Motion in a Plane Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A boat travels west at \( 6 \, \text{m/s} \) while a current flows south at \( 8 \, \text{m/s} \). What is the magnitude of the boat’s velocity relative to the ground?

Velocity components: \( v_x = -6 \, \text{m/s}, v_y = -8 \, \text{m/s} \).

Magnitude \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \).

6 m/s
10 m/s
14 m/s
8 m/s
2

In uniform circular motion, what is true about the direction of the net force relative to the velocity?

The net force (centripetal force) in uniform circular motion is perpendicular to the velocity vector, as it acts towards the center of the circle, while the velocity is tangential, ensuring the object follows a circular path.

It is parallel to velocity
It is perpendicular to velocity
It is opposite to velocity
It is at 45° to velocity
2

A particle moves in the x-y plane with a constant acceleration of (2 î + 3 ĵ) m/s². If it starts from rest at the origin, what is its speed after 2 s?

Velocity is given by v = v₀ + at, where v₀ = 0 (starts from rest).

Given: a = 2 î + 3 ĵ, t = 2 s.

v = 0 + (2 î + 3 ĵ) × 2 = 4 î + 6 ĵ m/s.

Speed v = √(v_x² + v_y²) = √(4² + 6²) = √(16 + 36) = √52 ≈ 7.21 m/s.

7.21 m/s
5 m/s
10 m/s
12 m/s
1

A projectile is launched with a speed of \( 32 \, \text{m/s} \) at \( 60^\circ \) to the horizontal. What is its maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 60^\circ = 0.866 \))

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

Given: \( v_0 = 32 \, \text{m/s}, \sin 60^\circ = 0.866, g = 10 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 32 \times 0.866 \approx 27.712 \, \text{m/s} \).

\( h_m = \frac{(27.712)^2}{2 \times 10} = \frac{767.95}{20} \approx 38.4 \, \text{m} \).

38.4 m
20 m
50 m
30 m
1

A boy throws a ball with a speed of 20 m/s at an angle of 30° to the horizontal. What is the maximum height reached by the ball? (Take g = 10 m/s²)

Maximum height is given by h_m = (v₀ sin θ₀)² / (2g).

Given: v₀ = 20 m/s, θ₀ = 30°, g = 10 m/s².

sin 30° = 0.5, so v₀ sin θ₀ = 20 × 0.5 = 10 m/s.

h_m = (10)² / (2 × 10) = 100 / 20 = 5 m.

5 m
10 m
15 m
20 m
1

In projectile motion, what is the effect of launching at complementary angles (e.g., 30° and 60°) on the range?

For complementary angles (\( \theta \) and \( 90^\circ - \theta \)), \( \sin 2\theta = \sin (180^\circ - 2\theta) \), so \( R = \frac{v_0^2 \sin 2\theta}{g} \) is the same (e.g., \( \sin 60^\circ = \sin 120^\circ \)). Thus, the range remains the same for complementary angles.

Range increases
Range decreases
Range becomes zero
Range remains the same
4

For an object in uniform circular motion, what is the direction of the velocity vector relative to the acceleration vector?

In uniform circular motion, the velocity vector is tangential to the circular path, while the centripetal acceleration is directed towards the center. Thus, the velocity and acceleration vectors are perpendicular to each other at every point.

Parallel
Opposite
Perpendicular
At an angle of 45°
3

A projectile is launched at \( 30 \, \text{m/s} \) at \( 45^\circ \). What is its time to reach maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 45^\circ = 0.707 \))

Time to maximum height \( t_m = \frac{v_0 \sin \theta_0}{g} \).

Given: \( v_0 = 30 \, \text{m/s}, \sin 45^\circ = 0.707, g = 10 \, \text{m/s}^2 \).

\( t_m = \frac{30 \times 0.707}{10} \approx \frac{21.21}{10} \approx 2.12 \, \text{s} \).

2 s
3 s
2.12 s
1.5 s
3

A projectile is launched with a speed of 50 m/s at 45°. What is its horizontal range? (Take g = 10 m/s²)

Range R = (v₀² sin 2θ₀) / g.

Given: v₀ = 50 m/s, θ₀ = 45°, sin 90° = 1, g = 10 m/s².

R = (50² × 1) / 10 = 2500 / 10 = 250 m.

250 m
125 m
500 m
100 m
1

In projectile motion, what is true about the horizontal component of velocity throughout the trajectory?

In projectile motion, assuming no air resistance, the horizontal component of velocity remains constant because there is no horizontal acceleration. Gravity acts only vertically, leaving the horizontal motion unaffected.

It remains constant
It increases with time
It decreases with time
It becomes zero at the peak
1

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