Motion in a Plane Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A projectile is launched at \( 18 \, \text{m/s} \) at \( 30^\circ \). What is its range? (Take \( g = 9.8 \, \text{m/s}^2 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 60^\circ, \sin 60^\circ = 0.866 \).

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

\( R = \frac{18^2 \times 0.866}{9.8} = \frac{324 \times 0.866}{9.8} \approx 28.63 \, \text{m} \).

20 m
28.63 m
35 m
15 m
2

A particle completes 10 revolutions in \( 5 \, \text{s} \) in a circular path of radius \( 1.5 \, \text{m} \). What is its linear speed?

Frequency \( f = \frac{n}{t} = \frac{10}{5} = 2 \, \text{Hz} \).

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

Linear speed \( v = \omega R = 4\pi \times 1.5 = 6\pi \approx 18.85 \, \text{m/s} \).

12 m/s
15 m/s
20 m/s
18.85 m/s
4

What is the shape of the path followed by a projectile under the influence of gravity alone?

The path of a projectile under gravity alone is a parabola. This results from the horizontal motion being uniform (constant velocity) and the vertical motion being uniformly accelerated due to gravity, combining to form a parabolic trajectory.

Parabola
Circle
Straight line
Ellipse
1

What is true about the angular speed in uniform circular motion if the period of revolution increases?

Angular speed is \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. If the period increases, \( \omega \) decreases because they are inversely proportional; a longer period means fewer revolutions per unit time.

It increases
It remains constant
It becomes zero
It decreases
4

In uniform circular motion, what causes the acceleration of the object?

The acceleration in uniform circular motion is centripetal, caused by a force (e.g., tension, gravity, friction) directed towards the center of the circle. This force continuously changes the direction of the velocity, maintaining circular motion.

Change in speed
Tangential force
Linear momentum
Centripetal force
4

In motion with constant acceleration in a plane, what happens if the acceleration is opposite to the initial velocity?

If the acceleration is opposite to the initial velocity, the object slows down, reverses direction, and then accelerates in the direction of the acceleration along a straight line. The path remains linear but with a change in direction.

The path becomes parabolic
The object moves in a circle
The object stops permanently
The object slows, reverses, and accelerates
4

In uniform circular motion, what happens to the object’s motion if its speed doubles while the radius remains constant?

If the speed doubles (\( v' = 2v \)) and the radius remains constant, the centripetal acceleration \( a_c = \frac{v^2}{R} \) increases by a factor of four (\( a_c' = \frac{(2v)^2}{R} = 4 \frac{v^2}{R} \)), as it is proportional to the square of the speed.

Centripetal acceleration doubles
Centripetal acceleration halves
Centripetal acceleration remains the same
Centripetal acceleration increases fourfold
4

A particle starts from rest with a constant acceleration of \( (4 \hat{i} - 3 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 5 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \), with \( \mathbf{v}_0 = 0 \).

Given: \( \mathbf{a} = 4 \hat{i} - 3 \hat{j}, t = 5 \, \text{s} \).

\( \mathbf{v} = (4 \hat{i} - 3 \hat{j}) \times 5 = 20 \hat{i} - 15 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \, \text{m/s} \).

25 m/s
20 m/s
30 m/s
15 m/s
1

What is the key difference between the velocity and acceleration vectors in uniform circular motion?

In uniform circular motion, the velocity vector is tangential to the path, while the acceleration vector (centripetal) is directed towards the center. This perpendicular relationship ensures the object maintains a circular path with constant speed.

They are parallel
They are perpendicular
They are opposite
They have the same magnitude
2

A particle starts with velocity \( 7 \hat{j} \, \text{m/s} \) and accelerates at \( (3 \hat{i} - 2 \hat{j}) \, \text{m/s}^2 \). What is its speed after \( 2 \, \text{s} \)?

Velocity \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \).

Given: \( \mathbf{v}_0 = 7 \hat{j}, \mathbf{a} = 3 \hat{i} - 2 \hat{j}, t = 2 \, \text{s} \).

\( \mathbf{v} = 7 \hat{j} + (3 \hat{i} - 2 \hat{j}) \times 2 = 6 \hat{i} + (7 - 4) \hat{j} = 6 \hat{i} + 3 \hat{j} \, \text{m/s} \).

Speed \( v = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 \, \text{m/s} \).

5 m/s
10 m/s
6.71 m/s
8 m/s
3

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