Mechanics Practice Questions — JAMB UTME — Page 7

Question 121Question →

A stone is thrown into the air with an initial velocity of $20\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. What is the magnitude of its velocity at the highest point of its trajectory?

0\text{ m/s}

10\text{ m/s}

17.3\text{ m/s}

20\text{ m/s}

Show answer & explanation

Answer: $10\text{ m/s}$

Answer

The magnitude of the velocity at the highest point is

10\text{ m/s}

.

At the peak of a projectile's trajectory, the vertical velocity component vanishes (

v_y = 0\text{ m/s}

), but horizontal motion continues at a constant speed (

v_x = u \cos \theta

). Substituting

u = 20\text{ m/s}

and

\theta = 60^\circ

yields

v_x = 20 \times 0.5 = 10\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the apex

At maximum height, the vertical velocity component is

v_y = 0\text{ m/s}

, while the horizontal component remains constant at

v_x = u_x = u \cos \theta

.

Horizontal acceleration is zero when air resistance is neglected.

2

Calculate the horizontal component of velocity

v_x = 20 \times \cos(60^\circ) = 20 \times 0.5 = 10\text{ m/s}

.

The trigonometric cosine function gives the horizontal projection of the initial velocity vector.

3

Determine total velocity magnitude at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = 10\text{ m/s}

.

Since the vertical component is zero, the total velocity at the peak equals the horizontal component.

Key Concept

Velocity components at maximum height in projectile motion

Question 122Question →

A body is projected from ground level with an initial speed of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the time taken, in seconds, for the body to reach its maximum height. (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 2

Answer

The time taken to reach maximum height is

2\text{ s}

.

The initial vertical velocity component is

u_y = u \sin(30^\circ) = 40 \times 0.5 = 20\text{ m/s}

. Under gravitational deceleration (

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

), the vertical speed drops to zero at maximum height after

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of initial velocity (

u_y

)

u_y = 40 \sin(30^\circ) = 20\text{ m/s}

Only the vertical component of initial velocity determines the time to reach maximum height.

2

Calculate the time to maximum height (

t

)

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

At maximum height, vertical velocity

v_y = 0

, giving

t = \frac{u_y}{g}

.

Key Concept

Time to reach maximum height in projectile motion

Question 123Question →

An autonomous drone navigating an obstacle course experiences three mutually perpendicular velocity vectors simultaneously: a horizontal forward velocity of $12\text{ m s}^{-1}$ due East, a horizontal crosswind drift velocity of $9\text{ m s}^{-1}$ due North, and a vertical downdraft velocity of $8\text{ m s}^{-1}$ directed straight downward. What is the magnitude of the resultant velocity vector of the drone in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 17

Answer

17 m s⁻¹

Because the three velocity components are mutually perpendicular, the magnitude of the overall resultant velocity is calculated using the 3D Pythagorean theorem:

\sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the perpendicular vector components

v_x = 12\text{ m s}^{-1}

,

v_y = 9\text{ m s}^{-1}

,

v_z = 8\text{ m s}^{-1}

The three given velocity vectors act along mutually orthogonal spatial axes (East, North, and Downward).

2

Apply the 3D vector resultant magnitude formula

v_r = \sqrt{v_x^2 + v_y^2 + v_z^2}

Since the vector components are perpendicular to one another, the magnitude of their resultant is given by the extension of the Pythagorean theorem to three dimensions.

3

Substitute values and evaluate

v_r = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

Squaring each component, adding them, and taking the principal square root yields the total magnitude of the velocity vector.

Key Concept

Magnitude of three mutually perpendicular vector components using the 3D Pythagorean theorem

Question 124Question →

A cannonball is fired from level ground with an initial speed of $20\text{ m/s}$ at an angle of $30^\circ$ above the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the total time of flight of the cannonball before it returns to ground level?

2.0\text{ s}

1.0\text{ s}

4.0\text{ s}

0.5\text{ s}

Show answer & explanation

Answer: $2.0\text{ s}$

Answer

2.0\text{ s}

The correct answer of

2.0\text{ s}

is determined by resolving the initial speed into its vertical component

u_y = 20 \sin 30^\circ = 10\text{ m/s}

and using the total time of flight formula

T = \frac{2 u \sin \theta}{g} = \frac{2(10)}{10} = 2.0\text{ s}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity (

u_y

).

u_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\text{ m/s}

Only the vertical component of velocity determines the time the projectile remains in the air.

2

Calculate the total time of flight (

T

) using the kinematic formula for full trajectory.

T = \frac{2 u_y}{g} = \frac{2 \times 10}{10} = 2.0\text{ s}

The total time of flight includes both the time to ascend to peak height and descend back to the launch height under gravity

g

.

Key Concept

Total Time of Flight in Projectile Motion

Question 125Question →

A particle is projected from ground level at an angle to the horizontal. At its highest point of trajectory, which of the following statements correctly describes its velocity and acceleration?

The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Both the total velocity and the acceleration of the particle are zero.

The total velocity of the particle is zero, but its acceleration is directed downward.

The vertical component of velocity is at its maximum value, and the acceleration is zero.

Show answer & explanation

Answer: The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Answer

At the highest point of projectile motion, the vertical velocity component is zero, the horizontal velocity component is non-zero, and acceleration due to gravity acts downward.

At the peak of a projectile trajectory, vertical upward velocity is fully decelerated by gravity to zero. However, horizontal velocity is maintained because there is no horizontal acceleration. Acceleration due to gravity continues to act constantly downward.

Step-by-Step Solution

1

Analyze vertical velocity at the highest point

v_y = 0

At maximum height, the upward vertical component of motion reduces to zero before the particle begins descending.

2

Analyze horizontal velocity throughout motion

v_x = u \cos\theta \neq 0

In the absence of air resistance, no horizontal force acts on the projectile, keeping horizontal velocity constant throughout flight.

3

Determine the direction of acceleration

a = g

acting vertically downward

Gravity is the only force acting on a projectile in motion, providing continuous downward acceleration.

Key Concept

Velocity components and acceleration at the apex of projectile motion

Estimated Time:45s

Question 126Question →

A research rocket is launched vertically upwards from rest with a constant acceleration of $5.0\text{ m/s}^2$ . At an altitude of $250\text{ m}$ , its engine suddenly fails and the rocket continues to move vertically upward under gravity alone. Calculate the total time, in seconds, taken by the rocket from launch until it reaches its maximum height. (Take acceleration due to gravity $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 15

Answer

The total time taken from launch to reach maximum height is

15\text{ s}

.

The motion occurs in two phases. In phase 1, accelerating uniformly from rest at

5.0\text{ m/s}^2

over

250\text{ m}

yields a velocity of

50\text{ m/s}

in

10\text{ s}

. In phase 2, moving upward under gravity alone (

10\text{ m/s}^2

) reduces the velocity from

50\text{ m/s}

to rest (

0\text{ m/s}

) in

5\text{ s}

. Adding the durations of both phases gives

10\text{ s} + 5\text{ s} = 15\text{ s}

.

Step-by-Step Solution

1

Calculate the rocket's velocity and elapsed time at the moment of engine failure.

Velocity

v_1 = 50\text{ m/s}

and time

t_1 = 10\text{ s}

.

The rocket accelerates uniformly from rest at

5.0\text{ m/s}^2

over a distance of

250\text{ m}

.

2

Calculate the duration of the unpowered upward motion until vertical velocity becomes zero.

Unpowered flight time

t_2 = 5\text{ s}

.

After engine failure, the rocket acts as a free projectile moving upward against gravity (

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

) with an initial velocity of

50\text{ m/s}

.

3

Sum the time intervals of both stages.

Total time

t_{\text{total}} = 10\text{ s} + 5\text{ s} = 15\text{ s}

.

The total motion consists of two distinct stages: powered acceleration followed by gravitational deceleration.

Key Concept

Multi-stage vertical motion under constant acceleration followed by free-fall under gravity

Question 127Question →

The energy density $u$ (defined as energy per unit volume) stored in an electrostatic field is related to the permittivity of free space $\epsilon_0$ and the electric field strength $E$ by the dimensional formula $u = k \epsilon_0^x E^y$ , where $k$ is a dimensionless constant. What is the value of the numerical exponent $y$ ?

Show answer & explanation

Answer: 2

Answer

The value of the exponent

y

is 2.

By writing the dimensions of energy density

[M L^{-1} T^{-2}]

, permittivity

[M^{-1} L^{-3} T^4 I^2]

, and electric field strength

[M L T^{-3} I^{-1}]

, equating powers of electric current

I

yields

2x - y = 0

(or

y = 2x

). Substituting this into the equation for powers of mass

M

,

-x + y = 1

, yields

-x + 2x = 1

, so

x = 1

and

y = 2

.

Step-by-Step Solution

1

Derive the dimensional formulas for energy density

u

, permittivity

\epsilon_0

, and electric field

E

.

[u] = M L^{-1} T^{-2}, [\epsilon_0] = M^{-1} L^{-3} T^4 I^2, [E] = M L T^{-3} I^{-1}.

Expressing quantities in terms of base dimensions (M, L, T, I) is required for dimensional homogeneity.

2

Form the dimensional equation

u = k \epsilon_0^x E^y

and combine powers.

M L^{-1} T^{-2} = M^{-x+y} L^{-3x+y} T^{4x-3y} I^{2x-y}.

Applies the principle of dimensional consistency across the formula.

3

Equate corresponding powers of base dimensions to set up equations for

x

and

y

.

For I: 2x - y = 0; for M: -x + y = 1.

Base unit exponents on both sides of a physically valid equation must match.

4

Solve the algebraic equations for the unknown exponent

y

.

x = 1, y = 2.

Substituting y = 2x into -x + y = 1 directly gives x = 1 and y = 2.

Key Concept

Dimensional Analysis and Dimensional Homogeneity

Estimated Time:1m 30s

Question 128Question →

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

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

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

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

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

Show answer & explanation

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

Answer

The recoil velocity of the rifle is

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

(or

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

in the direction opposite to the bullet).

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

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

. Substituting

m_{rifle} = 4.0\text{ kg}

,

m_{bullet} = 0.01\text{ kg}

, and

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

gives

4.0 v_{rifle} + 4.0 = 0

, which yields

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

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

Step-by-Step Solution

1

State the principle of conservation of linear momentum

Total Initial Momentum = Total Final Momentum = 0

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

2

Set up the linear momentum conservation equation

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

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

3

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

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

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

Key Concept

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

Estimated Time:45s

Question 129Question →

A shell is launched from level ground into the air. It reaches a maximum height of $45\text{ m}$ above the ground and has a total horizontal range of $240\text{ m}$ . Taking $g = 10\text{ m/s}^2$ , calculate the magnitude of the initial launch velocity of the shell in $\text{m/s}$ .

Show answer & explanation

Answer: 50

Answer

The initial launch velocity of the shell is

50\text{ m/s}

.

Combining the expressions for maximum height

H = \frac{u^2\sin^2\theta}{2g}

and range

R = \frac{2u^2\sin\theta\cos\theta}{g}

gives

\tan\theta = \frac{4H}{R}

. With

H = 45\text{ m}

and

R = 240\text{ m}

, we get

\tan\theta = 0.75 = \frac{3}{4}

, which yields

\sin\theta = 0.6

. Substituting these into the height equation yields

45 = \frac{u^2(0.6)^2}{20}

, solving to

u = 50\text{ m/s}

.

Step-by-Step Solution

1

Express the launch angle in terms of maximum height and horizontal range

\tan\theta = \frac{4H}{R} = \frac{4 \times 45}{240} = 0.75

Dividing the maximum height formula

H = \frac{u^2\sin^2\theta}{2g}

by the horizontal range formula

R = \frac{2u^2\sin\theta\cos\theta}{g}

yields

\frac{H}{R} = \frac{1}{4}\tan\theta

.

2

Find the sine of the launch angle from the tangent value

\sin\theta = 0.6

For a right-angled triangle with

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

, the hypotenuse is

\sqrt{3^2 + 4^2} = 5

, giving

\sin\theta = \frac{3}{5} = 0.6

.

3

Calculate the magnitude of the initial velocity

u

u = 50\text{ m/s}

Substituting values into

H = \frac{u^2\sin^2\theta}{2g}

gives

45 = \frac{u^2 (0.6)^2}{2(10)} \implies 900 = 0.36 u^2 \implies u = 50\text{ m/s}

.

Key Concept

Interdependence of Maximum Height, Range, and Launch Velocity in Projectile Motion

Estimated Time:1m 30s

Question 130Question →

A body starting from rest accelerates uniformly along a straight path at a rate of $2.5\text{ m/s}^2$ . What is the distance covered by the body in $4.0\text{ s}$ ?

10\text{ m}

20\text{ m}

40\text{ m}

5.0\text{ m}

Show answer & explanation

Answer: $20\text{ m}$

Answer

The distance covered by the body in

4.0\text{ s}

is

20\text{ m}

.

Using the kinematic relation

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

with

u = 0\text{ m/s}

,

a = 2.5\text{ m/s}^2

, and

t = 4.0\text{ s}

gives

s = 0 + \frac{1}{2} \times 2.5 \times (4.0)^2 = 20\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic values.

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 2.5\text{ m/s}^2

, and time

t = 4.0\text{ s}

.

The problem states the body starts from rest and undergoes uniform linear acceleration.

2

Select the appropriate formula for distance under uniform acceleration.

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

This equation connects initial velocity, acceleration, elapsed time, and total displacement.

3

Substitute the given values into the equation and compute displacement.

s = (0)(4.0) + \frac{1}{2}(2.5)(4.0)^2 = 0.5 \times 2.5 \times 16 = 20\text{ m}

.

Evaluating the expressions yields the distance traveled in meters.

Key Concept

Kinematic Equation for Linear Distance Under Uniform Acceleration

Estimated Time:45s

Question 131Question →

An object is projected from level ground with an initial speed of $30\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the object in meters.

Show answer & explanation

Answer: 11.25

Answer

The maximum height reached by the object is

11.25\text{ m}

.

The vertical component of initial velocity is

u_y = 30 \sin(30^\circ) = 15\text{ m/s}

. At maximum height, vertical velocity becomes zero, so

H = \frac{u_y^2}{2g} = \frac{15^2}{20} = 11.25\text{ m}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity.

u_y = u \sin\theta = 30 \times \sin(30^\circ) = 15\text{ m/s}

Only the vertical component of initial velocity determines the maximum height.

2

Apply the vertical motion equation at maximum height where vertical velocity is zero.

H = \frac{u_y^2}{2g} = \frac{15^2}{2 \times 10} = 11.25\text{ m}

Using

v_y^2 = u_y^2 - 2gH

with

v_y = 0

gives

H = \frac{u_y^2}{2g}

.

Key Concept

Maximum height of a projectile

Estimated Time:45s

Question 132Question →

The torque $\tau$ required to rotate a thin flat disk of radius $r$ at a constant angular velocity $\omega$ in a fluid of dynamic viscosity $\eta$ is expressed by the dimensional formula $\tau = k \eta^x \omega^y r^z$ , where $k$ is a dimensionless constant. What is the value of the sum of the exponents $x + y + z$ ?

Show answer & explanation

Answer: 5

Answer

The sum of the exponents

x + y + z

is 5.

By substituting the base dimensions into

\tau = k \eta^x \omega^y r^z

, we get

M L^2 T^{-2} = (M L^{-1} T^{-1})^x (T^{-1})^y L^z = M^x L^{-x+z} T^{-x-y}

. Equating exponents of

M

gives

x = 1

. Equating exponents of

T

gives

-1 - y = -2 \implies y = 1

. Equating exponents of

L

gives

-1 + z = 2 \implies z = 3

. Thus,

x + y + z = 1 + 1 + 3 = 5

.

Step-by-Step Solution

1

Determine the dimensions of torque, dynamic viscosity, angular velocity, and radius in base mechanical dimensions (M, L, T).

[\tau] = M L^2 T^{-2}

,

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

,

[\omega] = T^{-1}

, and

[r] = L

.

Dimensional analysis requires converting all parameters into base dimensions.

2

Apply the principle of dimensional homogeneity to set up exponential equations for each base dimension.

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

.

Both sides of a physically valid equation must share identical net dimensions.

3

Solve for each exponent individually by comparing indices.

x = 1

,

y = 1

,

z = 3

.

Matching powers of M yields

x=1

, matching powers of T yields

y=1

, and matching powers of L yields

z=3

.

4

Sum the three calculated exponent values.

1 + 1 + 3 = 5

.

The question asks specifically for the value of

x + y + z

.

Key Concept

Dimensional analysis and dimensional homogeneity

Estimated Time:1m 30s

Question 133Question →

A projectile is launched from level ground with an initial speed of $25\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.80$ . Calculate the maximum height reached by the projectile in meters. [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The maximum height reached by the projectile is

20\text{ m}

.

The vertical component of the initial launch velocity is

u_y = u \sin\theta = 25 \times 0.80 = 20\text{ m/s}

. Using the equation for maximum height

H = \frac{u_y^2}{2g}

, we substitute

u_y = 20\text{ m/s}

and

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

to obtain

H = \frac{400}{20} = 20\text{ m}

.

Step-by-Step Solution

1

Calculate the initial vertical velocity component (

u_y

)

u_y = 25\text{ m/s} \times 0.80 = 20\text{ m/s}

Only the vertical component of velocity determines the maximum height reached.

2

Calculate the maximum height (

H

) using kinematic equations

H = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = 20\text{ m}

At maximum height, the vertical component of velocity becomes zero.

Key Concept

Maximum height of a projectile depends entirely on its initial vertical component of velocity and acceleration due to gravity.

Question 134Question →

What is the length of a simple pendulum that has a period of oscillation of $2.0\text{ s}$ at a location where the acceleration due to gravity is $g = \pi^2\text{ m/s}^2$ ?

0.5\text{ m}

1.0\text{ m}

2.0\text{ m}

4.0\text{ m}

Show answer & explanation

Answer: $1.0\text{ m}$

Answer

The length of the simple pendulum is

1.0\text{ m}

.

Using the period equation

T = 2\pi \sqrt{l/g}

, squaring both sides yields

T^2 = 4\pi^2 l / g

. Rearranging gives

l = \frac{T^2 g}{4\pi^2}

. Substituting

T = 2.0\text{ s}

and

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

yields

l = \frac{4 \pi^2}{4 \pi^2} = 1.0\text{ m}

.

Step-by-Step Solution

1

State the formula for the period of a simple pendulum.

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

This formula relates the oscillation period

T

, pendulum length

l

, and gravitational acceleration

g

.

2

Square both sides of the equation to solve for

l

.

T^2 = 4\pi^2 \left(\frac{l}{g}\right) \implies l = \frac{T^2 \cdot g}{4\pi^2}

Isolating

l

allows direct evaluation using the given numerical values.

3

Substitute

T = 2.0\text{ s}

and

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

into the expression.

l = \frac{(2.0)^2 \cdot \pi^2}{4\pi^2} = \frac{4\pi^2}{4\pi^2} = 1.0\text{ m}

Simplifying by canceling

\pi^2

and

4

gives the exact length.

Key Concept

Simple Pendulum Period and Length Relationship

Question 135Question →

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

True

False

Show answer & explanation

Answer: True

Answer

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

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

Step-by-Step Solution

1

Determine the work done by a perpendicular force.

The work done is

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

.

Force perpendicular to displacement performs no work on the object.

2

Relate work done to speed and magnitude of linear momentum.

Zero work means zero change in kinetic energy, so speed

v

is constant, making magnitude

p = mv

constant.

Linear momentum magnitude depends solely on mass and speed.

3

Evaluate the effect on momentum direction.

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

\vec{p}

.

By Newton's Second Law (

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

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

Key Concept

Vector nature of linear momentum and perpendicular force action

Question 136Question →

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

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

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

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

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

Show answer & explanation

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

Answer

The new velocity of the cart is

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

in the forward direction.

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

50\text{ kg}

moving at

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

) equals

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

. When the

10\text{ kg}

package is ejected backward at

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

, its momentum is

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

. Setting

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

yields

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

.

Step-by-Step Solution

1

Calculate the total initial momentum of the system before ejection.

Total mass

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

. Initial momentum

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

.

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

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

.

2

Set up the expression for final momentum considering direction.

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

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

. Final momentum

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

.

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

3

Apply the law of conservation of linear momentum (

P_i = P_f

) to solve for the cart's final velocity.

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

.

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

Key Concept

Conservation of Linear Momentum

Estimated Time:2m 0s

Question 137Question →

A stone is projected horizontally with a speed of $15\text{ m/s}$ from the top of a vertical cliff of height $20\text{ m}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the magnitude of the velocity of the stone just before it strikes the ground?

15\text{ m/s}

20\text{ m/s}

25\text{ m/s}

35\text{ m/s}

Show answer & explanation

Answer: $25\text{ m/s}$

Answer

The magnitude of the velocity of the stone just before hitting the ground is

25\text{ m/s}

.

For a horizontally launched projectile, the horizontal velocity component remains constant at

v_x = 15\text{ m/s}

. The vertical component just before impact is found using

v_y^2 = u_y^2 + 2gh = 0 + 2(10)(20) = 400

, giving

v_y = 20\text{ m/s}

. Combining these perpendicular components yields a total speed of

v = \sqrt{15^2 + 20^2} = 25\text{ m/s}

.

Step-by-Step Solution

1

Identify horizontal velocity component

v_x = 15\text{ m/s}

Air resistance is neglected, so horizontal velocity remains constant throughout flight.

2

Calculate vertical velocity component just before impact using third equation of motion

v_y^2 = u_y^2 + 2gh = 0^2 + 2(10)(20) = 400 \implies v_y = 20\text{ m/s}

Initial vertical velocity

u_y = 0\text{ m/s}

for horizontal projection; stone falls through a vertical displacement of

20\text{ m}

under gravity.

3

Compute total resultant velocity magnitude using Pythagorean theorem

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m/s}

Horizontal and vertical velocity components are mutually perpendicular.

Key Concept

Horizontal Projection and Resultant Velocity Vector Synthesis

Estimated Time:2m 0s

Question 138Question →

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

Show answer & explanation

Answer: 60

Answer

The magnitude of the retarding force acting on the block is

60\text{ N}

.

During the impact, the bullet loses momentum equal to

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

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

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

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

0.25\text{ s}

is

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

.

Step-by-Step Solution

1

Calculate the momentum lost by the bullet during penetration.

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

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

2

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

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

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

3

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

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

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

Key Concept

Conservation of Linear Momentum and Newton's Second Law

Question 139Question →

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

True

False

Show answer & explanation

Answer: True

Answer

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

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

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

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

Step-by-Step Solution

1

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

The net impulse

\vec{J}

acting on a body equals its change in linear momentum:

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

.

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

2

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

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

\vec{v}_f

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

\vec{v}_i

.

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

3

Calculate the change in momentum

\Delta \vec{p}

.

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

. Thus, net impulse

\vec{J} = \vec{0}

.

Subtracting identical vectors yields zero vector magnitude.

Key Concept

Impulse-Momentum Theorem and Vector Nature of Linear Momentum

Estimated Time:1m 30s

Question 140Question →

An electric crane lifts a load of $250\text{ kg}$ vertically upwards through a height of $12\text{ m}$ in $10\text{ s}$ at a constant speed. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the useful output power of the crane in watts?

Show answer & explanation

Answer: 3000

Answer

The useful output power of the crane is

3000\text{ W}

.

The work done in lifting the load vertically is equal to the gravitational potential energy gained,

W = mgh = 250 \times 10 \times 12 = 30,000\text{ J}

. Power is the rate of doing work, so

P = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}

.

Step-by-Step Solution

1

Identify the given values

Mass

m = 250\text{ kg}

, height

h = 12\text{ m}

, time

t = 10\text{ s}

, acceleration due to gravity

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

.

Extract values needed for work and power calculations.

2

Calculate the work done in lifting the load

W = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}

.

The work done against gravity equals the gain in gravitational potential energy.

3

Calculate the power output

P = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}

.

Power is defined as the rate at which work is done (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

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