Motion in a Plane Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle starts from rest with an acceleration of \( (4 \hat{i} + 3 \hat{j}) \, \text{m/s}^2 \). 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: \( \mathbf{a} = 4 \hat{i} + 3 \hat{j}, t = 2 \, \text{s} \).

\( \mathbf{r} = \frac{1}{2} (4 \hat{i} + 3 \hat{j}) \times 2^2 = 2 (4 \hat{i} + 3 \hat{j}) = 8 \hat{i} + 6 \hat{j} \, \text{m} \).

Magnitude \( |\mathbf{r}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{m} \).

12 m
8 m
15 m
10 m
4

For a projectile launched at an angle, what is true about the horizontal and vertical motions?

The horizontal and vertical motions in projectile motion are independent of each other. The horizontal motion is uniform (constant velocity), while the vertical motion is uniformly accelerated due to gravity, allowing separate analysis of each component.

They are dependent
They are both uniform
They are independent
They are both accelerated
3

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

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

At \( t = 2 \, \text{s} \): \( v_x = 4 \times 2 + 3 = 11 \, \text{m/s}, v_y = 2 \times 2 - 2 = 2 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{11^2 + 2^2} = \sqrt{121 + 4} = \sqrt{125} \approx 11.18 \, \text{m/s} \).

10 m/s
11.18 m/s
15 m/s
8 m/s
2

In projectile motion, what is conserved throughout the trajectory if air resistance is negligible?

With negligible air resistance, the total mechanical energy (kinetic + potential) is conserved in projectile motion. Energy transforms between kinetic and potential forms, but the sum remains constant as gravity is a conservative force.

Horizontal velocity only
Total mechanical energy
Vertical velocity
Speed
2

A ball is thrown horizontally at \( 6 \, \text{m/s} \) from a height of \( 4.9 \, \text{m} \). What is its horizontal range? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

Horizontal range \( R = v_x t = 6 \times 1 = 6 \, \text{m} \).

3 m
9 m
12 m
6 m
4

A particle moves in a circular path of radius \( 3 \, \text{m} \) with a speed of \( 6 \, \text{m/s} \). What is its angular speed?

Angular speed \( \omega = \frac{v}{R} \).

Given: \( v = 6 \, \text{m/s}, R = 3 \, \text{m} \).

\( \omega = \frac{6}{3} = 2 \, \text{rad/s} \).

1 rad/s
2 rad/s
3 rad/s
6 rad/s
2

What is the primary factor affecting the time of flight of a projectile launched from level ground?

The time of flight of a projectile depends on the initial vertical velocity component (\( v_0 \sin \theta_0 \)) and gravity (\( g \)), as given by \( T_f = \frac{2 v_0 \sin \theta_0}{g} \). The vertical component determines how long the projectile stays in the air.

Horizontal velocity
Vertical velocity component
Total speed
Launch angle alone
1

A stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the ground? (Take g = 10 m/s²)

Horizontal velocity: v_x = 15 m/s (constant).

Vertical velocity: v_y = √(2gh) = √(2 × 10 × 45) = √900 = 30 m/s.

Speed v = √(v_x² + v_y²) = √(15² + 30²) = √(225 + 900) = √1125 ≈ 33.54 m/s.

15 m/s
33.54 m/s
45 m/s
30 m/s
2

What is the physical significance of the time of flight being zero for a projectile?

A time of flight of zero occurs when the initial vertical velocity component (\( v_0 \sin \theta_0 \)) is zero, meaning the projectile is launched horizontally (\( \theta_0 = 0^\circ \)) from ground level, resulting in no vertical displacement and immediate landing.

It reaches infinite height
It has maximum range
It is launched horizontally from ground
It is launched vertically upward
3

A particle moves in a circle of radius \( 1.5 \, \text{m} \) with a speed of \( 3 \, \text{m/s} \). What is its centripetal acceleration?

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

Given: \( v = 3 \, \text{m/s}, R = 1.5 \, \text{m} \).

\( a_c = \frac{3^2}{1.5} = \frac{9}{1.5} = 6 \, \text{m/s}^2 \).

4 m/s²
6 m/s²
9 m/s²
3 m/s²
2

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