Motion in a Plane Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In projectile motion, what is true about the total mechanical energy, assuming no air resistance?

Assuming no air resistance, the total mechanical energy (kinetic + potential) of a projectile remains conserved throughout its trajectory. Energy transforms between kinetic and potential forms, but the sum stays constant as no external non-conservative forces act.

It increases with height
It decreases with height
It remains conserved
It becomes zero at the peak
3

What is the significance of the horizontal component of velocity in determining the range of a projectile?

The horizontal component of velocity (\( v_0 \cos \theta_0 \)) determines how far the projectile travels horizontally during its time of flight. It remains constant and, combined with the time of flight (influenced by the vertical component), defines the range.

It affects only the height
It has no effect on range
It determines the horizontal distance
It changes with time
3

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

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

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

\( \mathbf{r} = \frac{1}{2} (3 \hat{i} + 5 \hat{j}) \times 2^2 = 2 (3 \hat{i} + 5 \hat{j}) = 6 \hat{i} + 10 \hat{j} \, \text{m} \).

Magnitude \( |\mathbf{r}| = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \approx 11.66 \, \text{m} \).

10 m
15 m
12 m
11.66 m
4

A swimmer moves north at \( 3 \, \text{m/s} \) in a river flowing west at \( 4 \, \text{m/s} \). What is the magnitude of the swimmer’s velocity relative to the ground?

Velocity components: \( v_x = -4 \, \text{m/s}, v_y = 3 \, \text{m/s} \).

Magnitude \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s} \).

3 m/s
5 m/s
7 m/s
4 m/s
2

A ball is thrown horizontally at \( 12 \, \text{m/s} \) from a height of \( 24.5 \, \text{m} \). What is its horizontal range? (Take \( g = 9.8 \, \text{m/s}^2 \))

Time to fall: \( h = \frac{1}{2} g t^2 \Rightarrow 24.5 = \frac{1}{2} \times 9.8 \times t^2 \Rightarrow 24.5 = 4.9 t^2 \Rightarrow t^2 = 5 \Rightarrow t = \sqrt{5} \approx 2.24 \, \text{s} \).

Horizontal range \( R = v_x t = 12 \times 2.24 \approx 26.88 \, \text{m} \).

20 m
30 m
25 m
26.88 m
4

A projectile is launched with a speed of \( 15 \, \text{m/s} \) at an angle of \( 60^\circ \) 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: \( v_0 = 15 \, \text{m/s}, \sin 60^\circ = 0.866, g = 10 \, \text{m/s}^2 \).

\( v_0 \sin \theta_0 = 15 \times 0.866 \approx 12.99 \, \text{m/s} \).

\( h_m = \frac{(12.99)^2}{2 \times 10} = \frac{168.74}{20} \approx 8.44 \, \text{m} \).

8.44 m
15 m
5 m
10 m
1

For an object moving in a plane with constant acceleration, what is true about its velocity vector?

In motion with constant acceleration in a plane, the velocity vector changes in magnitude and/or direction over time because the acceleration continuously alters it. Unlike uniform motion, the velocity is not constant unless the acceleration is zero.

It remains constant
It changes only in direction
It changes only in magnitude
It changes in magnitude and/or direction
4

In projectile motion, what is the condition for achieving maximum horizontal range on level ground?

The horizontal range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \) is maximized when \( \sin 2\theta_0 = 1 \), which occurs at \( 2\theta_0 = 90^\circ \) or \( \theta_0 = 45^\circ \). This angle balances the horizontal and vertical components for the greatest distance.

Launch angle of 30°
Launch angle of 60°
Launch angle of 90°
Launch angle of 45°
4

A projectile is launched at \( 15 \, \text{m/s} \) at \( 30^\circ \). What is its time of flight? (Take \( g = 10 \, \text{m/s}^2, \sin 30^\circ = 0.5 \))

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

Given: \( v_0 = 15 \, \text{m/s}, \sin 30^\circ = 0.5, g = 10 \, \text{m/s}^2 \).

\( T_f = \frac{2 \times 15 \times 0.5}{10} = \frac{15}{10} = 1.5 \, \text{s} \).

1 s
1.5 s
2 s
2.5 s
2

A stone is thrown horizontally at \( 10 \, \text{m/s} \) from a height of \( 80 \, \text{m} \). What is its time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Vertical motion: \( y = \frac{1}{2} g t^2 \), where \( y = -80 \, \text{m} \).

\( -80 = \frac{1}{2} \times 10 \times t^2 \Rightarrow -80 = 5 t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4 \, \text{s} \).

4 s
2 s
6 s
3 s
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