Motion in a Plane Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A projectile is launched at \( 50 \, \text{m/s} \) at \( 60^\circ \). What is its range? (Take \( g = 10 \, \text{m/s}^2, \sin 60^\circ = 0.866 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 120^\circ, \sin 120^\circ = 0.866 \).

Given: \( v_0 = 50 \, \text{m/s}, g = 10 \, \text{m/s}^2 \).

\( R = \frac{50^2 \times 0.866}{10} = \frac{2500 \times 0.866}{10} = 216.5 \, \text{m} \).

200 m
250 m
150 m
216.5 m
4

A projectile is launched with a speed of \( 25 \, \text{m/s} \) at an angle of \( 37^\circ \) to the horizontal. What is the time taken to reach its maximum height? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 37^\circ = 0.6 \))

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

Given: \( v_0 = 25 \, \text{m/s}, \theta_0 = 37^\circ, \sin 37^\circ = 0.6, g = 10 \, \text{m/s}^2 \).

\( t_m = \frac{25 \times 0.6}{10} = \frac{15}{10} = 1.5 \, \text{s} \).

1.5 s
2.5 s
3 s
1 s
1

What is the primary reason a projectile follows a curved path?

The curved (parabolic) path of a projectile results from the constant acceleration due to gravity acting vertically, while the horizontal motion remains uniform. This combination of uniform horizontal motion and accelerated vertical motion produces the curve.

Constant acceleration due to gravity
Variable horizontal velocity
Tangential force
Centripetal force
1

In uniform circular motion, what is the relationship between centripetal force and the object’s mass?

The centripetal force is given by \( F_c = m \frac{v^2}{R} = m a_c \), where \( m \) is the mass. Thus, the centripetal force is directly proportional to the mass of the object, meaning a greater mass requires a larger force to maintain the same motion.

It is inversely proportional
It is independent of mass
It is proportional to mass squared
It is directly proportional
4

A projectile is launched at \( 35 \, \text{m/s} \) at \( 60^\circ \). What is its range? (Take \( g = 10 \, \text{m/s}^2, \sin 120^\circ = 0.866 \))

Range \( R = \frac{v_0^2 \sin 2\theta_0}{g} \), where \( 2\theta_0 = 120^\circ \).

Given: \( v_0 = 35 \, \text{m/s}, \sin 120^\circ = 0.866, g = 10 \, \text{m/s}^2 \).

\( R = \frac{35^2 \times 0.866}{10} = \frac{1225 \times 0.866}{10} \approx \frac{1060.85}{10} \approx 106.09 \, \text{m} \).

100 m
120 m
90 m
106.09 m
4

What is the nature of the acceleration in uniform circular motion?

In uniform circular motion, the acceleration is centripetal, meaning it is directed towards the center of the circle. It changes direction continuously to remain radial but maintains a constant magnitude, as the speed is constant.

Tangential
Variable in magnitude
Constant in magnitude and direction
Constant in magnitude but not direction
3

A ball is thrown horizontally at \( 9 \, \text{m/s} \) from a height of \( 44.1 \, \text{m} \). What is its speed on hitting the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

Vertical velocity: \( v_y = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 44.1} = \sqrt{863.28} \approx 29.38 \, \text{m/s} \).

Horizontal velocity: \( v_x = 9 \, \text{m/s} \).

Speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{9^2 + 29.38^2} = \sqrt{81 + 863.28} = \sqrt{944.28} \approx 30.73 \, \text{m/s} \).

25 m/s
30.73 m/s
35 m/s
20 m/s
2

In projectile motion, at what point is the speed of the projectile minimum?

The speed of a projectile is minimum at its maximum height because the vertical component of velocity (\( v_y \)) is zero, leaving only the horizontal component (\( v_x \)), which is constant. At all other points, the speed includes both components.

At the point of launch
At landing
At maximum height
Halfway up its trajectory
3

A projectile is launched at \( 45 \, \text{m/s} \) at \( 60^\circ \). What is its speed at the maximum height? (Take \( g = 10 \, \text{m/s}^2, \cos 60^\circ = 0.5 \))

At maximum height, \( v_y = 0 \), speed = \( v_x = v_0 \cos \theta_0 \).

Given: \( v_0 = 45 \, \text{m/s}, \cos 60^\circ = 0.5 \).

\( v_x = 45 \times 0.5 = 22.5 \, \text{m/s} \).

30 m/s
45 m/s
15 m/s
22.5 m/s
4

A projectile is launched at \( 50 \, \text{m/s} \) 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: \( v_0 = 50 \, \text{m/s}, \cos 37^\circ = 0.8 \).

\( v_x = 50 \times 0.8 = 40 \, \text{m/s} \).

30 m/s
50 m/s
20 m/s
40 m/s
4

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