Motion in a Plane Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle starts with velocity \( 3 \hat{i} \, \text{m/s} \) and accelerates at \( (2 \hat{i} + 5 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 2 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \).

Given: \( \mathbf{v}_0 = 3 \hat{i}, \mathbf{a} = 2 \hat{i} + 5 \hat{j}, t = 2 \, \text{s} \).

\( \mathbf{v} = 3 \hat{i} + (2 \hat{i} + 5 \hat{j}) \times 2 = 3 \hat{i} + 4 \hat{i} + 10 \hat{j} = 7 \hat{i} + 10 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{7^2 + 10^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21 \, \text{m/s} \).

10 m/s
15 m/s
12.21 m/s
8 m/s
3

A projectile is launched at \( 25 \, \text{m/s} \) at \( 60^\circ \). What is its range? (Take \( g = 10 \, \text{m/s}^2, \sin 120^\circ = 0.866 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 120^\circ \).

Given: \( v_0 = 25 \, \text{m/s}, \sin 120^\circ = 0.866, g = 10 \, \text{m/s}^2 \).

\( R = \frac{25^2 \times 0.866}{10} = \frac{625 \times 0.866}{10} = \frac{541.25}{10} = 54.125 \, \text{m} \).

50 m
54.13 m
60 m
40 m
2

In uniform circular motion, what is the relationship between centripetal acceleration (\( a_c \)) and angular speed (\( \omega \))?

The centripetal acceleration in uniform circular motion is given by \( a_c = \omega^2 R \), where \( R \) is the radius. This shows that \( a_c \) is proportional to the square of the angular speed, reflecting the dependence on the rate of change of direction.

\( a_c \propto \omega \)
\( a_c \propto 1/\omega \)
\( a_c \propto \omega^3 \)
\( a_c \propto \omega^2 \)
4

A ball is thrown at 30 m/s at 60° to the horizontal. What is the time of flight? (Take g = 10 m/s²)

Time of flight T_f = (2 v₀ sin θ₀) / g.

Given: v₀ = 30 m/s, θ₀ = 60°, sin 60° = √3/2, g = 10 m/s².

T_f = (2 × 30 × (√3/2)) / 10 = (30 √3) / 10 = 3 √3 ≈ 5.2 s.

3 s
5.2 s
6 s
4 s
2

A particle starts from rest with a constant acceleration of \( (6 \hat{i} - 4 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 2 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \), with \( \mathbf{v}_0 = 0 \).

Given: \( \mathbf{a} = 6 \hat{i} - 4 \hat{j}, t = 2 \, \text{s} \).

\( \mathbf{v} = (6 \hat{i} - 4 \hat{j}) \times 2 = 12 \hat{i} - 8 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.42 \, \text{m/s} \).

14.42 m/s
12 m/s
16 m/s
10 m/s
1

A particle starts with velocity \( 8 \hat{i} \, \text{m/s} \) and accelerates at \( (2 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 4 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \).

Given: \( \mathbf{v}_0 = 8 \hat{i}, \mathbf{a} = 2 \hat{j}, t = 4 \, \text{s} \).

\( \mathbf{v} = 8 \hat{i} + 2 \hat{j} \times 4 = 8 \hat{i} + 8 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \approx 11.31 \, \text{m/s} \).

10 m/s
12 m/s
15 m/s
11.31 m/s
4

In projectile motion, what happens to the kinetic energy at the maximum height compared to the launch point?

At the maximum height, the vertical velocity is zero, so the kinetic energy is due only to the horizontal velocity, which remains constant. Thus, the kinetic energy decreases compared to the launch point, where both components contribute.

It increases
It decreases
It remains the same
It becomes zero
2

A plane flies east at \( 40 \, \text{m/s} \) while a wind blows north at \( 30 \, \text{m/s} \). What is the magnitude of the plane’s velocity relative to the ground?

Velocity components: \( v_x = 40 \, \text{m/s}, v_y = 30 \, \text{m/s} \).

Magnitude \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{m/s} \).

40 m/s
50 m/s
70 m/s
30 m/s
2

What is the relationship between angular speed (\( \omega \)) and linear speed (\( v \)) in uniform circular motion?

In uniform circular motion, the linear speed \( v \) is related to the angular speed \( \omega \) by the equation \( v = \omega R \), where \( R \) is the radius of the circular path. This shows that linear speed is directly proportional to angular speed.

\( v = \omega / R \)
\( v = \omega R \)
\( v = \omega^2 R \)
\( v = R / \omega \)
2

A stone is projected horizontally with a speed of 10 m/s from a height of 20 m. How long does it take to reach the ground? (Take g = 10 m/s²)

For vertical motion: y = v₀y t + (1/2) a_y t², where v₀y = 0, a_y = -g.

Given: y = -20 m, g = 10 m/s².

-20 = 0 - (1/2) × 10 × t² ⇒ -20 = -5 t² ⇒ t² = 4 ⇒ t = 2 s.

2 s
1 s
3 s
4 s
1

A ball is thrown at \( 35 \, \text{m/s} \) at \( 37^\circ \). What is its time of flight? (Take \( g = 10 \, \text{m/s}^2, \sin 37^\circ = 0.6 \))

Time of flight \( T_f = \frac{2 v_0 \sin \theta_0}{g} \).

Given: \( v_0 = 35 \, \text{m/s}, \sin 37^\circ = 0.6, g = 10 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 35 \times 0.6}{10} = \frac{42}{10} = 4.2 \, \text{s} \).

3 s
6 s
4.2 s
5 s
3

A stone tied to a string moves in a circle of radius \( 1.5 \, \text{m} \) with a frequency of \( 2 \, \text{Hz} \). What is its centripetal acceleration?

Angular speed \( \omega = 2\pi f = 2\pi \times 2 = 4\pi \, \text{rad/s} \).

Centripetal acceleration \( a_c = \omega^2 R = (4\pi)^2 \times 1.5 = 16\pi^2 \times 1.5 \approx 236.87 \, \text{m/s}^2 \).

150 m/s²
236.87 m/s²
100 m/s²
300 m/s²
2

A particle’s position is given by \( x = 3t^2 - 4t \) and \( y = 5t \) (in meters and seconds). What is its speed at \( t = 1 \, \text{s} \)?

Velocity: \( v_x = \frac{dx}{dt} = 6t - 4, v_y = \frac{dy}{dt} = 5 \).

At \( t = 1 \, \text{s} \): \( v_x = 6 \times 1 - 4 = 2 \, \text{m/s}, v_y = 5 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39 \, \text{m/s} \).

4 m/s
5.39 m/s
7 m/s
3 m/s
2

A particle in circular motion has a centripetal acceleration of \( 81 \, \text{m/s}^2 \) and radius \( 9 \, \text{m} \). What is its speed?

Centripetal acceleration \( a_c = \frac{v^2}{R} \).

Given: \( a_c = 81 \, \text{m/s}^2, R = 9 \, \text{m} \).

\( 81 = \frac{v^2}{9} \Rightarrow v^2 = 729 \Rightarrow v = \sqrt{729} = 27 \, \text{m/s} \).

18 m/s
30 m/s
27 m/s
24 m/s
3

A projectile is launched with a speed of \( 10 \, \text{m/s} \) at an angle of \( 45^\circ \) to the horizontal. What is its maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 45^\circ = 0.707 \))

Maximum height \( h_m = \frac{(v_0 \sin \theta_0)^2}{2g} \).

Given: \( v_0 = 10 \, \text{m/s}, \sin 45^\circ = 0.707, g = 10 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 10 \times 0.707 \approx 7.07 \, \text{m/s} \).

\( h_m = \frac{(7.07)^2}{2 \times 10} = \frac{49.98}{20} \approx 2.5 \, \text{m} \).

2.5 m
5 m
7.5 m
10 m
1

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