Mechanics Practice Questions — JAMB UTME — Page 11

Question 201Question →

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

Show answer & explanation

Answer: 0.05

Answer

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

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

and kinetic energy is

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

. Setting

E_k = 3E_p

gives

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

, which simplifies to

A^2 = 4x^2

, or

x = \frac{A}{2}

. Given an amplitude

A = 0.10\text{ m}

, the displacement is

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

.

Step-by-Step Solution

1

Set up the relation between kinetic energy and potential energy using the given condition.

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

In simple harmonic motion, energy is partitioned between kinetic and potential forms based on displacement

x

.

2

Solve the algebraic equation for displacement

x

in terms of amplitude

A

.

A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x = \frac{A}{2}

Canceling the common factor

\frac{1}{2}k

isolates the geometric parameters

A

and

x

.

3

Substitute the known amplitude value into the expression for

x

.

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

Plugging in

A = 0.10\text{ m}

yields the required displacement.

Key Concept

Energy Conservation in Simple Harmonic Motion

Question 202Question →

Two point masses, each of mass $125\text{ kg}$ , are fixed at positions $(-3.0\text{ m}, 0)$ and $(3.0\text{ m}, 0)$ in the $xy$ -plane. What is the magnitude of the net gravitational field strength at a point $P$ located at $(0, 4.0\text{ m})$ , in terms of the universal gravitational constant $G$ ?

8.0G\text{ N/kg}

10.0G\text{ N/kg}

6.0G\text{ N/kg}

4.0G\text{ N/kg}

Show answer & explanation

Answer: $8.0G\text{ N/kg}$

Answer

The magnitude of the net gravitational field strength at point

P

is

8.0G\text{ N/kg}

.

The distance from each mass to the target point is 5.0 m, yielding an individual field magnitude of 5.0G N/kg. Resolving vectors shows that horizontal components cancel out while vertical components add to give 2 × (5.0G × 4/5) = 8.0G N/kg.

Step-by-Step Solution

1

Calculate the distance

r

from each mass to point

P(0, 4.0\text{ m})

.

r = \sqrt{(-3.0 - 0)^2 + (0 - 4.0)^2} = \sqrt{9 + 16} = 5.0\text{ m}

.

The distance formula in two dimensions gives the hypotenuse of the right triangle formed by the coordinates.

2

Determine the magnitude of the gravitational field

E

created by each individual mass at point

P

.

E = \frac{G M}{r^2} = \frac{G \times 125}{5.0^2} = \frac{125G}{25} = 5.0G\text{ N/kg}

.

Newton's law of universal gravitation defines field strength as

E = \frac{GM}{r^2}

.

3

Resolve the field vectors into horizontal and vertical components.

By symmetry, the horizontal components

E_x = E \sin\theta

are equal in magnitude and opposite in direction, so

E_{x,\text{net}} = 0

. The vertical components

E_y = E \cos\theta

point downwards towards the origin.

Gravitational field is a vector quantity, so opposite components cancel while aligned components add together.

4

Calculate the total vertical component of the net gravitational field.

\cos\theta = \frac{4.0}{5.0} = 0.8

. Thus,

E_{\text{net}} = 2 \times E_y = 2 \times (5.0G \times 0.8) = 8.0G\text{ N/kg}

.

Summing the vertical contributions from both identical masses gives the net magnitude.

Key Concept

Vector Addition of Gravitational Field Strengths

Question 203Question →

An electric engine delivers a constant power of $600\text{ W}$ to move a $50\text{ kg}$ box at a constant speed of $4\text{ m s}^{-1}$ along a horizontal surface. What is the coefficient of kinetic friction between the box and the surface? (Take $g = 10\text{ m s}^{-2}$ )

0.30

1.20

3.00

4.80

Show answer & explanation

Answer: $0.30$

Answer

The coefficient of kinetic friction between the box and the surface is 0.30.

The option stating 0.30 is correct because the power delivered by the engine

P = F v

yields an applied force of

150\text{ N}

. Since velocity is constant, the frictional force equals

150\text{ N}

. Dividing by the normal force

N = mg = 500\text{ N}

gives

\mu_k = 0.30

.

Step-by-Step Solution

1

Calculate the pulling force exerted by the engine using the relationship between power, force, and velocity.

F = \frac{P}{v} = \frac{600\text{ W}}{4\text{ m s}^{-1}} = 150\text{ N}

Power is defined as work per unit time, or force multiplied by constant speed.

2

Determine the frictional force acting on the box.

f_k = F = 150\text{ N}

Since the box moves at a constant velocity, net horizontal force is zero, meaning pulling force equals kinetic friction force.

3

Calculate the normal reaction force exerted by the surface on the box.

N = m g = 50\text{ kg} \times 10\text{ m s}^{-2} = 500\text{ N}

On a horizontal plane, normal reaction equals the weight of the object.

4

Compute the coefficient of kinetic friction.

\mu_k = \frac{f_k}{N} = \frac{150\text{ N}}{500\text{ N}} = 0.30

The coefficient of kinetic friction is the ratio of kinetic friction force to normal reaction force.

Key Concept

Power, Work against Friction, and Coefficient of Kinetic Friction

Question 204Question →

A planet has a mass equal to $8$ times the mass of the Earth and a radius equal to $2$ times the radius of the Earth. If the acceleration due to gravity on the surface of the Earth is $10\text{ m/s}^2$ , what is the acceleration due to gravity on the surface of the planet in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 20

Answer

The acceleration due to gravity on the surface of the planet is

20\text{ m/s}^2

.

The acceleration due to gravity at the surface of a spherical body is given by

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

. When the mass is multiplied by

8

and the radius is multiplied by

2

, the new acceleration becomes

g_p = \frac{G(8M)}{(2R)^2} = \frac{8}{4}\frac{GM}{R^2} = 2g

. Substituting

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

gives

20\text{ m/s}^2

.

Step-by-Step Solution

1

Write the expression for acceleration due to gravity at the surface of Earth.

g_e = \frac{GM_e}{R_e^2} = 10\text{ m/s}^2

Gravitational field strength at the surface depends directly on body mass and inversely on the square of radius.

2

Set up the ratio equation for the planet's gravitational acceleration using relative mass and radius values.

g_p = \frac{G(8M_e)}{(2R_e)^2} = \frac{8}{4} \left(\frac{GM_e}{R_e^2}\right) = 2g_e

Squaring the radius multiplier of

2

yields

4

in the denominator, while the numerator increases by a factor of

8

.

3

Calculate the final numeric value.

g_p = 2 \times 10\text{ m/s}^2 = 20\text{ m/s}^2

Multiplying Earth's value by the combined ratio of

2

gives the answer.

Key Concept

Gravitational field strength on planetary surfaces (

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

)

Question 205Question →

An inclined plane of length $10\text{ m}$ is used to raise a heavy crate of load $900\text{ N}$ to a height of $2\text{ m}$ . If an effort force of $250\text{ N}$ is applied parallel to the inclined surface to push the crate up at constant speed, what is the efficiency of the simple machine?

$72\%

$139\%

$36\%

$5.6\%

Show answer & explanation

Answer: $72\%

Answer

The efficiency of the inclined plane is

72\%

.

The mechanical advantage of the machine is

MA = \frac{900\text{ N}}{250\text{ N}} = 3.6

, and its velocity ratio is

VR = \frac{10\text{ m}}{2\text{ m}} = 5

. Dividing

MA

by

VR

and multiplying by

100\%

gives an efficiency of

72\%

.

Step-by-Step Solution

1

Calculate the Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{900\text{ N}}{250\text{ N}} = 3.6

Mechanical advantage is defined as the ratio of load force to effort force.

2

Calculate the Velocity Ratio (VR) of the inclined plane

VR = \frac{\text{Length of plane}}{\text{Height of plane}} = \frac{10\text{ m}}{2\text{ m}} = 5

Velocity ratio for an inclined plane is the ratio of distance moved by effort along the incline to distance moved by load vertically.

3

Calculate the Efficiency (\eta)

\eta = \left(\frac{MA}{VR}\right) \times 100\% = \left(\frac{3.6}{5}\right) \times 100\% = 72\%

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

Key Concept

Efficiency of an Inclined Plane

Question 206Question →

The resistive force $F$ experienced by a small sphere of radius $r$ moving at velocity $v$ through a viscous fluid is given by Stokes' law: $F = 6\pi \eta r v$ , where $\eta$ is the coefficient of viscosity. What is the dimensional formula of $\eta$ ?

\text{M L}^{-1}\text{T}^{-1}

\text{M L T}^{-1}

\text{M L}^{-1}\text{T}^{-2}

\text{M L}^{-2}\text{T}^{-1}

Show answer & explanation

Answer: $\text{M L}^{-1}\text{T}^{-1}$

Answer

The dimensional formula of the coefficient of viscosity

\eta

is

\text{M L}^{-1}\text{T}^{-1}

.

Rearranging Stokes' law gives

\eta = \frac{F}{6\pi r v}

. Since the constant

6\pi

has no dimensions, substituting

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

,

[r] = \text{L}

, and

[v] = \text{L T}^{-1}

yields

[\eta] = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1}\text{T}^{-1}

.

Step-by-Step Solution

1

Isolate the coefficient of viscosity

\eta

from Stokes' formula

\eta = \frac{F}{6\pi r v}

Numerical constants like

6\pi

are dimensionless, so

[\eta] = \frac{[F]}{[r][v]}

.

2

Substitute the fundamental dimensions for force, radius, and velocity

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

,

[r] = \text{L}

, and

[v] = \text{L T}^{-1}

Force is mass times acceleration, radius is length, and velocity is displacement per unit time.

3

Perform exponent simplification for like base dimensions

[\eta] = \frac{\text{M L T}^{-2}}{\text{L} \cdot \text{L T}^{-1}} = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{1-2} \text{T}^{-2-(-1)} = \text{M L}^{-1}\text{T}^{-1}

Subtract indices of denominator dimensions from those in the numerator.

Key Concept

Dimensional Analysis of Viscosity

Alternative Method

Alternatively, unit derivation can be used: the SI unit of viscosity is

\text{N}\cdot\text{s}\cdot\text{m}^{-2}

. Replacing Newtons with fundamental SI units (

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

) gives

(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{s} \cdot \text{m}^{-2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-1}

, which corresponds directly to

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

.

Estimated Time:1m 0s

Question 207Question →

A particle is subjected to two mutually perpendicular horizontal forces of magnitudes $15\text{ N}$ acting due East and $20\text{ N}$ acting due South. What is the magnitude of the resultant force acting on the particle?

25\text{ N}

35\text{ N}

5\text{ N}

13.2\text{ N}

Show answer & explanation

Answer: $25\text{ N}$

Answer

The magnitude of the resultant force acting on the particle is

25\text{ N}

.

Because the two forces act at right angles (

90^\circ

) relative to each other, vector addition requires applying the Pythagorean theorem. Squaring both component magnitudes (

15^2 = 225

and

20^2 = 400

), summing them to obtain

625

, and taking the square root gives the resultant magnitude of

25\text{ N}

.

Step-by-Step Solution

1

Identify the force components and their geometric orientation

The East force component

F_x = 15\text{ N}

and the South force component

F_y = 20\text{ N}

meet at an angle of

90^\circ

.

Perpendicular vectors form the adjacent and opposite sides of a right-angled vector triangle.

2

Apply the Pythagorean theorem to find the hypotenuse representing the resultant force

R = \sqrt{F_x^2 + F_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ N}

.

The resultant of two orthogonal vector quantities equals the square root of the sum of their individual squares.

Key Concept

Vector Addition of Perpendicular Forces

Question 208Question →

Match each physical quantity or scenario on the left with its correct scalar or vector classification and physical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) Displacement on a closed track matches Vector quantity depending on initial and final positions (evaluating to zero); (2) Work done matches Scalar quantity defined as the dot product of force and displacement; (3) Impulse matches Vector quantity equal to change in momentum with spatial direction; (4) Hydrostatic pressure matches Scalar quantity defined as normal force per unit area acting equally in all directions.

Each physical quantity is categorized accurately based on vector/scalar definition: Displacement is a vector that evaluates to zero over a closed loop. Work done is a scalar derived from the vector dot product. Impulse is a vector defined by momentum change direction. Hydrostatic pressure is a scalar acting equally in all directions at a fluid depth.

Step-by-Step Solution

1

Analyze item 1: Displacement of a particle moving in a complete circular track of radius

r

Displacement is a vector pointing from start to end position. Since the path returns to the start, the magnitude is

0\text{ m}

.

Vectors require direction and position change; returning to origin yields zero vector displacement.

2

Analyze item 2: Work done on a wooden box pushed horizontally

Work is calculated using

W = \vec{F} \cdot \vec{s} = F s \cos\theta

, which yields a scalar value measured in Joules.

The dot product of two vectors produces a scalar quantity.

3

Analyze item 3: Impulse imparted by a vertical wall to a bouncing rubber ball

Impulse is defined as

\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}

. It is a vector pointing away from the wall.

Impulse has the direction of the applied average force or change in momentum vector.

4

Analyze item 4: Hydrostatic pressure exerted by a fluid at depth

h

Pressure is calculated as

P = \rho g h

. It acts omnidirectionally at any point in fluid statics.

Because pressure at a point has no preferred direction, it is classified as a scalar physical quantity.

Key Concept

Classification of Physical Quantities into Scalars and Vectors based on Directional Properties

Question 209Question →

A projectile is launched from ground level with an initial speed of $40\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.8$ and $\cos\theta = 0.6$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile in $\text{m/s}$ when it reaches a height of $35\text{ m}$ above the ground?

Show answer & explanation

Answer: 30

Answer

The speed of the projectile at a height of

35\text{ m}

above the ground is

30\text{ m/s}

.

The correct answer is

30\text{ m/s}

. The vertical component of velocity at height

35\text{ m}

is found using

v_y^2 = u_y^2 - 2gh = 32^2 - 2(10)(35) = 324

, giving

v_y = 18\text{ m/s}

. Since horizontal velocity component remains constant at

v_x = 24\text{ m/s}

, the overall speed magnitude is given by

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = \sqrt{900} = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial velocity into horizontal and vertical components

u_x = 24\text{ m/s}

and

u_y = 32\text{ m/s}

Horizontal component

u_x = u\cos\theta = 40 \times 0.6 = 24\text{ m/s}

and vertical component

u_y = u\sin\theta = 40 \times 0.8 = 32\text{ m/s}

.

2

Calculate vertical velocity component at height

h = 35\text{ m}

v_y = 18\text{ m/s}

Using

v_y^2 = u_y^2 - 2gh

, we get

v_y^2 = 32^2 - 2(10)(35) = 1024 - 700 = 324

, yielding

v_y = 18\text{ m/s}

.

3

Calculate the magnitude of total velocity at height

h = 35\text{ m}

v = 30\text{ m/s}

Because air resistance is neglected, horizontal velocity remains constant (

v_x = u_x = 24\text{ m/s}

). Total speed is

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = 30\text{ m/s}

.

Key Concept

Independence of perpendicular velocity components and calculation of instantaneous speed in projectile motion.

Question 210Question →

The impulse imparted to an object by a net force is equal to the time rate of change of the object's linear momentum.

True

False

Show answer & explanation

Answer: False

Answer

False. Impulse is equal to the total change in linear momentum (

J = \Delta p

), while the time rate of change of linear momentum (

\frac{\Delta p}{\Delta t}

) represents the net force.

The statement is false. By definition, impulse is the product of net force and the time interval during which it acts (

J = F \Delta t

), which equals the total change in linear momentum (

\Delta p

). In contrast, the time rate of change of momentum (

\frac{\Delta p}{\Delta t}

) is equal to the net applied force.

Step-by-Step Solution

1

Define impulse according to the Impulse-Momentum Theorem

Impulse (

J

) is defined as

J = F \Delta t = \Delta p

, which represents the total change in momentum.

Establishing the mathematical definition of impulse.

2

Identify the physical quantity equal to the time rate of change of momentum

Newton's Second Law states that force

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

, which is the time rate of change of linear momentum.

Distinguishing between force and impulse.

3

Evaluate the statement

Because impulse equals total change in momentum (

\Delta p

) rather than rate of change (

\frac{\Delta p}{\Delta t}

), the statement is false.

Concluding the true/false verification.

Key Concept

Impulse-Momentum Theorem vs Newton's Second Law

Question 211Question →

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

True

False

Show answer & explanation

Answer: True

Answer

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

\Delta p = F \Delta t

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

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

J = F \Delta t

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

\Delta p = J

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

Step-by-Step Solution

1

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

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

, where

F

is the net force and

\Delta p

is the change in linear momentum over time

\Delta t

.

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

2

Rearrange the equation to isolate the change in momentum (

\Delta p

).

\Delta p = F \Delta t

.

Multiplying both sides by

\Delta t

gives the impulse-momentum relationship.

3

Evaluate the dependence of

\Delta p

on the body's mass.

The expression

\Delta p = F \Delta t

contains only force and time, with no mass term (

m

).

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

Key Concept

Impulse-Momentum Theorem and Newton's Second Law

Question 212Question →

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

5.0\text{ N s}

1.0\text{ N s}

3.0\text{ N s}

2.0\text{ N s}

Show answer & explanation

Answer: $5.0\text{ N s}$

Answer

5.0\text{ N s}

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

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

), the rebound velocity is negative (

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

). The change in velocity is

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

. Multiplying by mass (

0.2\text{ kg}

) gives an impulse magnitude of

5.0\text{ N s}

.

Step-by-Step Solution

1

Assign direction signs to the initial and final velocity vectors

Initial velocity

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

, final velocity after rebound

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

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

2

Apply the impulse-momentum theorem (

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

)

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

Impulse is equal to the net change in linear momentum.

3

Determine the magnitude of the impulse

|I| = 5.0\text{ N s}

Magnitude is the absolute value of the vector quantity.

Key Concept

Impulse-Momentum Theorem in One-Dimensional Rebound Scenarios

Question 213Question →

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

Show answer & explanation

Answer: 90

Answer

The magnitude of the linear momentum is

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

.

Linear momentum (

p

) is the product of an object's mass (

m

) and its velocity (

v

). Substituting

m = 6\text{ kg}

and

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

yields

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

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 6\text{ kg}

and velocity

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

These are the parameters provided in the problem statement.

2

Apply the linear momentum formula

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

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

Key Concept

Linear Momentum Definition

Estimated Time:30s

Question 214Question →

An electric cart moving along a straight horizontal track at an initial speed of $10\text{ m/s}$ accelerates uniformly at $3\text{ m/s}^2$ for a duration of $4\text{ s}$ . Immediately after this phase, it applies its brakes and decelerates uniformly at $2\text{ m/s}^2$ until it comes to a complete stop. What is the total distance, in meters, traveled by the cart during the entire motion?

Show answer & explanation

Answer: 185

Answer

The total distance traveled by the cart during the entire motion is

185\text{ m}

.

The total distance is calculated by analyzing the two distinct stages of motion. In the first phase, traveling at an initial velocity of

10\text{ m/s}

with an acceleration of

3\text{ m/s}^2

for

4\text{ s}

yields a distance of

64\text{ m}

and a peak velocity of

22\text{ m/s}

. In the second phase, decelerating from

22\text{ m/s}

to rest at

2\text{ m/s}^2

requires a distance of

121\text{ m}

. Adding both distances (

64\text{ m} + 121\text{ m}

) yields the total distance of

185\text{ m}

.

Step-by-Step Solution

1

Determine the distance (

s_1

) and final velocity (

v_1

) during the uniform acceleration phase

s_1 = 64\text{ m}

and

v_1 = 22\text{ m/s}

Using kinematic formulas

s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2

and

v_1 = u_1 + a_1 t_1

for constant acceleration.

2

Determine the stopping distance (

s_2

) during the uniform deceleration phase

s_2 = 121\text{ m}

Using

v_2^2 = v_1^2 - 2 a_2 s_2

with final speed

v_2 = 0\text{ m/s}

.

3

Sum the distances from both stages

S_{\text{total}} = 185\text{ m}

Total displacement for multi-stage motion along a straight line is the sum of displacements in each stage.

Key Concept

Multi-stage linear motion with constant acceleration and deceleration

Question 215Question →

A projectile is launched from level ground with an initial velocity of $50\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile at the highest point of its trajectory?

0\text{ m/s}

25\text{ m/s}

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

50\text{ m/s}

Show answer & explanation

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

Answer

The speed of the projectile at its highest point is

25\text{ m/s}

.

In 2D projectile motion, the horizontal component of velocity remains constant throughout flight because no horizontal force acts on the object. At the highest point, vertical velocity reduces to zero, making the total speed equal exclusively to the horizontal velocity component:

v = u \cos 60^\circ = 50 \times 0.5 = 25\text{ m/s}

.

Step-by-Step Solution

1

Resolve the initial velocity into horizontal and vertical components

u_x = u \cos\theta = 50 \cos 60^\circ = 25\text{ m/s}

and

u_y = u \sin\theta = 50 \sin 60^\circ = 25\sqrt{3}\text{ m/s}

Projectile motion decomposes into independent horizontal (constant velocity) and vertical (uniform acceleration) components.

2

Determine the vertical component of velocity at maximum height

v_y = 0\text{ m/s}

At the apex of the parabolic path, the upward vertical motion momentarily stops before descending.

3

Calculate total speed at the apex using vector magnitude

v = \sqrt{v_x^2 + v_y^2} = \sqrt{25^2 + 0^2} = 25\text{ m/s}

Because horizontal acceleration is zero (ignoring air resistance),

v_x = u_x = 25\text{ m/s}

throughout the flight.

Key Concept

Horizontal Component of Velocity in Projectile Motion

Question 216Question →

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

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

due east

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

due east

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

due east

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

due east

Show answer & explanation

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

Answer

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

due east

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

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

and trolley Q is

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

, yielding a net initial momentum of

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

. After impact, the total mass is

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

. The common velocity is

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

, where the positive sign denotes a direction due east.

Step-by-Step Solution

1

Assign a directional coordinate system

Let the eastward direction be positive (

+

) and the westward direction be negative (

-

)

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

2

Calculate total initial momentum (

p_i

)

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

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

3

Calculate final common velocity (

v

)

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

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

5.0\text{ kg}

.

Key Concept

Conservation of Linear Momentum in 1D Inelastic Collisions

Question 217Question →

A hiker walks $12\text{ km}$ due East, then $9\text{ km}$ due South, and finally $4\text{ km}$ due North. What is the magnitude of the hiker's total displacement from the starting point?

13\text{ km}

25\text{ km}

15\text{ km}

7\text{ km}

Show answer & explanation

Answer: $13\text{ km}$

Answer

The magnitude of the hiker's total displacement is

13\text{ km}

.

Displacement is a vector quantity requiring directional vector addition. The horizontal Eastward component is

12\text{ km}

. The net vertical component along the North-South line is

9\text{ km \text{ South}} - 4\text{ km \text{ North}} = 5\text{ km \text{ South}}

. Because the East and South directions are perpendicular (

90^\circ

), the resultant displacement magnitude is

\sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ km}

.

Step-by-Step Solution

1

Set up a coordinate system for horizontal (East-West) and vertical (North-South) components.

East is

+x

direction and North is

+y

direction.

Resolving vectors into orthogonal components allows simple independent summation along each axis.

2

Calculate net displacement along the x-axis and y-axis.

R_x = +12\text{ km}

and

R_y = -9\text{ km} + 4\text{ km} = -5\text{ km}

.

North and South act in opposite directions along the y-axis, so their magnitudes subtract.

3

Calculate the magnitude of the resultant displacement vector using Pythagoras' theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ km}

.

Perpendicular components combine geometrically to give the magnitude of the resultant vector.

Key Concept

Vector resolution and addition of non-collinear displacement vectors.

Estimated Time:1m 30s

Question 218Question →

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

True

False

Show answer & explanation

Answer: True

Answer

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

F \propto v^2

).

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

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

). For a continuous fluid stream,

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

. Since each unit mass loses speed

v

upon impact,

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

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

Step-by-Step Solution

1

Determine the mass of water striking the wall in time interval

\Delta t

.

\Delta m = \rho A v \Delta t

, where

\rho

is fluid density,

A

is cross-sectional area, and

v

is speed.

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

A v

.

2

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

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

.

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

v

.

3

Analyze the dependence of force

F

on speed

v

.

Since density

\rho

and cross-sectional area

A

are constant,

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

.

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

v

.

Key Concept

Newton's Second Law and continuous mass flow momentum change

Question 219Question →

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

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

due west

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

due east

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

due west

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

due west

Show answer & explanation

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

Answer

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

due west

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

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

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

2.0\text{ kg}

fragment has a momentum of

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

. To balance this, the

3.0\text{ kg}

fragment must possess a momentum of

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

, yielding

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

, which means a speed of

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

directed due west.

Step-by-Step Solution

1

Determine the initial momentum of the system

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

The object is initially at rest.

2

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

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

Linear momentum is conserved in an isolated system.

3

Equate initial momentum to final momentum and solve for

v_2

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

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

Key Concept

Conservation of Linear Momentum in Explosions

Estimated Time:1m 30s

Question 220Question →

Match each physical scenario or vector operation on the left with its corresponding physical property or resultant classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Particle in uniform circular motion pairs with a vector quantity of constant magnitude but changing direction; Dot product of force and displacement pairs with a scalar quantity representing work done; Two equal forces at 120 degrees pair with a resultant magnitude equal to P; Slope of displacement-time graph pairs with a vector quantity representing velocity.

Each physical scenario correctly corresponds to its fundamental vector/scalar classification or calculated resultant based on vector algebra.

Step-by-Step Solution

1

Analyze circular motion at constant speed

The velocity vector changes direction at every point, giving rise to centripetal acceleration directed toward the center.

Any physical quantity with direction is a vector quantity.

2

Evaluate the product of force and displacement vectors

W = \mathbf{F} \cdot \mathbf{d} = F d \cos\theta

, producing a scalar quantity (work done).

The dot product of two vectors yields a scalar.

3

Calculate the resultant of two equal vectors

P

at

120^\circ

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

.

Vector addition follows the law of cosines taking spatial angle into account.

4

Determine the physical quantity represented by the gradient of a displacement-time graph

\text{Slope} = \frac{\Delta s}{\Delta t} = v \text{ (velocity)}.

Displacement per unit time is velocity, a vector quantity.

Key Concept

Scalars, Vectors, and Vector Addition Laws

Mechanics

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution