Motion in a Plane Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A ball is thrown at \( 40 \, \text{m/s} \) at \( 30^\circ \). What is its vertical velocity at the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

At maximum height, the vertical velocity \( v_y = 0 \, \text{m/s} \) (constant property of projectile motion).

20 m/s
10 m/s
40 m/s
0 m/s
4

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

Velocity: \( v_x = \frac{dx}{dt} = 2 + 2t, v_y = \frac{dy}{dt} = 3 - 4t \).

At \( t = 2 \, \text{s} \): \( v_x = 2 + 2 \times 2 = 6 \, \text{m/s}, v_y = 3 - 4 \times 2 = -5 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \approx 7.81 \, \text{m/s} \).

6 m/s
7.81 m/s
10 m/s
5 m/s
2

A particle moves in a circular path of radius \( 2 \, \text{m} \) with a period of \( 1 \, \text{s} \). What is its centripetal acceleration?

Angular speed \( \omega = \frac{2\pi}{T} = \frac{2\pi}{1} = 2\pi \, \text{rad/s} \).

Centripetal acceleration \( a_c = \omega^2 R = (2\pi)^2 \times 2 = 4\pi^2 \times 2 \approx 78.96 \, \text{m/s}^2 \).

78.96 m/s²
40 m/s²
100 m/s²
20 m/s²
1

A body is projected with a speed of 20 m/s at 45°. What is the time to reach maximum height? (Take g = 10 m/s²)

Time to max height t_m = (v₀ sin θ₀) / g.

Given: v₀ = 20 m/s, sin 45° = 1/√2, g = 10 m/s².

t_m = (20 × (1/√2)) / 10 = (20 / √2) / 10 = √2 ≈ 1.41 s.

2 s
1 s
1.41 s
3 s
3

In projectile motion, what is true about the acceleration at all points in the trajectory?

In projectile motion, assuming no air resistance, the acceleration is due to gravity and remains constant in magnitude and direction (downward) throughout the trajectory, regardless of the object’s position or velocity.

It is constant in magnitude and direction
It varies with height
It is zero at the peak
It changes direction with velocity
1

A projectile is launched at \( 28 \, \text{m/s} \) at \( 45^\circ \). What is its time of flight? (Take \( g = 9.8 \, \text{m/s}^2, \sin 45^\circ = 0.707 \))

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

Given: \( v_0 = 28 \, \text{m/s}, \sin 45^\circ = 0.707, g = 9.8 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 28 \times 0.707}{9.8} = \frac{39.592}{9.8} \approx 4.04 \, \text{s} \).

4.04 s
3 s
5 s
2 s
1

A projectile is launched at \( 20 \, \text{m/s} \) at \( 53^\circ \). What is its maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 53^\circ = 0.8 \))

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

Given: \( v_0 = 20 \, \text{m/s}, \sin 53^\circ = 0.8, g = 10 \, \text{m/s}^2 \).

\( h_m = \frac{(20 \times 0.8)^2}{2 \times 10} = \frac{16^2}{20} = \frac{256}{20} = 12.8 \, \text{m} \).

10 m
16 m
12.8 m
20 m
3

In projectile motion, what remains constant throughout the trajectory, assuming no air resistance?

In projectile motion without air resistance, the horizontal component of velocity remains constant because there is no horizontal acceleration. The vertical component changes due to gravity, but the horizontal motion is uniform.

Vertical velocity
Horizontal velocity
Total speed
Vertical acceleration
2

What is the primary factor affecting the maximum height of a projectile launched from level ground?

The maximum height \( h_m = \frac{(v_0 \sin \theta_0)^2}{2g} \) depends on the initial vertical velocity component (\( v_0 \sin \theta_0 \)). This component, influenced by both speed and angle, determines how high the projectile rises against gravity.

Horizontal velocity
Total speed alone
Launch angle alone
Initial vertical velocity component
4

A projectile is launched at \( 18 \, \text{m/s} \) at \( 53^\circ \). What is its maximum height? (Take \( g = 9.8 \, \text{m/s}^2, \sin 53^\circ = 0.8 \))

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

Given: \( v_0 = 18 \, \text{m/s}, \sin 53^\circ = 0.8, g = 9.8 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 18 \times 0.8 = 14.4 \, \text{m/s} \).

\( h_m = \frac{14.4^2}{2 \times 9.8} = \frac{207.36}{19.6} \approx 10.58 \, \text{m} \).

8 m
12 m
15 m
10.58 m
4

A projectile is launched at \( 42 \, \text{m/s} \) at \( 30^\circ \). What is its range? (Take \( g = 9.8 \, \text{m/s}^2, \sin 60^\circ = 0.866 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 60^\circ \).

Given: \( v_0 = 42 \, \text{m/s}, \sin 60^\circ = 0.866, g = 9.8 \, \text{m/s}^2 \).

\( R = \frac{42^2 \times 0.866}{9.8} = \frac{1764 \times 0.866}{9.8} \approx \frac{1527.62}{9.8} \approx 155.88 \, \text{m} \).

100 m
200 m
150 m
155.88 m
4

A particle moves in a circular path of radius \( 6 \, \text{m} \) 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 \( a_c = \omega^2 R = (4\pi)^2 \times 6 = 16\pi^2 \times 6 \approx 947.52 \, \text{m/s}^2 \).

947.52 m/s²
500 m/s²
1000 m/s²
750 m/s²
1

A ball is thrown horizontally at \( 5 \, \text{m/s} \) from a height of \( 20 \, \text{m} \). What is its horizontal range? (Take \( g = 10 \, \text{m/s}^2 \))

Time to fall: \( h = \frac{1}{2} g t^2 \Rightarrow 20 = \frac{1}{2} \times 10 \times t^2 \Rightarrow 20 = 5 t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2 \, \text{s} \).

Horizontal range \( R = v_x t = 5 \times 2 = 10 \, \text{m} \).

5 m
15 m
20 m
10 m
4

In motion with constant acceleration in a plane, how are the x- and y-components of the motion related?

For motion with constant acceleration in a plane, the x- and y-components are independent of each other. The motion can be treated as two separate one-dimensional motions along perpendicular directions, with each component governed by its own constant acceleration.

They are independent
They are directly proportional
They are inversely proportional
They are always equal
1

A particle completes 6 revolutions in \( 3 \, \text{s} \) in a circular path of radius \( 1 \, \text{m} \). What is its linear speed?

Frequency \( f = \frac{n}{t} = \frac{6}{3} = 2 \, \text{Hz} \).

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

Linear speed \( v = \omega R = 4\pi \times 1 = 4\pi \approx 12.57 \, \text{m/s} \).

8 m/s
10 m/s
15 m/s
12.57 m/s
4

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