Motion In A Plane Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the significance of the angle of projection in determining the maximum height of a projectile?

The maximum height \( h_m = \frac{(v_0 \sin \theta_0)^2}{2g} \) depends on the vertical component of the initial velocity (\( v_0 \sin \theta_0 \)). The angle of projection determines this component, with larger angles (up to 90°) increasing the height by maximizing the vertical contribution.

It affects only the range
It has no effect
It determines the horizontal velocity
It influences the vertical velocity component
4

A projectile is launched at \( 14 \, \text{m/s} \) at \( 53^\circ \). What is its time of flight? (Take \( g = 9.8 \, \text{m/s}^2, \sin 53^\circ = 0.8 \))

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

Given: \( v_0 = 14 \, \text{m/s}, \sin 53^\circ = 0.8, g = 9.8 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 14 \times 0.8}{9.8} = \frac{22.4}{9.8} \approx 2.29 \, \text{s} \).

1.5 s
2.29 s
3 s
2 s
2

A ball is thrown at \( 22 \, \text{m/s} \) at \( 60^\circ \). What is its vertical velocity at \( t = 1.5 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2, \sin 60^\circ = 0.866 \))

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

Given: \( v_0 = 22 \, \text{m/s}, \sin 60^\circ = 0.866, t = 1.5 \, \text{s}, g = 10 \, \text{m/s}^2 \).

\( v_y = 22 \times 0.866 - 10 \times 1.5 = 19.052 - 15 = 4.05 \, \text{m/s} \).

5 m/s
10 m/s
4.05 m/s
0 m/s
3

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

Displacement \( \mathbf{r} = \frac{1}{2} \mathbf{a} t^2 \) (since \( \mathbf{v}_0 = 0 \)).

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

\( \mathbf{r} = \frac{1}{2} (5 \hat{i} - 2 \hat{j}) \times 3^2 = \frac{9}{2} (5 \hat{i} - 2 \hat{j}) = 22.5 \hat{i} - 9 \hat{j} \, \text{m} \).

Magnitude \( |\mathbf{r}| = \sqrt{22.5^2 + (-9)^2} = \sqrt{506.25 + 81} = \sqrt{587.25} \approx 24.23 \, \text{m} \).

20 m
25 m
30 m
24.23 m
4

In uniform circular motion, which quantity is constant in direction?

In uniform circular motion, the acceleration is centripetal and always directed towards the center of the circle. Its magnitude is constant, but its direction changes to remain radial. The speed (magnitude of velocity) is constant, while the velocity direction and acceleration direction are not.

Velocity
Speed
Acceleration
Angular velocity direction
2

A ball is projected at \( 30 \, \text{m/s} \) at \( 53^\circ \). What is the time to reach maximum height? (Take \( g = 10 \, \text{m/s}^2, \sin 53^\circ = 0.8 \))

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

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

\( t_m = \frac{30 \times 0.8}{10} = \frac{24}{10} = 2.4 \, \text{s} \).

2 s
3 s
2.4 s
1.5 s
3

A projectile is launched at \( 12 \, \text{m/s} \) at \( 37^\circ \). What is its time of flight? (Take \( g = 10 \, \text{m/s}^2, \sin 37^\circ = 0.6 \))

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

Given: \( v_0 = 12 \, \text{m/s}, \sin 37^\circ = 0.6, g = 10 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 12 \times 0.6}{10} = \frac{14.4}{10} = 1.44 \, \text{s} \).

1 s
1.44 s
2 s
3 s
2

A projectile is launched at \( 45 \, \text{m/s} \) at \( 30^\circ \). What is its speed at maximum height? (Take \( g = 10 \, \text{m/s}^2, \cos 30^\circ = 0.866 \))

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

Given: \( v_0 = 45 \, \text{m/s}, \cos 30^\circ = 0.866 \).

\( v_x = 45 \times 0.866 \approx 38.97 \, \text{m/s} \).

30 m/s
45 m/s
35 m/s
38.97 m/s
4

A projectile is launched at \( 36 \, \text{m/s} \) at \( 37^\circ \). What is its horizontal range? (Take \( g = 12 \, \text{m/s}^2, \sin 74^\circ = 0.96 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 74^\circ \).

Given: \( v_0 = 36 \, \text{m/s}, \sin 74^\circ = 0.96, g = 12 \, \text{m/s}^2 \).

\( R = \frac{36^2 \times 0.96}{12} = \frac{1296 \times 0.96}{12} = \frac{1244.16}{12} \approx 103.68 \, \text{m} \).

103.68 m
50 m
150 m
80 m
1

For an object moving in a plane with constant acceleration, what can be said about its path if the acceleration is perpendicular to the initial velocity?

If the constant acceleration is perpendicular to the initial velocity, the path becomes a parabola. This is because the acceleration affects only one component (e.g., vertical in projectile motion), while the other (e.g., horizontal) remains uniform, resulting in a parabolic trajectory.

It is parabolic
It is a straight line
It is circular
It is elliptical
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!