Motion in a Plane Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A projectile is launched at \( 24 \, \text{m/s} \) at \( 60^\circ \) to the horizontal. What is its time of flight? (Take \( g = 9.6 \, \text{m/s}^2, \sin 60^\circ = 0.866 \))

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

Given: \( v_0 = 24 \, \text{m/s}, \sin 60^\circ = 0.866, g = 9.6 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 24 \times 0.866}{9.6} = \frac{41.568}{9.6} \approx 4.33 \, \text{s} \).

4.33 s
2 s
5 s
3 s
1

In projectile motion, what is the direction of acceleration throughout the trajectory?

The acceleration in projectile motion, due to gravity, is always directed downward towards the Earth. It remains constant in magnitude and direction, acting vertically regardless of the projectile’s position or velocity.

Downward
Horizontal
Along the velocity vector
Upward
1

A stone is thrown horizontally at \( 14 \, \text{m/s} \) from a height of \( 39.2 \, \text{m} \). What is its time of flight? (Take \( g = 9.8 \, \text{m/s}^2 \))

Vertical motion: \( y = \frac{1}{2} g t^2 \), where \( y = -39.2 \, \text{m} \).

\( -39.2 = \frac{1}{2} \times 9.8 \times t^2 \Rightarrow -39.2 = 4.9 t^2 \Rightarrow t^2 = 8 \Rightarrow t = \sqrt{8} \approx 2.83 \, \text{s} \).

2.83 s
2 s
3.5 s
4 s
1

A ball is thrown horizontally at \( 11 \, \text{m/s} \) from a height of \( 19.6 \, \text{m} \). What is its speed on hitting the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

Vertical velocity: \( v_y = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 19.6} = \sqrt{384.16} \approx 19.6 \, \text{m/s} \).

Horizontal velocity: \( v_x = 11 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{11^2 + 19.6^2} = \sqrt{121 + 384.16} = \sqrt{505.16} \approx 22.47 \, \text{m/s} \).

20 m/s
22.47 m/s
25 m/s
15 m/s
2

A projectile is launched at \( 40 \, \text{m/s} \) at \( 37^\circ \). What is its speed at maximum height? (Take \( g = 10 \, \text{m/s}^2, \cos 37^\circ = 0.8 \))

At maximum height, speed = \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 40 \, \text{m/s}, \cos 37^\circ = 0.8 \).

\( v_x = 40 \times 0.8 = 32 \, \text{m/s} \).

20 m/s
40 m/s
25 m/s
32 m/s
4

For an object in uniform circular motion, what is the direction of its acceleration?

In uniform circular motion, the acceleration is centripetal, always directed towards the center of the circular path. This is because the velocity vector changes direction continuously, and the acceleration is perpendicular to the velocity, pointing inward.

Tangential to the path
Towards the center of the circle
Away from the center
Along the velocity vector
1

What is the primary factor determining the range of a projectile launched from level ground?

The range of a projectile depends on the launch angle and initial speed, as given by \( R = \frac{v_0^2 \sin 2\theta_0}{g} \). The angle \( \theta_0 \) maximizes the range when \( 2\theta_0 = 90^\circ \) (i.e., \( \theta_0 = 45^\circ \)), making it a critical factor alongside speed.

Vertical velocity alone
Launch angle and initial speed
Height of launch
Horizontal velocity alone
2

A particle in circular motion has a centripetal acceleration of \( 64 \, \text{m/s}^2 \) and radius \( 4 \, \text{m} \). What is its speed?

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

Given: \( a_c = 64 \, \text{m/s}^2, R = 4 \, \text{m} \).

\( 64 = \frac{v^2}{4} \Rightarrow v^2 = 256 \Rightarrow v = \sqrt{256} = 16 \, \text{m/s} \).

12 m/s
20 m/s
16 m/s
8 m/s
3

What is the primary reason an object in uniform circular motion does not move towards the center?

The object does not move towards the center because its tangential velocity provides the inertia to maintain the circular path. The centripetal force only changes the direction of this velocity, not its magnitude, balancing the tendency to move outward.

Centripetal force is too weak
Acceleration is tangential
Gravity prevents it
Tangential velocity keeps it in orbit
4

An object completes 5 revolutions in 10 s in a circular path of radius 3 m. What is its linear speed?

Angular speed ω = (2π n) / t = (2π × 5) / 10 = π rad/s.

Linear speed v = ω R = π × 3 ≈ 3.14 × 3 ≈ 9.42 m/s.

3 m/s
6 m/s
12 m/s
9.42 m/s
4

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