Mechanics Practice Questions — JAMB UTME — Page 3

Question 41Question →

A projectile is launched from ground level with a horizontal velocity component of $15\text{ m/s}$ and a vertical velocity component of $20\text{ m/s}$ . Neglecting air resistance, what is the magnitude of the velocity of the projectile at its maximum height?

15\text{ m/s}

0\text{ m/s}

20\text{ m/s}

25\text{ m/s}

Show answer & explanation

Answer: $15\text{ m/s}$

Answer

The magnitude of the velocity of the projectile at its maximum height is

15\text{ m/s}

.

In projectile motion under gravity without air resistance, the horizontal component of velocity remains constant throughout flight (

v_x = 15\text{ m/s}

). At the highest point (apex), the vertical component of velocity momentarily becomes zero (

v_y = 0\text{ m/s}

). Therefore, the magnitude of the velocity at the maximum height is equal to the horizontal component, which is

15\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the maximum height of a projectile

Vertical component

v_y = 0\text{ m/s}

and horizontal component

v_x = u_x = 15\text{ m/s}

.

Gravity acts vertically, reducing vertical velocity to zero at the peak, while horizontal velocity remains constant in the absence of air resistance.

2

Calculate the total magnitude of velocity at maximum height

v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 0^2} = 15\text{ m/s}

.

The magnitude of the resultant velocity vector is derived using the Pythagorean theorem.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 42Question →

A spherical particle of radius $3.0 \times 10^{-3}\text{ m}$ and density $8.0 \times 10^3\text{ kg/m}^3$ is released from rest and falls vertically through a tall column of a viscous liquid of density $2.0 \times 10^3\text{ kg/m}^3$ . If the coefficient of viscosity of the fluid is $0.40\text{ Pa}\cdot\text{s}$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the magnitude of its terminal velocity in $\text{m/s}$ .

Show answer & explanation

Answer: 0.3

Answer

The magnitude of the terminal velocity of the falling sphere is

0.3\text{ m/s}

.

When a body falls at terminal velocity through a viscous medium, its weight is balanced by the sum of buoyancy upthrust and Stokes' viscous drag force. Applying

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with sphere radius

r = 0.003\text{ m}

, sphere density

\rho_s = 8000\text{ kg/m}^3

, fluid density

\rho_f = 2000\text{ kg/m}^3

, viscosity

\eta = 0.40\text{ Pa}\cdot\text{s}

, and

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

yields

v_t = 0.3\text{ m/s}

.

Step-by-Step Solution

1

Formulate the dynamic equilibrium condition at terminal velocity.

At terminal velocity, the net acceleration is zero, leading to the force balance equation

W = U + F_v

, where

W

is the gravitational weight of the sphere,

U

is the buoyant upthrust, and

F_v

is the retarding viscous force.

Terminal velocity occurs when the downward force of gravity is precisely balanced by the sum of upward resistive and buoyancy forces.

2

Substitute algebraic expressions for weight, upthrust, and Stokes' viscous drag.

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_t

.

Archimedes' principle defines the upthrust force equal to the weight of displaced liquid, while Stokes' law governs viscous resistance on spherical bodies.

3

Solve the equilibrium equation for terminal velocity

v_t

.

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

Equating

\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi \eta r v_t

and simplifying cancels common factors of

\pi

and

r

.

4

Substitute the specified numerical parameters into the derived expression.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 2000) \times 10}{9 \times 0.40} = \frac{2 \times 9.0 \times 10^{-6} \times 6000 \times 10}{3.6} = \frac{1.08}{3.6} = 0.3\text{ m/s}

.

Direct calculation yields the exact value of terminal velocity.

Key Concept

Terminal Velocity, Stokes' Law, and Archimedes' Principle

Question 43Question →

A uniform spherical planet has twice the mass and twice the radius of the Earth. If the acceleration due to gravity at the Earth's surface is $g$ , what is the acceleration due to gravity at the surface of this planet?

0.25g

0.5g

1.0g

2.0g

Show answer & explanation

Answer: $0.5g$

Answer

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

0.5g

.

The acceleration due to gravity at a planet's surface is given by

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

. Doubling the mass doubles the field strength, but doubling the radius reduces the field strength by a factor of

2^2 = 4

due to the inverse-square law. Combining these changes results in

\frac{2}{4}g = 0.5g

.

Step-by-Step Solution

1

Write the general expression for surface acceleration due to gravity.

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

, where

G

is the gravitational constant,

M

is planetary mass, and

R

is planetary radius.

Establishes the fundamental formula relating gravity to mass and radius.

2

Substitute the given parameters for the new planet (

M_p = 2M

and

R_p = 2R

).

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

.

Applies the proportional changes to mass and radius into the formula.

3

Simplify the expression to find

g_p

in terms of

g

.

g_p = \frac{2}{4} \left(\frac{GM}{R^2}\right) = 0.5g

.

Evaluates the fractional change relative to Earth's surface gravity.

Key Concept

Dependence of surface gravitational field strength on planetary mass and radius via the inverse-square law.

Estimated Time:1m 15s

Question 44Question →

The speed $v$ of a transverse wave traveling along a stretched string under tension $T$ with mass per unit length $\mu$ is given by $v = k T^x \mu^y$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $x$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

x

is 0.5.

Applying the principle of dimensional homogeneity, the dimensions on both sides must match. Speed

[v] = L T^{-1}

, tension force

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

, and mass per unit length

[\mu] = M L^{-1}

. Substituting these into

v = k T^x \mu^y

yields

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

. Comparing the exponents of

T

gives

-2x = -1

, leading to

x = 0.5

.

Step-by-Step Solution

1

Determine the dimensions of speed

v

, tension force

T

, and linear density

\mu

.

[v] = L T^{-1}

,

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

,

[\mu] = M L^{-1}

Tension is a force (

F=ma

) with dimensions

[M L T^{-2}]

, and

\mu

is mass per unit length (

m/l

) with dimensions

[M L^{-1}]

.

2

Substitute the dimensional formulas into the equation

v = k T^x \mu^y

and collect powers of base dimensions.

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

Combining exponents for base dimensions

M

,

L

, and

T

allows applying the principle of dimensional homogeneity.

3

Equate the exponent of

T

on both sides to solve for

x

.

-2x = -1 \implies x = 0.5

The exponent of

T

on the left side is

-1

and on the right side is

-2x

.

Key Concept

Dimensional Analysis and Determination of Exponents

Estimated Time:1m 30s

Question 45Question →

A solid cylinder of length $10\text{ cm}$ floats vertically at the interface of two immiscible liquids: oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If $4\text{ cm}$ of its length is submerged in water while the remaining $6\text{ cm}$ is submerged in oil, what is the density of the cylinder?

880\text{ kg/m}^3

920\text{ kg/m}^3

900\text{ kg/m}^3

400\text{ kg/m}^3

Show answer & explanation

Answer: $880\text{ kg/m}^3$

Answer

The density of the cylinder is

880\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of the fluids in which it floats. For a body submerged across two immiscible liquids, the total buoyant force is the sum of the upthrusts from both fluids:

U = (\rho_w V_w + \rho_o V_o)g

. Equating this to the total weight

W = \rho_c V_{total} g

yields

\rho_c (0.10\text{ m}) = 1000(0.04\text{ m}) + 800(0.06\text{ m}) = 88

, which gives

\rho_c = 880\text{ kg/m}^3

.

Step-by-Step Solution

1

Set up the condition for flotation.

Weight of the cylinder = Total upthrust exerted by both liquids.

For a body floating in equilibrium, its total weight is balanced by the sum of buoyant forces from all surrounding fluids.

2

Express the weight and upthrusts in terms of density, cross-sectional area

A

, length

L

, and acceleration due to gravity

g

.

\rho_c A L g = (\rho_w A h_w + \rho_o A h_o) g

The weight of the cylinder is

\rho_c V_{total} g

and upthrust from each liquid is

\rho_{liquid} V_{submerged} g

.

3

Cancel common factors

A

and

g

from both sides.

\rho_c L = \rho_w h_w + \rho_o h_o

Since the cylinder has a uniform cross-sectional area, volume ratio simplifies to length ratio.

4

Substitute the given values (

L = 0.10\text{ m}

,

h_w = 0.04\text{ m}

,

h_o = 0.06\text{ m}

,

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

,

\rho_o = 800\text{ kg/m}^3

).

\rho_c (0.10) = 1000(0.04) + 800(0.06) = 40 + 48 = 88\text{ kg/m}^2

Evaluating the weighted contribution of buoyancy from each fluid.

5

Solve for the density of the cylinder

\rho_c

.

\rho_c = \frac{88}{0.10} = 880\text{ kg/m}^3

Dividing both sides by the total length of

0.10\text{ m}

gives the density.

Key Concept

Archimedes' Principle for Floating Bodies in Immiscible Liquids

Question 46Question →

A screw jack with a thread pitch of $0.5\text{ cm}$ is operated using a Tommy bar of length $35\text{ cm}$ . If the machine has an efficiency of $40\%$ , what is the minimum effort force required to raise a load of mass $880\text{ kg}$ ? (Take $g = 10\text{ m/s}^2$ and $\pi = \frac{22}{7}$ )

50\text{ N}

8\text{ N}

20\text{ N}

100\text{ N}

Show answer & explanation

Answer: $50\text{ N}$

Answer

The minimum effort force required is

50\text{ N}

.

The effort distance per turn is the circumference

2\pi r = 2 \times \frac{22}{7} \times 35\text{ cm} = 220\text{ cm}

. Dividing by the pitch (

0.5\text{ cm}

) yields a Velocity Ratio of

440

. Applying the

40\%

efficiency gives a Mechanical Advantage of

\text{MA} = 0.40 \times 440 = 176

. Finally, dividing the load force (

8800\text{ N}

) by

\text{MA}

gives an effort force of

50\text{ N}

.

Step-by-Step Solution

1

Calculate the total load force in Newtons

L = m \cdot g = 880\text{ kg} \times 10\text{ m/s}^2 = 8800\text{ N}

Effort overcomes weight force, which is mass multiplied by gravitational acceleration.

2

Calculate the Velocity Ratio (VR) of the screw jack

\text{VR} = \frac{2 \pi r}{p} = \frac{2 \times \frac{22}{7} \times 35\text{ cm}}{0.5\text{ cm}} = \frac{220\text{ cm}}{0.5\text{ cm}} = 440

The effort travels around the circumference of a circle of radius

r

, while the load moves vertically by one pitch length

p

per revolution.

3

Determine the Mechanical Advantage (MA) from efficiency

\text{MA} = \eta \times \text{VR} = 0.40 \times 440 = 176

Efficiency is defined as

\eta = \frac{\text{MA}}{\text{VR}}

, so

\text{MA} = \eta \cdot \text{VR}

.

4

Calculate the effort force E

E = \frac{L}{\text{MA}} = \frac{8800\text{ N}}{176} = 50\text{ N}

Mechanical advantage is the ratio of load to effort force

(\text{MA} = \frac{L}{E})

.

Key Concept

Relationship between Velocity Ratio, Mechanical Advantage, Efficiency, and Pitch in a Screw Jack

Question 47Question →

A constant net force of $20\text{ N}$ acts on an object of mass $4\text{ kg}$ that is initially at rest on a frictionless horizontal surface. What is the magnitude of the linear momentum of the object after $3\text{ s}$ ?

Show answer & explanation

Answer: 60

Answer

The magnitude of the linear momentum of the object after

3\text{ s}

is

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

.

According to Newton's Second Law in terms of momentum, force is the rate of change of linear momentum (

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

). Rearranging gives

\Delta p = F \Delta t

. Since the object is initially at rest (

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

), the final momentum is

p_f = F \Delta t = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

.

Step-by-Step Solution

1

Apply the impulse-momentum theorem.

J = F \Delta t = \Delta p

The impulse of the net force acting on an object equals the change in its linear momentum.

2

Calculate the final linear momentum.

p_f = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

Since the object starts from rest, its initial momentum is zero (

p_i = 0

), so the final momentum is equal to the total impulse supplied.

Key Concept

Impulse-Momentum Theorem

Question 48Question →

The viscous drag force $F$ acting on a spherical body moving through a fluid at speed $v$ is given by Stokes' law, $F = 6 \pi \eta r v$ , where $r$ is the radius of the sphere and $\eta$ is the coefficient of viscosity. What is the dimensional formula of the coefficient of viscosity $\eta$ ?

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

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

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

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

Show answer & explanation

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

Answer

The dimensional formula of the coefficient of viscosity is

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

.

Isolating the coefficient of viscosity from Stokes' law yields

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

. Substituting the dimensions

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

,

[r] = \text{L}

, and

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

gives

[\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

Identify the dimensional formulas for force, radius, and velocity.

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

,

[r] = \text{L}

, and

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

.

Base physical quantities must be represented by their fundamental dimensions.

2

Rearrange Stokes' law to express

\eta

in terms of the other variables, treating

6\pi

as a dimensionless constant.

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

.

Numerical constants do not carry physical dimensions.

3

Substitute the base dimensions into the formula and simplify using index rules.

[\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} \text{T}^{-1}

.

Subtracting denominator indices from numerator indices yields the final dimensional expression.

Key Concept

Dimensional analysis of physical constants and equations

Question 49Question →

An electric train starts from rest and accelerates uniformly at $2\text{ m/s}^2$ for a time interval $t$ . It then maintains the maximum velocity attained for a time interval of $t$ . Finally, it decelerates uniformly at $4\text{ m/s}^2$ until it comes to a complete stop. If the total distance covered during the entire journey is $350\text{ m}$ , what is the value of $t$ ?

5\text{ s}

7\text{ s}

10\text{ s}

14\text{ s}

Show answer & explanation

Answer: $10\text{ s}$

Answer

10\text{ s}

The motion consists of three stages: acceleration from rest (

s_1 = t^2

), uniform velocity (

s_2 = 2t^2

), and deceleration to rest (

s_3 = 0.5t^2

). Adding these gives a total distance of

S = 3.5t^2

. Setting

3.5t^2 = 350\text{ m}

yields

t^2 = 100

, so

t = 10\text{ s}

.

Step-by-Step Solution

1

Analyze Stage 1 (Acceleration phase)

Maximum velocity

v = 2t

, displacement

s_1 = \frac{1}{2}(2)t^2 = t^2

Starting from rest (

u = 0

) with acceleration

a_1 = 2\text{ m/s}^2

for duration

t

,

v = u + a_1 t = 2t

and

s_1 = \frac{1}{2} a_1 t^2 = t^2

.

2

Analyze Stage 2 (Constant velocity phase)

Displacement

s_2 = (2t)(t) = 2t^2

The train moves at constant velocity

v = 2t

for duration

t

, so distance is velocity multiplied by time.

3

Analyze Stage 3 (Deceleration phase)

Deceleration time

t_3 = 0.5t

, displacement

s_3 = 0.5t^2

Decelerating from

v = 2t

to

0

at

a_2 = 4\text{ m/s}^2

takes time

t_3 = \frac{v}{a_2} = \frac{2t}{4} = 0.5t

. Distance

s_3 = \frac{1}{2} v t_3 = \frac{1}{2} (2t)(0.5t) = 0.5t^2

.

4

Calculate total displacement and solve for

t

Total distance

S = 3.5t^2 = 350 \implies t^2 = 100 \implies t = 10\text{ s}

Summing the displacements:

S = t^2 + 2t^2 + 0.5t^2 = 3.5t^2

. Equating to

350\text{ m}

gives

3.5t^2 = 350

, so

t^2 = 100

, giving

t = 10\text{ s}

.

Key Concept

Multi-stage linear motion and displacement calculation

Estimated Time:2m 0s

Question 50Question →

A spherical planet has a radius of $7.2 \times 10^{6} \text{ m}$ and an acceleration due to gravity of $10 \text{ m/s}^2$ at its surface. What is the escape velocity for an object launched from the surface of this planet, in $\text{km/s}$ ?

Show answer & explanation

Answer: 12

Answer

The escape velocity from the surface of the planet is 12 km/s.

The escape velocity

v_e

from the surface of a spherical planet of radius

R

with surface gravity

g

is given by

v_e = \sqrt{2gR}

. Substituting

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

and

R = 7.2 \times 10^6 \text{ m}

yields

v_e = \sqrt{2 \times 10 \times 7.2 \times 10^6} = \sqrt{1.44 \times 10^8} = 12000 \text{ m/s}

, which equals

12 \text{ km/s}

.

Step-by-Step Solution

1

Identify the relationship between surface gravity, planetary radius, and escape velocity

The escape velocity formula is

v_e = \sqrt{2gR}

.

Escape velocity is the minimum initial speed required for an object to overcome the gravitational pull of a celestial body.

2

Substitute the given numerical values into the formula

v_e = \sqrt{2 \times 10 \text{ m/s}^2 \times 7.2 \times 10^6 \text{ m}} = \sqrt{144 \times 10^6} \text{ m/s}

.

Plugging in the given values allows direct calculation of the velocity in standard SI units.

3

Simplify the square root and convert units to km/s

v_e = 12000 \text{ m/s} = 12 \text{ km/s}

.

Taking the square root of

144 \times 10^6

gives

12000 \text{ m/s}

, which corresponds to

12 \text{ km/s}

.

Key Concept

Escape Velocity from a Planet's Surface

Question 51Question →

A toy truck of mass $2.5\text{ kg}$ moving at a velocity of $6.0\text{ m s}^{-1}$ collides head-on with a toy car of mass $1.5\text{ kg}$ moving in the opposite direction at $2.0\text{ m s}^{-1}$ . If the two vehicles stick together upon impact, what is their combined velocity immediately after the collision?

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

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

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

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

Show answer & explanation

Answer: $3.0\text{ m s}^{-1}$

Answer

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

in the direction of the initial motion of the truck

According to the principle of conservation of linear momentum, total initial momentum equals total final momentum. Taking the direction of the toy truck as positive gives an initial momentum of

(2.5\text{ kg} \times 6.0\text{ m s}^{-1}) + (1.5\text{ kg} \times -2.0\text{ m s}^{-1}) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}

. Dividing this net momentum by the combined mass

(2.5\text{ kg} + 1.5\text{ kg} = 4.0\text{ kg})

yields a common final velocity of

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

.

Step-by-Step Solution

1

Assign direction signs to the velocity vectors

Let the initial direction of the truck be positive (

u_1 = +6.0\text{ m s}^{-1}

). The car moves in the opposite direction, so its velocity is negative (

u_2 = -2.0\text{ m s}^{-1}

).

Linear momentum is a vector quantity, so direction must be accounted for in one-dimensional motion.

2

Calculate the total initial momentum of the system

p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.5 \times 6.0) + (1.5 \times -2.0) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}

The total momentum before collision is the vector sum of individual momenta.

3

Apply the law of conservation of linear momentum to solve for common final velocity

v

p_{\text{final}} = (m_1 + m_2) v = (2.5 + 1.5) v = 4.0 v

. Setting

p_{\text{final}} = p_{\text{initial}} \Rightarrow 4.0 v = 12.0 \Rightarrow v = 3.0\text{ m s}^{-1}

.

Since the vehicles stick together, they move as a single combined mass.

Key Concept

Conservation of Linear Momentum in Inelastic Collisions

Estimated Time:50s

Question 52Question →

An aircraft flies due East at a velocity of $120\text{ m/s}$ while a crosswind blows due North at $50\text{ m/s}$ . What is the magnitude of the resultant velocity of the aircraft?

130\text{ m/s}

170\text{ m/s}

70\text{ m/s}

110\text{ m/s}

Show answer & explanation

Answer: $130\text{ m/s}$

Answer

The magnitude of the resultant velocity of the aircraft is

130\text{ m/s}

.

Because the aircraft's velocity and the wind's velocity are perpendicular to each other, their vector sum is found using the Pythagorean theorem

R = \sqrt{v_1^2 + v_2^2}

. Substituting

120\text{ m/s}

and

50\text{ m/s}

gives

R = \sqrt{120^2 + 50^2} = \sqrt{16900} = 130\text{ m/s}

.

Step-by-Step Solution

1

Identify the given velocity vectors and their relative direction.

Velocity due East

\vec{v}_e = 120\text{ m/s}

and velocity due North

\vec{v}_n = 50\text{ m/s}

are perpendicular to each other (

90^\circ

angle).

East and North directions are orthogonal to each other.

2

Apply the Pythagorean theorem to calculate the magnitude of the resultant vector

R

.

R = \sqrt{v_e^2 + v_n^2} = \sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\text{ m/s}

.

The resultant of two perpendicular vectors forms the hypotenuse of a right-angled triangle.

Key Concept

Vector Addition of Perpendicular Vectors

Estimated Time:1m 0s

Question 53Question →

A projectile is launched from ground level over flat terrain with a constant horizontal velocity component of $10\text{ m/s}$ . At a height of $40\text{ m}$ above the ground, the magnitude of its vertical velocity component is equal to its horizontal velocity component. Taking $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the projectile in meters.

Show answer & explanation

Answer: 45

Answer

The maximum height reached by the projectile is

45\text{ m}

.

At the height of

40\text{ m}

, the vertical velocity is equal to the horizontal velocity of

10\text{ m/s}

. Applying the vertical kinematic relation

v_y^2 = u_y^2 - 2gh

yields

10^2 = u_y^2 - 2(10)(40)

, which gives

u_y^2 = 900\text{ m}^2/\text{s}^2

. The maximum height attained above ground level occurs when

v_y = 0

, calculated as

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Step-by-Step Solution

1

Identify the vertical component of velocity at the given height

v_y = 10\text{ m/s}

at

h = 40\text{ m}

The problem states that at

h = 40\text{ m}

, the vertical velocity component equals the constant horizontal component

u_x = 10\text{ m/s}

.

2

Determine the initial vertical launch velocity component

u_y

u_y^2 = 900\text{ m}^2/\text{s}^2 \implies u_y = 30\text{ m/s}

Applying the vertical motion equation

v_y^2 = u_y^2 - 2gh

gives

10^2 = u_y^2 - 2(10)(40) \implies u_y^2 = 100 + 800 = 900

.

3

Calculate the maximum height

H_{\text{max}}

H_{\text{max}} = 45\text{ m}

At maximum height, the vertical velocity becomes zero. Using

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Key Concept

Vertical Kinematics and Maximum Height of a Projectile

Question 54Question →

A satellite orbits the Earth at an altitude equal to three times the radius of the Earth, $R$ . If the acceleration due to gravity at the Earth's surface is $g$ , what is the gravitational field strength experienced by the satellite at its orbit?

\frac{g}{16}

\frac{g}{9}

\frac{g}{4}

\frac{g}{3}

Show answer & explanation

Answer: $\frac{g}{16}$

Answer

The gravitational field strength at the orbital position is

\frac{g}{16}

.

The total distance from the center of the Earth to the satellite is

r = R + 3R = 4R

. Since gravitational field strength is inversely proportional to the square of the distance from the center of mass, increasing the distance by a factor of

4

reduces the field strength by a factor of

4^2 = 16

, giving

\frac{g}{16}

.

Step-by-Step Solution

1

Determine the total distance from the center of the Earth to the satellite.

The orbital radius is

r = R + h = R + 3R = 4R

.

Gravitational calculations must be measured from the center of mass of the attracting body, not its surface.

2

Apply Newton's law of universal gravitation for gravitational field strength.

At the surface (

r = R

),

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

. At orbit (

r = 4R

),

g' = \frac{GM}{(4R)^2} = \frac{GM}{16R^2}

.

Gravitational field strength follows an inverse-square law with respect to distance from the center of mass.

3

Express the orbital gravitational field strength in terms of the surface gravity

g

.

g' = \frac{1}{16} \left(\frac{GM}{R^2}\right) = \frac{g}{16}

.

Substituting the surface value

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

yields the simplified ratio.

Key Concept

Gravitational Field Strength and Inverse-Square Law

Question 55Question →

A body of mass $4\text{ kg}$ moves with a constant velocity of $5\text{ m s}^{-1}$ . Calculate the kinetic energy of the body in Joules.

Show answer & explanation

Answer: 50

Answer

The kinetic energy of the body is

50\text{ J}

.

The kinetic energy of an object in linear motion is calculated using

E_k = \frac{1}{2} m v^2

. Substituting

m = 4\text{ kg}

and

v = 5\text{ m s}^{-1}

gives

E_k = \frac{1}{2} \times 4 \times 5^2 = 50\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 4\text{ kg}

and velocity

v = 5\text{ m s}^{-1}

Extract values provided in the problem statement.

2

Apply the formula for kinetic energy

E_k = \frac{1}{2} m v^2

Kinetic energy is defined as half the product of mass and the square of velocity.

3

Substitute values and solve

E_k = \frac{1}{2} \times 4 \times (5)^2 = 50\text{ J}

Perform arithmetic evaluation to get the final energy in Joules.

Key Concept

Kinetic Energy

Question 56Question →

A particle is projected from horizontal ground at an initial angle $\theta$ to the horizontal. At its maximum height $H = 20\text{ m}$ , its kinetic energy is exactly half of its initial kinetic energy at launch. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the horizontal range of the projectile?

40\text{ m}

80\text{ m}

20\text{ m}

0\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

The horizontal range of the projectile is

80\text{ m}

.

At the trajectory's highest point, the vertical component of velocity becomes zero while the horizontal component

u\cos\theta

remains unchanged. The kinetic energy at the apex is therefore

\frac{1}{2}m(u\cos\theta)^2 = E_{k,i}\cos^2\theta

. Setting

\cos^2\theta = \frac{1}{2}

gives

\theta = 45^\circ

. For a

45^\circ

launch angle, the horizontal range

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

is related to the maximum height

H = \frac{u^2}{4g}

by

R = 4H

. Substituting

H = 20\text{ m}

yields

R = 80\text{ m}

.

Step-by-Step Solution

1

Relate kinetic energy at maximum height to initial kinetic energy.

Initial kinetic energy

E_k(0) = \frac{1}{2}m u^2

. At peak height, vertical velocity component

v_y = 0

, so velocity is

v_x = u\cos\theta

. Thus,

E_k(\text{peak}) = \frac{1}{2}m (u\cos\theta)^2 = E_k(0)\cos^2\theta

.

At maximum height, only the horizontal component of velocity remains.

2

Calculate the launch angle

\theta

.

Given

E_k(\text{peak}) = \frac{1}{2} E_k(0)

, we have

\cos^2\theta = \frac{1}{2}

, giving

\theta = 45^\circ

.

Setting the energy expression equal to the given condition enables solving for the angle.

3

Relate maximum height

H

to horizontal range

R

for

\theta = 45^\circ

.

For

\theta = 45^\circ

,

H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2}{4g}

and

R = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g}

. Therefore,

R = 4H

.

Standard formulas for maximum height and range express both quantities in terms of launch speed

u

and acceleration due to gravity

g

.

4

Substitute the maximum height value into the range formula.

R = 4 \times 20\text{ m} = 80\text{ m}

.

Multiplying the given peak height of

20\text{ m}

by

4

yields the exact horizontal range.

Key Concept

Kinetic Energy Conservation and Range-Height Relationship in Projectile Motion

Estimated Time:2m 0s

Question 57Question →

A uniform wooden cube of edge length $0.20\text{ m}$ floats in water of density $1000\text{ kg/m}^3$ with $0.15\text{ m}$ of its vertical height submerged. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what minimum mass, in kilograms, must be placed on the top surface of the cube so that its upper face becomes just flush with the water surface?

Show answer & explanation

Answer: 2

Answer

The minimum mass required to submerge the cube completely flush with the water surface is

2.0\text{ kg}

.

By the Law of Flotation, a floating body displaces its own weight of fluid. Initially, the cube displaces a volume of

0.20\text{ m} \times 0.20\text{ m} \times 0.15\text{ m} = 0.006\text{ m}^3

of water, corresponding to an upthrust of

60\text{ N}

(or mass of

6.0\text{ kg}

). When completely submerged, the total volume displaced is

0.20^3 = 0.008\text{ m}^3

, providing a total upthrust of

80\text{ N}

(or mass equivalent of

8.0\text{ kg}

). The additional mass required on top is therefore the difference:

8.0\text{ kg} - 6.0\text{ kg} = 2.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the cube

A = (0.20\text{ m})^2 = 0.04\text{ m}^2

The base area is needed to find the volume of the block submerged and unsubmerged.

2

Calculate the volume of the cube above the water surface

V_{\text{above}} = 0.04\text{ m}^2 \times (0.20\text{ m} - 0.15\text{ m}) = 0.002\text{ m}^3

To push the cube level with the surface, the additional weight added on top must balance the extra upthrust created by submerging this remaining volume.

3

Calculate the additional mass required

m = \rho_{\text{water}} \times V_{\text{above}} = 1000\text{ kg/m}^3 \times 0.002\text{ m}^3 = 2.0\text{ kg}

By Archimedes' principle, the additional downward mass must equal the mass of the extra water displaced when fully submerged.

Key Concept

Archimedes' Principle and Law of Flotation

Question 58Question →

A shell of total mass $5.0\text{ kg}$ is moving horizontally with a velocity of $20\text{ m s}^{-1}$ when an internal explosion splits it into two fragments. One fragment of mass $2.0\text{ kg}$ is propelled backward along the original path at a speed of $10\text{ m s}^{-1}$ . What is the magnitude of the velocity, in $\text{m s}^{-1}$ , of the second fragment immediately after the explosion?

Show answer & explanation

Answer: 40

Answer

The magnitude of the velocity of the second fragment immediately after the explosion is

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

.

According to the Law of Conservation of Linear Momentum, the total momentum of a system remains constant when no net external horizontal force acts on it. Taking the original direction of motion as positive, the initial momentum is

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

. Since the

2.0\text{ kg}

fragment moves backward at

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

, its momentum is

-20\text{ kg m s}^{-1}

. For total momentum to remain

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

, the remaining

3.0\text{ kg}

fragment must carry a momentum of

+120\text{ kg m s}^{-1}

, which corresponds to a velocity of

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

.

Step-by-Step Solution

1

Calculate the initial momentum of the shell prior to the explosion.

p_i = 5.0\text{ kg} \times 20\text{ m s}^{-1} = 100\text{ kg m s}^{-1}

in the initial forward direction.

Before the internal explosion, the system consists of a single mass moving with a constant velocity.

2

Apply the Law of Conservation of Linear Momentum taking vector direction into account.

p_i = m_1 v_1 + m_2 v_2 \implies 100 = 2.0(-10) + 3.0 v_2

In the absence of external forces, total momentum is conserved. The fragment propelled backward takes a negative sign relative to the initial forward motion.

3

Solve the algebraic equation for the unknown velocity

v_2

.

100 + 20 = 3.0 v_2 \implies 120 = 3.0 v_2 \implies v_2 = 40\text{ m s}^{-1}

Isolating

v_2

gives the forward velocity magnitude of the remaining

3.0\text{ kg}

piece.

Key Concept

Conservation of Linear Momentum in Explosions (1D Vector Sign Convention)

Question 59Question →

A block is pulled along a smooth horizontal table by a horizontal force of $30\text{ N}$ through a displacement of $4\text{ m}$ . At the same time, a vertical upward force of $40\text{ N}$ acts on the block as it moves. What is the total work done on the block?

120\text{ J}

280\text{ J}

160\text{ J}

7.5\text{ J}

Show answer & explanation

Answer: $120\text{ J}$

Answer

The total work done on the block is

120\text{ J}

.

Work done is given by

W = F d \cos\theta

. The horizontal force acts along the line of motion (

\theta = 0^\circ

), performing

30\text{ N} \times 4\text{ m} = 120\text{ J}

of work. The vertical force acts at an angle of

90^\circ

to the horizontal displacement, performing zero work because

\cos(90^\circ) = 0

. Thus, the total work done is

120\text{ J}

.

Step-by-Step Solution

1

Identify the formula for work done by a constant force

W = F \cdot d \cdot \cos(\theta)

Work is defined as the scalar product of force and displacement vectors.

2

Calculate work done by the horizontal force

W_{\text{horizontal}} = 30\text{ N} \times 4\text{ m} \times \cos(0^\circ) = 120\text{ J}

The horizontal force is in the exact direction of motion (

\theta = 0^\circ

).

3

Calculate work done by the vertical force

W_{\text{vertical}} = 40\text{ N} \times 4\text{ m} \times \cos(90^\circ) = 0\text{ J}

The vertical force is perpendicular to the horizontal displacement (

\theta = 90^\circ

), so

\cos(90^\circ) = 0

.

4

Sum the work done by all forces

W_{\text{total}} = 120\text{ J} + 0\text{ J} = 120\text{ J}

Work is a scalar quantity, so total work is the algebraic sum of individual work values.

Key Concept

Work done by perpendicular forces is zero

Estimated Time:45s

Question 60Question →

Match each displacement vector combination on the left with its corresponding resultant displacement magnitude or vector on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairs correspond as follows:

3\text{ m}

East and

4\text{ m}

North matches a resultant magnitude of

5\text{ m}

;

5\text{ m}

East and

12\text{ m}

South matches a resultant magnitude of

13\text{ m}

;

8\text{ m}

East and

6\text{ m}

West matches a resultant displacement of

2\text{ m}

East; and

9\text{ m}

North and

12\text{ m}

East matches a resultant magnitude of

15\text{ m}

.

Each vector combination is resolved according to its directional alignment: perpendicular displacements require the Pythagorean theorem (

R = \sqrt{A^2 + B^2}

), whereas anti-parallel collinear displacements require vector subtraction.

Step-by-Step Solution

1

Identify orthogonal vector scenarios

Perpendicular displacement vectors form right-angled triangles.

Directions such as East-North, East-South, and North-East are at

90^\circ

relative to one another.

2

Calculate magnitudes for orthogonal pairs using the Pythagorean theorem

For

3\text{ m}

and

4\text{ m}

:

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

. For

5\text{ m}

and

12\text{ m}

:

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

. For

9\text{ m}

and

12\text{ m}

:

\sqrt{9^2 + 12^2} = 15\text{ m}

.

The magnitude of two perpendicular vectors

\vec{A}

and

\vec{B}

is given by

R = \sqrt{A^2 + B^2}

.

3

Calculate net displacement for opposite collinear vectors

For

8\text{ m}

East and

6\text{ m}

West:

8 - 6 = 2\text{ m}

East.

Vectors pointing in opposite directions along the same axis subtract algebraically, retaining the direction of the vector with the greater magnitude.

Key Concept

Addition of Perpendicular and Collinear Displacement Vectors

Estimated Time:1m 30s

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