Motion in a Plane Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle starts with velocity \( 5 \hat{i} \, \text{m/s} \) and accelerates at \( (4 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 3 \, \text{s} \)?

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

Given: \( \mathbf{v}_0 = 5 \hat{i}, \mathbf{a} = 4 \hat{j}, t = 3 \, \text{s} \).

\( \mathbf{v} = 5 \hat{i} + 4 \hat{j} \times 3 = 5 \hat{i} + 12 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{m/s} \).

10 m/s
15 m/s
12 m/s
13 m/s
4

A particle starts from rest with a constant acceleration of \( (-2 \hat{i} + 5 \hat{j}) \, \text{m/s}^2 \). What is the magnitude of its velocity after \( 3 \, \text{s} \)?

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

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

\( \mathbf{v} = (-2 \hat{i} + 5 \hat{j}) \times 3 = -6 \hat{i} + 15 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-6)^2 + 15^2} = \sqrt{36 + 225} = \sqrt{261} \approx 16.16 \, \text{m/s} \).

16.16 m/s
12 m/s
20 m/s
10 m/s
1

A ball is thrown horizontally at \( 8 \, \text{m/s} \) from a height of \( 12.5 \, \text{m} \). What is its speed on hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Vertical velocity: \( v_y = \sqrt{2gh} = \sqrt{2 \times 10 \times 12.5} = \sqrt{250} \approx 15.81 \, \text{m/s} \).

Horizontal velocity: \( v_x = 8 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{8^2 + 15.81^2} = \sqrt{64 + 250} = \sqrt{314} \approx 17.72 \, \text{m/s} \).

15 m/s
17.72 m/s
20 m/s
12 m/s
2

A particle moves in a circle of radius \( 4 \, \text{m} \) with an angular speed of \( 3 \, \text{rad/s} \). What is its centripetal acceleration?

Centripetal acceleration \( a_c = \omega^2 R \).

Given: \( \omega = 3 \, \text{rad/s}, R = 4 \, \text{m} \).

\( a_c = 3^2 \times 4 = 9 \times 4 = 36 \, \text{m/s}^2 \).

12 m/s²
36 m/s²
48 m/s²
24 m/s²
2

A ball is thrown horizontally from a tower with a speed of \( 8 \, \text{m/s} \). It lands \( 24 \, \text{m} \) from the base. What is the height of the tower? (Take \( g = 10 \, \text{m/s}^2 \))

Horizontal distance \( x = v_x t \), so \( t = \frac{x}{v_x} = \frac{24}{8} = 3 \, \text{s} \).

Vertical height \( h = \frac{1}{2} g t^2 \).

\( h = \frac{1}{2} \times 10 \times 3^2 = 5 \times 9 = 45 \, \text{m} \).

45 m
20 m
30 m
60 m
1

A stone is thrown horizontally at 12 m/s from a 80 m high cliff. How far from the base does it land? (Take g = 10 m/s²)

Time to fall: y = (1/2) g t², -80 = (1/2) × 10 × t² ⇒ t² = 16 ⇒ t = 4 s.

Horizontal distance x = v_x t = 12 × 4 = 48 m.

24 m
36 m
60 m
48 m
4

A particle completes 8 revolutions in \( 4 \, \text{s} \) in a circular path of radius \( 2 \, \text{m} \). What is its linear speed?

Frequency \( f = \frac{n}{t} = \frac{8}{4} = 2 \, \text{Hz} \).

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

Linear speed \( v = \omega R = 4\pi \times 2 = 8\pi \approx 25.13 \, \text{m/s} \).

16 m/s
20 m/s
10 m/s
25.13 m/s
4

A projectile is launched at \( 20 \, \text{m/s} \) at \( 53^\circ \). What is its range? (Take \( g = 10 \, \text{m/s}^2, \sin 106^\circ = 0.96 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 106^\circ \).

Given: \( v_0 = 20 \, \text{m/s}, \sin 106^\circ = 0.96, g = 10 \, \text{m/s}^2 \).

\( R = \frac{20^2 \times 0.96}{10} = \frac{400 \times 0.96}{10} = \frac{384}{10} = 38.4 \, \text{m} \).

38.4 m
20 m
50 m
30 m
1

In projectile motion, what is true about the velocity vector just before landing compared to just after launch, assuming level ground?

For a projectile on level ground, the velocity just before landing has the same magnitude as just after launch but with the vertical component reversed in direction (downward instead of upward) due to symmetry, while the horizontal component remains unchanged.

It is greater in magnitude
It is less in magnitude
It is identical
It has the same magnitude but opposite vertical direction
4

In projectile motion, what is the direction of the velocity vector at the highest point of its trajectory?

In projectile motion, the velocity has both horizontal (\( v_x \)) and vertical (\( v_y \)) components. At the highest point, the vertical component \( v_y = 0 \) due to the symmetry of the parabolic path, while the horizontal component \( v_x \) remains constant. Thus, the velocity vector is purely horizontal.

Horizontal
Vertical
At an angle to the horizontal
Zero
1

A projectile is launched at \( 20 \, \text{m/s} \) at \( 37^\circ \). What is its horizontal velocity component? (Take \( \cos 37^\circ = 0.8 \))

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

Given: \( v_0 = 20 \, \text{m/s}, \cos 37^\circ = 0.8 \).

\( v_x = 20 \times 0.8 = 16 \, \text{m/s} \).

12 m/s
20 m/s
16 m/s
10 m/s
3

In uniform circular motion, what is the role of the centripetal acceleration?

The centripetal acceleration in uniform circular motion changes the direction of the velocity vector to keep the object moving in a circular path. It does not affect the speed, which remains constant.

It increases the speed
It decreases the speed
It changes the direction of velocity
It changes the magnitude of velocity
3

A particle moves with a constant acceleration of \( (5 \hat{i} + 2 \hat{j}) \, \text{m/s}^2 \) from rest. What is its speed after \( 3 \, \text{s} \)?

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

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

\( \mathbf{v} = (5 \hat{i} + 2 \hat{j}) \times 3 = 15 \hat{i} + 6 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 6^2} = \sqrt{225 + 36} = \sqrt{261} \approx 16.16 \, \text{m/s} \).

16.16 m/s
12 m/s
20 m/s
10 m/s
1

A ball is thrown at \( 30 \, \text{m/s} \) at \( 53^\circ \). What is its vertical velocity at \( t = 2 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2, \sin 53^\circ = 0.8 \))

Vertical velocity \( v_y = v_0 \sin \theta_0 - g t \).

Given: \( v_0 = 30 \, \text{m/s}, \sin 53^\circ = 0.8, t = 2 \, \text{s}, g = 10 \, \text{m/s}^2 \).

\( v_y = 30 \times 0.8 - 10 \times 2 = 24 - 20 = 4 \, \text{m/s} \).

10 m/s
20 m/s
4 m/s
0 m/s
3

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