Motion in a Plane Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An object moves in a circular path of radius \( 4 \, \text{m} \) with a frequency of \( 5 \, \text{Hz} \). What is the magnitude of its centripetal acceleration?

Angular speed \( \omega = 2\pi f = 2\pi \times 5 = 10\pi \, \text{rad/s} \).

Centripetal acceleration \( a_c = \omega^2 R = (10\pi)^2 \times 4 = 100\pi^2 \times 4 \approx 3947.8 \, \text{m/s}^2 \).

3947.8 m/s²
200 m/s²
1000 m/s²
500 m/s²
1

A projectile is launched at \( 18 \, \text{m/s} \) at \( 45^\circ \). What is its horizontal velocity component?

Horizontal velocity \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 18 \, \text{m/s}, \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \).

\( v_x = 18 \times 0.707 \approx 12.73 \, \text{m/s} \).

10 m/s
18 m/s
12.73 m/s
9 m/s
3

A particle’s position is given by \( x = 5t - t^2 \) and \( y = 2t^2 \) (in meters and seconds). What is the magnitude of its acceleration?

Velocity: \( v_x = \frac{dx}{dt} = 5 - 2t, v_y = \frac{dy}{dt} = 4t \).

Acceleration: \( a_x = \frac{dv_x}{dt} = -2, a_y = \frac{dv_y}{dt} = 4 \).

Magnitude \( a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47 \, \text{m/s}^2 \).

2 m/s²
4.47 m/s²
6 m/s²
3 m/s²
2

An object in uniform circular motion has a radius of 5 m and completes 10 revolutions in 20 s. What is the magnitude of its centripetal acceleration?

Angular speed ω = (2π n) / t, where n = 10, t = 20 s.

ω = (2π × 10) / 20 = π rad/s.

Centripetal acceleration a_c = ω² R = (π)² × 5 ≈ 9.87 × 5 ≈ 49.35 m/s².

49.35 m/s²
25 m/s²
10 m/s²
100 m/s²
1

In motion with constant acceleration in a plane, what is true if the acceleration is zero?

If the acceleration is zero, the velocity remains constant in both magnitude and direction, resulting in uniform rectilinear (straight-line) motion. Without acceleration, the object continues in its initial state per Newton’s first law.

The path is parabolic
The object stops
The motion is circular
The motion is uniform and rectilinear
4

A projectile is launched with a speed of \( 16 \, \text{m/s} \) at an angle of \( 30^\circ \) to the horizontal. What is its maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 30^\circ = 0.5 \))

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

Given: \( v_0 = 16 \, \text{m/s}, \sin 30^\circ = 0.5, g = 10 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 16 \times 0.5 = 8 \, \text{m/s} \).

\( h_m = \frac{8^2}{2 \times 10} = \frac{64}{20} = 3.2 \, \text{m} \).

3.2 m
6 m
10 m
4 m
1

A particle starts with velocity \( 4 \hat{j} \, \text{m/s} \) and accelerates at \( (3 \hat{i} - 2 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 2 \, \text{s} \)?

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

Given: \( \mathbf{v}_0 = 4 \hat{j}, \mathbf{a} = 3 \hat{i} - 2 \hat{j}, t = 2 \, \text{s} \).

\( \mathbf{v} = 4 \hat{j} + (3 \hat{i} - 2 \hat{j}) \times 2 = 6 \hat{i} + (4 - 4) \hat{j} = 6 \hat{i} \, \text{m/s} \).

Speed \( v = \sqrt{6^2 + 0^2} = \sqrt{36} = 6 \, \text{m/s} \).

4 m/s
8 m/s
6 m/s
10 m/s
3

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

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

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

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

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160} \approx 12.65 \, \text{m/s} \).

12.65 m/s
10 m/s
15 m/s
8 m/s
1

A boat moves north at 20 km/h while a river current flows east at 15 km/h. What is the magnitude of the resultant velocity?

Resultant velocity R = √(v_x² + v_y²).

Given: v_x = 15 km/h (east), v_y = 20 km/h (north).

R = √(15² + 20²) = √(225 + 400) = √625 = 25 km/h.

20 km/h
25 km/h
35 km/h
15 km/h
2

A ball is thrown horizontally at \( 15 \, \text{m/s} \) from a height of \( 98 \, \text{m} \). What is its horizontal range? (Take \( g = 9.8 \, \text{m/s}^2 \))

Time to fall: \( h = \frac{1}{2} g t^2 \Rightarrow 98 = \frac{1}{2} \times 9.8 \times t^2 \Rightarrow 98 = 4.9 t^2 \Rightarrow t^2 = 20 \Rightarrow t = \sqrt{20} \approx 4.47 \, \text{s} \).

Horizontal range \( R = v_x t = 15 \times 4.47 \approx 67.05 \, \text{m} \).

60 m
70 m
75 m
67.05 m
4

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