)(5^2) = 75 + 25 = 100\text{ m}

.

2

Calculate the distance covered during the constant speed phase.

Distance

s_2 = 250\text{ m}

.

The car maintains the acquired speed of

25\text{ m/s}

for

10\text{ s}

, giving

s_2 = v \cdot t_2 = 25 \times 10 = 250\text{ m}

.

3

Calculate the distance covered during the retardation phase.

Distance

s_3 = 50\text{ m}

.

Using average velocity for uniform retardation to rest:

s_3 = \frac{v + 0}{2} t_3 = \frac{25}{2} \times 4 = 50\text{ m}

.

4

Sum the distances from all three stages to determine total distance.

Total distance

S = 400\text{ m}

.

S = s_1 + s_2 + s_3 = 100 + 250 + 50 = 400\text{ m}

.

Key Concept

Multi-stage linear motion and equations of uniform acceleration

Question 73Question →

A wheel and axle machine having a wheel radius of $25\text{ cm}$ and an axle radius of $5\text{ cm}$ is used to raise a load of mass $80\text{ kg}$ vertically through a height of $10\text{ m}$ . If the efficiency of the machine is $80\%$ , what is the work done against friction during this operation? (Take $g = 10\text{ m s}^{-2}$ )

1,600\text{ J}

2,000\text{ J}

8,000\text{ J}

10,000\text{ J}

Show answer & explanation

Answer: $2,000\text{ J}$

Answer

2,000\text{ J}

The useful work done on the load is

W_{\text{out}} = mgh = 80 \times 10 \times 10 = 8000\text{ J}

. Since efficiency is

80\%

, the total work input required is

W_{\text{in}} = \frac{8000}{0.80} = 10,000\text{ J}

. Therefore, the work lost to friction is

W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

.

Step-by-Step Solution

1

Calculate the useful work output of the machine

W_{\text{out}} = mgh = 80\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 8,000\text{ J}

Useful work output is the gravitational potential energy gained by raising the load.

2

Calculate the total work input using the efficiency equation

W_{\text{in}} = \frac{W_{\text{out}}}{\text{Efficiency}} = \frac{8,000\text{ J}}{0.80} = 10,000\text{ J}

Efficiency is defined as

\frac{\text{Work Output}}{\text{Work Input}}

, so Work Input =

\frac{\text{Work Output}}{\text{Efficiency}}

.

3

Determine the work done against friction

W_{\text{friction}} = W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

The energy wasted as heat and friction is the difference between total work input and useful work output.

Key Concept

Efficiency of Simple Machines and Work Done Against Friction

Question 74Question →

A traffic officer on a stationary motorcycle spots a car passing at a constant speed of $15\text{ m/s}$ . The officer immediately pursues the car, accelerating uniformly from rest at $4\text{ m/s}^2$ for $5\text{ s}$ , after which the motorcycle continues at the constant speed attained. How long after setting off does the motorcycle overtake the car?

10\text{ s}

7.5\text{ s}

5\text{ s}

15\text{ s}

Show answer & explanation

Answer: $10\text{ s}$

Answer

The total time required for the motorcycle to overtake the car is

10\text{ s}

.

During the first

5\text{ s}

, the motorcycle accelerates from rest at

4\text{ m/s}^2

to

20\text{ m/s}

, covering

50\text{ m}

. In that same interval, the car moving at

15\text{ m/s}

covers

75\text{ m}

, creating a

25\text{ m}

separation gap. Beyond

5\text{ s}

, the motorcycle maintains

20\text{ m/s}

and closes in on the car (

15\text{ m/s}

) at a relative speed of

5\text{ m/s}

. It takes

5\text{ s}

to close the

25\text{ m}

gap, making the total pursuit time

10\text{ s}

.

Step-by-Step Solution

1

Calculate the state of both vehicles at the end of the motorcycle's acceleration phase (

t_1 = 5\text{ s}

).

Motorcycle speed

v = u + at = 0 + 4(5) = 20\text{ m/s}

. Motorcycle distance

s_1 = \frac{1}{2} a t^2 = \frac{1}{2}(4)(5^2) = 50\text{ m}

. Car distance

s_{\text{car}} = v_{\text{car}} \times t = 15 \times 5 = 75\text{ m}

.

Determine the position gap and relative speed after the acceleration phase ends.

2

Determine the remaining distance gap and relative speed between the vehicles.

Separation gap

d = 75\text{ m} - 50\text{ m} = 25\text{ m}

. Relative speed

v_{\text{rel}} = 20\text{ m/s} - 15\text{ m/s} = 5\text{ m/s}

.

Both vehicles now move at constant velocities, so relative speed determines how quickly the gap closes.

3

Calculate the time taken in the second phase and sum for total time.

Phase 2 time

t_2 = \frac{d}{v_{\text{rel}}} = \frac{25}{5} = 5\text{ s}

. Total time

t = t_1 + t_2 = 5\text{ s} + 5\text{ s} = 10\text{ s}

.

The total duration of pursuit is the acceleration duration plus the constant velocity catch-up duration.

Key Concept

Multi-stage linear motion involving uniform acceleration followed by constant velocity.

Estimated Time:1m 30s

Question 75Question →

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 76Question →

A cannonball is fired from ground level with an initial speed of $50\text{ m/s}$ at an angle $\theta$ above the horizontal such that $\tan \theta = \frac{4}{3}$ . Assuming acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the magnitude of the velocity of the cannonball at the highest point of its trajectory?

30\text{ m/s}

0\text{ m/s}

40\text{ m/s}

50\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

30\text{ m/s}

At the maximum height of a projectile's flight, the vertical velocity component drops to zero due to gravity, while the horizontal component remains unchanged because horizontal acceleration is zero. For a launch speed of

50\text{ m/s}

at an angle with

\tan \theta = \frac{4}{3}

,

\cos \theta = \frac{3}{5}

. The magnitude of the velocity at apex is therefore equal to

u_x = 50 \times \frac{3}{5} = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve the initial velocity vector into orthogonal horizontal and vertical components.

Given

\tan \theta = \frac{4}{3}

, the trigonometric ratios are

\cos \theta = \frac{3}{5}

and

\sin \theta = \frac{4}{5}

. Thus,

u_x = 50 \times \frac{3}{5} = 30\text{ m/s}

and

u_y = 50 \times \frac{4}{5} = 40\text{ m/s}

.

Resolving 2D projectile motion into independent orthogonal components simplifies kinematic analysis.

2

Determine the velocity components at the apex (highest point) of the trajectory.

At maximum height, the vertical velocity component

v_y = 0\text{ m/s}

. Since there is no horizontal acceleration, the horizontal velocity component remains

v_x = u_x = 30\text{ m/s}

.

Gravity acts purely vertically, reducing vertical velocity to zero at the apex while leaving horizontal velocity unchanged.

3

Calculate the magnitude of total velocity at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

.

The magnitude of a 2D velocity vector is determined by combining its orthogonal components using the Pythagorean formula.

Key Concept

Velocity components at maximum height in projectile motion

Question 77Question →

A stone is thrown vertically downwards from the top of a $60\text{ m}$ high tower with an initial speed of $5\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the time taken, in seconds, for the stone to reach the ground.

Show answer & explanation

Answer: 3

Answer

The stone takes

3\text{ s}

to reach the ground.

Applying the equation

h = ut + \frac{1}{2}gt^2

with

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

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

yields

60 = 5t + 5t^2

. Dividing by

5

gives

t^2 + t - 12 = 0

, which factorizes into

(t + 4)(t - 3) = 0

. Rejecting

t = -4\text{ s}

leaves the correct time of

3\text{ s}

.

Step-by-Step Solution

1

Set up the vertical motion equation

Using

h = ut + \frac{1}{2}gt^2

, substitute

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

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

.

This equation directly relates displacement, initial speed, acceleration, and time.

2

Form and solve the quadratic equation for time

60 = 5t + 5t^2 \Rightarrow t^2 + t - 12 = 0 \Rightarrow (t - 3)(t + 4) = 0

, giving

t = 3\text{ s}

.

Time must be positive, so the physically meaningful solution is

t = 3\text{ s}

.

Key Concept

Vertical motion under gravity with non-zero initial downward velocity

Question 78Question →

A projectile is launched from horizontal ground with an initial speed of $60\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the speed of the projectile at its maximum height?

0\text{ m/s}

30\text{ m/s}

30\sqrt{3}\text{ m/s}

60\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

The speed of the projectile at its maximum height is

30\text{ m/s}

.

In two-dimensional projectile motion, gravity acts purely in the vertical direction. At the apex (maximum height), the vertical velocity component

v_y

drops to zero. However, the horizontal velocity component

v_x = u \cos \theta

remains constant throughout the entire flight because air resistance is neglected. Consequently, the speed at maximum height equals

u \cos 60^\circ = 60 \times 0.5 = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial launch velocity into orthogonal horizontal and vertical components.

u_x = u \cos \theta = 60 \cos 60^\circ = 30\text{ m/s}

and

u_y = u \sin \theta = 60 \sin 60^\circ = 30\sqrt{3}\text{ m/s}

.

Projectile motion consists of independent horizontal and vertical motions.

2

Determine the velocity components at the apex (maximum height).

At the peak, vertical velocity

v_y = 0\text{ m/s}

while horizontal velocity remains

v_x = u_x = 30\text{ m/s}

.

Gravity acts vertically causing

v_y

to become zero at peak height, whereas no horizontal force acts, leaving

v_x

unchanged.

3

Calculate the magnitude of total velocity (speed) at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

.

Speed is the resultant magnitude of orthogonal velocity components.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 79Question →

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 80Question →

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

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