Mechanics Practice Questions — JAMB UTME — Page 2

Question 21Question →

Two point masses are separated by a distance $r$ and exert a gravitational force $F$ on each other. If the distance between them is doubled while keeping their masses constant, what is the new gravitational force between them?

\frac{F}{4}

\frac{F}{2}

2F

4F

Show answer & explanation

Answer: $\frac{F}{4}$

Answer

The new gravitational force between the two masses is

\frac{F}{4}

.

Gravitational force obeys an inverse-square law with respect to distance (

F \propto \frac{1}{r^2}

). When separation distance is multiplied by 2, the force decreases by a factor of

2^2 = 4

, yielding

\frac{F}{4}

.

Step-by-Step Solution

1

Write down Newton's Law of Universal Gravitation

F = \frac{G m_1 m_2}{r^2}

Establish the mathematical relationship governing gravitational force and separation distance.

2

Substitute the new distance

r' = 2r

into the gravitational force equation

F' = \frac{G m_1 m_2}{(2r)^2} = \frac{G m_1 m_2}{4r^2}

Evaluate how doubling the separation distance affects the magnitude of the force.

3

Express the new force

F'

in terms of the initial force

F

F' = \frac{1}{4}\left(\frac{G m_1 m_2}{r^2}\right) = \frac{F}{4}

Relate the calculated force directly to the original force

F

.

Key Concept

Inverse-Square Law of Gravitation

Estimated Time:45s

Question 22Question →

A simple pendulum suspended in a terrestrial laboratory has a period of oscillation $T$ when fitted with a bob of mass $m$ . If the bob is replaced by another bob of mass $4m$ and the entire setup is moved to a high-altitude station where the acceleration due to gravity is $\frac{g}{4}$ , what is the new period of oscillation of the pendulum?

2T

4T

8T

\frac{T}{2}

Show answer & explanation

Answer: $2T$

Answer

The new period of oscillation is

2T

.

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. It depends strictly on the length of the string

L

and the local gravitational acceleration

g

, making it independent of the bob's mass

m

. Replacing mass

m

with

4m

does not alter the period. When gravity decreases to

g' = \frac{g}{4}

, the new period becomes

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2 \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum

The period

T

of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

, where

L

is the pendulum length and

g

is the acceleration due to gravity.

This formula defines how physical parameters affect the oscillatory period of a pendulum under simple harmonic motion.

2

Analyze the effect of changing the bob's mass

Changing the mass of the bob from

m

to

4m

has no effect on the period.

The equation for the period of a simple pendulum contains no mass term, demonstrating that mass does not influence the period of small-angle oscillations.

3

Calculate the new period

T'

under the altered gravitational field

g' = \frac{g}{4}

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Reducing the gravitational acceleration to one-fourth increases the square-root term

\sqrt{\frac{1}{1/4}} = 2

, thereby doubling the period.

Key Concept

Mass Independence and Gravity Dependence of Simple Pendulum Period

Question 23Question →

A particle executes simple harmonic motion along a straight line with an amplitude of $0.10\text{ m}$ . At what displacement from the equilibrium position is the kinetic energy of the particle equal to three times its potential energy?

0.05\text{ m}

0.087\text{ m}

0.071\text{ m}

0.025\text{ m}

Show answer & explanation

Answer: $0.05\text{ m}$

Answer

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

U = \frac{1}{2}kx^2

and kinetic energy is

K = \frac{1}{2}k(A^2 - x^2)

. Equating

K = 3U

yields

A^2 - x^2 = 3x^2

, which simplifies to

4x^2 = A^2

or

x = \frac{A}{2}

. For an amplitude of

0.10\text{ m}

, the displacement is

0.05\text{ m}

.

Step-by-Step Solution

1

Write the expressions for kinetic energy

K

and potential energy

U

in simple harmonic motion.

U = \frac{1}{2} k x^2

and

K = \frac{1}{2} k (A^2 - x^2)

, where

A

is amplitude and

x

is displacement.

These equations express the energy distribution at any displacement

x

.

2

Set up the condition given in the problem,

K = 3U

.

\frac{1}{2} k (A^2 - x^2) = 3 \times \left(\frac{1}{2} k x^2\right) \implies A^2 - x^2 = 3x^2

.

Canceling common factor

\frac{1}{2} k

simplifies the relationship between amplitude and displacement.

3

Solve for displacement

x

in terms of amplitude

A

.

A^2 = 4x^2 \implies x = \frac{A}{2}

.

Taking the square root of both sides gives the position where kinetic energy is three times potential energy.

4

Substitute the given amplitude

A = 0.10\text{ m}

to find

x

.

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

.

Carrying out the calculation yields the final numerical displacement.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Estimated Time:1m 15s

Question 24Question →

A body accelerates uniformly from rest at a rate of $4\text{ m/s}^2$ for $6\text{ s}$ . What is the distance covered by the body during this time interval?

72\text{ m}

144\text{ m}

24\text{ m}

12\text{ m}

Show answer & explanation

Answer: $72\text{ m}$

Answer

The distance covered by the body is

72\text{ m}

.

Applying the equation of motion

s = ut + \frac{1}{2}at^2

with

u = 0\text{ m/s}

,

a = 4\text{ m/s}^2

, and

t = 6\text{ s}

yields

s = 0 + \frac{1}{2}(4)(6^2) = 72\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic parameters

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 4\text{ m/s}^2

, and time interval

t = 6\text{ s}

.

Since the body starts from rest, its initial velocity is zero.

2

Select and set up the equation of motion for displacement

s = ut + \frac{1}{2}at^2 = (0)(6) + \frac{1}{2}(4)(6)^2

This formula directly relates displacement to initial velocity, acceleration, and time under constant acceleration.

3

Calculate the total distance

s = \frac{1}{2} \times 4 \times 36 = 72\text{ m}

Squaring

6\text{ s}

yields

36\text{ s}^2

, and multiplying by

2\text{ m/s}^2

gives

72\text{ m}

.

Key Concept

Linear motion under uniform acceleration

Estimated Time:45s

Question 25Question →

A body is released from rest from the top of a cliff of height $h$ . If it covers a distance equal to $\frac{7}{16}h$ in the final second of its motion before hitting the ground, what is the total height $h$ of the cliff? (Take acceleration due to gravity $g = 10\text{ m s}^{-2}$ )

45\text{ m}

80\text{ m}

100\text{ m}

160\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

80\text{ m}

Using the equation of motion under constant acceleration, the total distance fallen from rest in time

t

is

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

. The distance fallen up to time

(t-1)

is

h_{t-1} = \frac{1}{2}g(t-1)^2

. The distance traveled during the final second is

\Delta h = h - h_{t-1} = \frac{1}{2}g(2t-1)

. Equating this to

\frac{7}{16}h

yields

\frac{1}{2}g(2t-1) = \frac{7}{16}\left(\frac{1}{2}gt^2\right)

, which simplifies to

7t^2 - 32t + 16 = 0

. Factoring gives

t = 4\text{ s}

(rejecting

t = \frac{4}{7}\text{ s}

since time must exceed

1\text{ s}

). Substituting

t = 4\text{ s}

into

h = \frac{1}{2}(10)(4)^2

gives

80\text{ m}

.

Step-by-Step Solution

1

Express total height

h

in terms of total fall time

t

.

h = \frac{1}{2} g t^2 = 5t^2

Since the body starts from rest (

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

), displacement under uniform acceleration

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

is given by

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

.

2

Express the height fallen in the first

(t - 1)

seconds.

h' = \frac{1}{2} g (t - 1)^2 = 5(t - 1)^2

The distance covered up to one second before impact is the total distance fallen minus the distance covered in the final second.

3

Calculate the distance fallen in the final second and set up the equation.

\Delta h = h - h' = 5t^2 - 5(t - 1)^2 = 5(2t - 1)

. Given

\Delta h = \frac{7}{16}h

, we have

5(2t - 1) = \frac{7}{16}(5t^2)

.

The distance fallen during the last second is the difference between total height and height fallen up to

(t-1)

seconds.

4

Solve the quadratic equation for

t

.

7t^2 - 32t + 16 = 0 \implies (7t - 4)(t - 4) = 0 \implies t = 4\text{ s}

(since

t > 1\text{ s}

).

Simplifying

2t - 1 = \frac{7}{16}t^2

gives

7t^2 - 32t + 16 = 0

. The root

t = 4/7\text{ s}

is discarded as

t

must be greater than

1\text{ s}

.

5

Substitute

t = 4\text{ s}

back into the total height formula.

h = 5(4)^2 = 80\text{ m}

.

Calculating total height using

h = 5t^2

for

t = 4\text{ s}

gives

80\text{ m}

.

Key Concept

Free Fall under Gravity and Motion in the

n

-th Second

Question 26Question →

A solid object of volume $0.002\text{ m}^3$ is completely immersed in water of density $1000\text{ kg/m}^3$ . What is the magnitude of the upthrust exerted on the object by the water? [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The magnitude of the upthrust exerted on the object is

20\text{ N}

.

According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using

U = V \cdot \rho \cdot g

. Substituting

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

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

yields

U = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement.

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

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

.

These are the essential inputs required to calculate the weight of the displaced liquid.

2

Apply Archimedes' principle to find upthrust force.

U = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Key Concept

Archimedes' Principle and Upthrust

Question 27Question →

The volume flow rate

Q

of a viscous liquid flowing through a pipe of radius

r

under a pressure gradient

\frac{\Delta P}{l}

is modeled by the equation:

Q = k \eta^x r^y \left(\frac{\Delta P}{l}\right)^z

where

\eta

is the coefficient of dynamic viscosity and

k

is a dimensionless constant. What are the values of the exponents

x

,

y

, and

z

respectively?

x = -1, y = 4, z = 1

x = 1, y = 4, z = -1

x = -1, y = 3, z = 1

x = 1, y = 2, z = 1

Show answer & explanation

Answer: $x = -1, y = 4, z = 1$

Answer

x = -1

,

y = 4

, and

z = 1

The dimensional representation of volume flow rate is

[L^3 T^{-1}]

, dynamic viscosity is

[M L^{-1} T^{-1}]

, radius is

[L]

, and pressure gradient is

[M L^{-2} T^{-2}]

. Equating the powers of

M

,

L

, and

T

gives

x + z = 0

,

-x - 2z = -1

, and

-x + y - 2z = 3

. Solving these simultaneously yields

x = -1

,

y = 4

, and

z = 1

.

Step-by-Step Solution

1

Determine the fundamental dimensions of each physical quantity

Volume flow rate

Q = \frac{\text{Volume}}{\text{Time}} = [L^3 T^{-1}]

;
Dynamic viscosity

\eta = [M L^{-1} T^{-1}]

;
Radius

r = [L]

;
Pressure gradient

\frac{\Delta P}{l} = \frac{\text{Pressure}}{\text{Length}} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}]

.

Correct base dimensions are required for dimensional analysis.

2

Set up the dimensional equation by substituting the base dimensions into the formula

[L^3 T^{-1}] = [M L^{-1} T^{-1}]^x [L]^y [M L^{-2} T^{-2}]^z = M^{x+z} L^{-x + y - 2z} T^{-x - 2z}

.

The principle of dimensional homogeneity requires both sides of the equation to have matching exponents for

M

,

L

, and

T

.

3

Equate exponents for

M

,

T

, and

L

to form algebraic equations

For

M

:

x + z = 0 \implies z = -x

For

T

:

-x - 2z = -1

For

L

:

-x + y - 2z = 3

.

This creates a linear system of equations for the exponents

x

,

y

, and

z

.

4

Solve the system of linear equations

Substituting

z = -x

into the

T

equation:

-x - 2(-x) = -1 \implies x = -1

.
Hence

z = -(-1) = 1

.
Substituting

x = -1

and

z = 1

into the

L

equation:

-(-1) + y - 2(1) = 3 \implies 1 + y - 2 = 3 \implies y = 4

.

Yields the unique set of exponents

x = -1, y = 4, z = 1

.

Key Concept

Dimensional Analysis and Homogeneity

Estimated Time:2m 0s

Question 28Question →

A $0.50\text{ kg}$ mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is $0.16\text{ J}$ and the force constant of the spring is $32\text{ N/m}$ . What is the speed of the mass, in $\text{m/s}$ , at the instant when the magnitude of its acceleration is $3.84\text{ m/s}^2$ ?

Show answer & explanation

Answer: 0.64

Answer

The speed of the mass at that instant is

0.64\text{ m/s}

.

Using the relation for total mechanical energy

E = \frac{1}{2}kA^2

, the amplitude is

A = 0.10\text{ m}

. The angular frequency is

\omega = \sqrt{k/m} = 8.0\text{ rad/s}

. From

|a| = \omega^2 |x|

, the displacement magnitude when acceleration is

3.84\text{ m/s}^2

is

|x| = 0.06\text{ m}

. Substituting these values into

v = \omega \sqrt{A^2 - x^2}

yields

v = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the simple harmonic motion

\omega = 8.0\text{ rad/s}

The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.

2

Calculate the amplitude of oscillation from total energy

A = 0.10\text{ m}

The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.

3

Find the magnitude of displacement corresponding to the given acceleration

|x| = 0.06\text{ m}

In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.

4

Calculate the speed at this displacement using the SHM velocity-displacement relation

v = 0.64\text{ m/s}

Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Key Concept

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion

Question 29Question →

Match each vector scenario on the left with its correct resultant magnitude or value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match perpendicular forces of 6 N and 8 N to 10 N; minimum resultant of 7 N and 12 N forces to 5 N; two 15 N forces at 120 degrees to 15 N; and 3D displacement vector to 13 m.

Each vector calculation correctly applies the geometric or algebraic properties of vectors: orthogonal vector resolution, opposite-direction subtraction, law of cosines for equal magnitudes at 120 degrees, and 3D component magnitude synthesis.

Step-by-Step Solution

1

Calculate the magnitude of perpendicular vectors

R = \sqrt{6^2 + 8^2} = 10\text{ N}

Perpendicular vectors form a right-angled triangle, so the Pythagorean theorem applies.

2

Find the minimum resultant magnitude of two vectors

R_{\text{min}} = 12\text{ N} - 7\text{ N} = 5\text{ N}

Minimum resultant occurs when vectors act collinear in opposite directions.

3

Determine the resultant of two equal vectors at

120^\circ

R = 15\text{ N}

Using the cosine rule

R = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ}

, since

\cos 120^\circ = -0.5

,

R = F

.

4

Compute the magnitude of the 3D displacement vector

|\mathbf{r}| = \sqrt{3^2 + 4^2 + 12^2} = 13\text{ m}

3D magnitude is calculated using the square root of the sum of squared orthogonal components.

Key Concept

Vector Addition, Resolution, and Magnitude Evaluation

Question 30Question →

A projectile is launched from ground level over flat terrain. At time $t = 2\text{ s}$ after launch, the projectile passes through a point located $60\text{ m}$ horizontally and $60\text{ m}$ vertically from its launch point. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the total horizontal range of the projectile in meters?

Show answer & explanation

Answer: 240

Answer

The total horizontal range of the projectile is

240\text{ m}

.

The horizontal motion occurs at a constant velocity of

30\text{ m/s}

calculated from

\frac{60\text{ m}}{2\text{ s}}

. Substituting the vertical position (

60\text{ m}

) and time (

2\text{ s}

) into

y = u_y t - 5t^2

yields an initial vertical velocity of

40\text{ m/s}

. The total duration in the air is

T = \frac{2(40)}{10} = 8\text{ s}

. The total horizontal range is therefore

30\text{ m/s} \times 8\text{ s} = 240\text{ m}

.

Step-by-Step Solution

1

Determine the horizontal component of velocity

v_x = 30\text{ m/s}

Horizontal velocity remains constant throughout flight because there is no horizontal acceleration.

2

Determine the initial vertical component of velocity

u_y = 40\text{ m/s}

Applying the vertical displacement equation

y = u_y t - \frac{1}{2}g t^2

with

y = 60\text{ m}

,

t = 2\text{ s}

, and

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

.

3

Calculate the total time of flight

T = 8\text{ s}

The projectile completes its full parabolic trajectory when vertical displacement returns to zero, given by

T = \frac{2 u_y}{g}

.

4

Calculate the total horizontal range

R = 240\text{ m}

The total range is the product of the constant horizontal velocity component and total time of flight (

R = v_x \times T

).

Key Concept

Independence of horizontal and vertical components of projectile motion

Question 31Question →

A train accelerates uniformly along a straight track from an initial velocity of $10\text{ m/s}$ to a final velocity of $30\text{ m/s}$ over a distance of $100\text{ m}$ . What is the magnitude of the acceleration of the train?

0.2\text{ m/s}^2

4\text{ m/s}^2

5\text{ m/s}^2

8\text{ m/s}^2

Show answer & explanation

Answer: $4\text{ m/s}^2$

Answer

4\text{ m/s}^2

Using the equation of motion

v^2 = u^2 + 2as

, substitute

u = 10\text{ m/s}

,

v = 30\text{ m/s}

, and

s = 100\text{ m}

. This gives

30^2 = 10^2 + 2(a)(100)

, which simplifies to

900 - 100 = 200a

, or

800 = 200a

. Solving for acceleration yields

a = 4\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the given kinematic values from the problem statement.

Initial velocity

u = 10\text{ m/s}

, final velocity

v = 30\text{ m/s}

, and displacement

s = 100\text{ m}

.

Listing known values helps in selecting the appropriate equation of motion.

2

Select the linear motion formula relating

u

,

v

,

s

, and acceleration

a

.

v^2 = u^2 + 2as

This formula connects initial velocity, final velocity, distance, and acceleration without requiring time

t

.

3

Substitute the values into the equation and solve for

a

.

30^2 = 10^2 + 2(a)(100) \implies 900 = 100 + 200a \implies 800 = 200a \implies a = 4\text{ m/s}^2

Algebraic rearrangement yields the magnitude of acceleration.

Key Concept

Equations of Uniformly Accelerated Motion

Estimated Time:45s

Question 32Question →

A water pump driven by an engine with an efficiency of $80\%$ raises water from an underground tank of depth $20\text{ m}$ and discharges it through a nozzle of cross-sectional area $10\text{ cm}^2$ at a steady speed of $10\text{ m/s}$ . Taking the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the minimum input power rating (in $\text{W}$ ) required for the engine?

Show answer & explanation

Answer: 3125

Answer

The minimum input power rating required for the engine is

3125\text{ W}

.

The engine must supply power to lift

10\text{ kg}

of water per second through a vertical height of

20\text{ m}

while accelerating it to

10\text{ m/s}

. The useful power output is

2000\text{ W}

(potential)

+ 500\text{ W}

(kinetic)

= 2500\text{ W}

. Accounting for an engine efficiency of

80\%

, the total input power is

\frac{2500}{0.80} = 3125\text{ W}

.

Step-by-Step Solution

1

Determine the mass of water discharged per unit time (mass flow rate).

\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}

Water is moving through a cross-sectional area at a constant velocity.

2

Calculate the useful output power required to lift the water and impart kinetic energy.

P_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}

The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.

3

Calculate the total input power using engine efficiency.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}

Efficiency is the ratio of useful power output to total power input.

Key Concept

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations

Question 33Question →

An object of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a maximum acceleration of $20\text{ m/s}^2$ . What is the speed of the object when its displacement from the equilibrium position is $0.03\text{ m}$ ?

0.16\text{ m/s}

0.60\text{ m/s}

0.80\text{ m/s}

1.00\text{ m/s}

Show answer & explanation

Answer: $0.80\text{ m/s}$

Answer

The speed of the object at a displacement of

0.03\text{ m}

is

0.80\text{ m/s}

.

The maximum acceleration in simple harmonic motion is given by

a_{\text{max}} = \omega^2 A

. Substituting

a_{\text{max}} = 20\text{ m/s}^2

and

A = 0.05\text{ m}

gives

\omega^2 = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

. The speed at any displacement

x

is given by

v = \omega \sqrt{A^2 - x^2}

. For

x = 0.03\text{ m}

,

v = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}

.

Step-by-Step Solution

1

Determine the angular frequency (

\omega

) of the simple harmonic motion from the maximum acceleration formula.

\omega = 20\text{ rad/s}

Maximum acceleration is given by

a_{\text{max}} = \omega^2 A

. Rearranging gives

\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

.

2

Calculate the speed (

v

) at the given displacement (

x = 0.03\text{ m}

) using the SHM velocity formula.

v = 0.80\text{ m/s}

The speed at displacement

x

is

v = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}

.

Key Concept

Simple Harmonic Motion Velocity and Acceleration Relationships

Estimated Time:1m 30s

Question 34Question →

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 35Question →

A stone is projected from ground level with an initial velocity of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. What is the total time of flight of the stone, in seconds? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 4

Answer

The total time of flight of the stone is

4\text{ s}

.

The total time of flight

T

for a projectile launched over level ground is calculated using

T = \frac{2 u \sin \theta}{g}

. Substituting

u = 40\text{ m/s}

,

\theta = 30^\circ

, and

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

yields

T = \frac{2 \times 40 \times 0.5}{10} = 4\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of the launch velocity

u_y = 20\text{ m/s}

The vertical motion determines the time the projectile remains in the air.

2

Calculate the total time of flight

T = 4\text{ s}

Applying

T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20}{10} = 4\text{ s}

gives the total duration before landing back at ground level.

Key Concept

Time of Flight in Projectile Motion

Estimated Time:45s

Question 36Question →

A satellite of mass $m$ moves in a circular orbit around a uniform spherical planet of radius $R$ . If the height of the satellite above the planet's surface is $h = 2R$ and the acceleration due to gravity at the planet's surface is $g$ , which of the following expressions represents the orbital speed of the satellite?

\sqrt{\frac{gR}{3}}

\sqrt{\frac{gR}{2}}

\sqrt{\frac{2gR}{3}}

\sqrt{3gR}

Show answer & explanation

Answer: $\sqrt{\frac{gR}{3}}$

Answer

The orbital speed of the satellite is

\sqrt{\frac{gR}{3}}

.

The total orbital radius from the center of the planet is

r = R + h = R + 2R = 3R

. Since surface gravity is

g = \frac{GM}{R^2}

, we have

GM = gR^2

. Substituting these into the orbital velocity expression

v = \sqrt{\frac{GM}{r}}

yields

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

, which makes the expression

\sqrt{\frac{gR}{3}}

correct.

Step-by-Step Solution

1

Determine the total orbital radius from the center of the planet.

r = R + h = R + 2R = 3R

Gravitational attraction and circular orbital radii are always measured from the center of mass of the primary body, not its surface.

2

Relate the gravitational constant

G

and planet mass

M

to surface gravity

g

.

g = \frac{GM}{R^2} \implies GM = gR^2

At the surface of a spherical planet of radius

R

, the gravitational field strength is

g

.

3

Substitute

r = 3R

and

GM = gR^2

into the circular orbital velocity equation

v = \sqrt{\frac{GM}{r}}

.

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

Equating centripetal force to gravitational force

\frac{m v^2}{r} = \frac{G M m}{r^2}

yields

v = \sqrt{\frac{GM}{r}}

.

Key Concept

Orbital velocity of a satellite in terms of surface gravitational acceleration and orbital radius

Question 37Question →

A motorist traveling along a straight highway at a constant speed of $30\text{ m/s}$ observes a road hazard ahead. The driver experiences a reaction delay of $0.5\text{ s}$ before applying the brakes, after which the vehicle decelerates uniformly at a rate of $5\text{ m/s}^2$ . What is the total distance traveled by the vehicle from the instant the hazard is observed until the vehicle comes to a complete stop?

90\text{ m}

97.5\text{ m}

105\text{ m}

195\text{ m}

Show answer & explanation

Answer: $105\text{ m}$

Answer

The total distance traveled by the vehicle before coming to a stop is

105\text{ m}

.

The motion consists of two distinct stages: a constant-speed phase during the

0.5\text{ s}

reaction time (

15\text{ m}

) and a uniformly decelerating phase until rest (

90\text{ m}

). Summing both components yields

105\text{ m}

.

Step-by-Step Solution

1

Calculate the reaction distance (

s_1

) covered during the driver's reaction delay.

s_1 = u \times t_r = 30\text{ m/s} \times 0.5\text{ s} = 15\text{ m}

Before the brakes are applied, the vehicle continues moving at its initial constant speed of

30\text{ m/s}

.

2

Calculate the braking distance (

s_2

) using the third equation of motion.

v^2 = u^2 + 2as_2 \implies 0^2 = 30^2 + 2(-5)s_2 \implies 10s_2 = 900 \implies s_2 = 90\text{ m}

The vehicle decelerates from

30\text{ m/s}

to a final velocity of

0\text{ m/s}

at a uniform deceleration rate of

a = -5\text{ m/s}^2

.

3

Sum the reaction distance and the braking distance to find the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 15\text{ m} + 90\text{ m} = 105\text{ m}

The total stopping distance is the sum of the distance covered before braking begins and the distance covered while braking.

Key Concept

Two-phase linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Question 38Question →

A metallic sphere weighs $5.0\text{ N}$ in air. When completely immersed in water, its apparent weight is $3.0\text{ N}$ . When completely immersed in an unknown liquid $X$ , its apparent weight is $3.4\text{ N}$ . What is the density of liquid $X$ in $\text{kg/m}^3$ ? (Take the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 800

Answer

The density of liquid

X

is

800\text{ kg/m}^3

.

The upthrust in water (

2.0\text{ N}

) gives the volume of the sphere as

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

. Using the upthrust in liquid X (

1.6\text{ N}

), the density of liquid X is calculated as

\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3

.

Step-by-Step Solution

1

Calculate upthrust in water

U_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}

Upthrust equals the loss in weight of the submerged body in water.

2

Determine the volume of the metallic sphere

V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3

According to Archimedes' principle, upthrust in water equals the weight of displaced water.

3

Calculate upthrust in liquid X

U_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}

Loss of weight in liquid X gives the upthrust exerted by liquid X.

4

Calculate the density of liquid X

\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3

Rearranging

U_X = \rho_X V g

allows solving for the unknown fluid density.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:2m 0s

Question 39Question →

The maximum acceleration of a body oscillating in simple harmonic motion is $8\text{ m/s}^2$ . If the period of oscillation is $\pi\text{ s}$ , calculate the amplitude of the oscillation in meters.

Show answer & explanation

Answer: 2

Answer

The amplitude of the oscillation is

2.0\text{ m}

.

The correct answer of

2.0\text{ m}

is obtained by first deriving the angular frequency

\omega = \frac{2\pi}{T} = 2\text{ rad/s}

, and then using the relation

a_{\text{max}} = \omega^2 A

to solve for amplitude:

A = \frac{8}{2^2} = 2.0\text{ m}

.

Step-by-Step Solution

1

Calculate angular frequency (

\omega

) from the given period (

T

).

\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}

Angular frequency specifies the rate of phase change in oscillations.

2

Apply the maximum acceleration formula for simple harmonic motion to determine amplitude (

A

).

a_{\text{max}} = \omega^2 A \implies 8 = 2^2 \times A \implies A = 2.0\text{ m}

In simple harmonic motion, maximum acceleration occurs at the extreme position and equals

\omega^2 A

.

Key Concept

Simple Harmonic Motion Acceleration and Period Relationship

Question 40Question →

A stone is projected vertically upwards from the top edge of a cliff $80\text{ m}$ high with an initial speed of $30\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the total time, in seconds, taken by the stone to reach the ground at the base of the cliff.

Show answer & explanation

Answer: 8

Answer

The total time taken by the stone to reach the ground at the base of the cliff is

8\text{ s}

.

Using the equation of motion

s = ut - \frac{1}{2}gt^2

with

s = -80\text{ m}

,

u = 30\text{ m/s}

, and

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

yields the quadratic equation

t^2 - 6t - 16 = 0

. Solving gives

t = 8\text{ s}

(ignoring the unphysical negative root

t = -2\text{ s}

). Alternatively, breaking the motion into two parts: time to reach maximum height (

30\text{ m/s} / 10\text{ m/s}^2 = 3\text{ s}

, covering

45\text{ m}

) plus time to fall from maximum height of

125\text{ m}

to the ground (

t = \sqrt{2(125)/10} = 5\text{ s}

), giving a total time of

3 + 5 = 8\text{ s}

.

Step-by-Step Solution

1

Set up the kinematic equation with appropriate vector signs

Displacement

s = -80\text{ m}

, initial velocity

u = +30\text{ m/s}

, acceleration

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

Since the ground is below the release point, displacement is negative when taking the upward direction as positive.

2

Substitute values into

s = ut + \frac{1}{2}at^2

-80 = 30t - 5t^2

Relates displacement, initial speed, time, and constant gravitational acceleration.

3

Form and solve the quadratic equation

5t^2 - 30t - 80 = 0 \implies t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0

Simplifies the algebraic expression to find the time roots.

4

Select the physical root

t = 8\text{ s}

Time elapsed must be a positive quantity.

Key Concept

Kinematics of Vertical Motion under Gravity with Displacement from Elevation

Estimated Time:2m 0s

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