Laws of Motion Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 4.8 \, \text{kg} \) block on a \( 37^\circ \) incline (\( \mu_k = 0.2 \)) is pulled upward by a \( 6.8 \, \text{kg} \) mass over a pulley. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

For \( 6.8 \, \text{kg} \): \( 6.8g - T = 6.8a \Rightarrow 68 - T = 6.8a \).

For \( 4.8 \, \text{kg} \): \( T - mg \sin 37^\circ - f_k = 4.8a \).

\( N = mg \cos 37^\circ = 4.8 \times 10 \times 0.8 = 38.4 \, \text{N} \).

Friction: \( f_k = 0.2 \times 38.4 = 7.68 \, \text{N} \).

\( mg \sin 37^\circ = 48 \times 0.6 = 28.8 \, \text{N} \).

Net force: \( T - 28.8 - 7.68 = 4.8a \Rightarrow T - 36.48 = 4.8a \).

Solve: \( 68 - T = 6.8a \), \( T - 36.48 = 4.8a \).

Substitute: \( 68 - (4.8a + 36.48) = 6.8a \Rightarrow 68 - 36.48 - 4.8a = 6.8a \Rightarrow 31.52 = 11.6a \).

\( a \approx 2.72 \, \text{m/s}^2 \), \( T - 36.48 = 4.8 \times 2.72 \Rightarrow T - 36.48 \approx 13.06 \Rightarrow T \approx 49.54 \, \text{N} \).

45 N
49.5 N
53 N
57 N
2

A force of \( 10 \, \text{N} \) acts on a body of mass \( 2 \, \text{kg} \) initially at rest. What is the velocity of the body after \( 5 \, \text{s} \)?

First, calculate acceleration using Newton’s Second Law: \( a = \frac{F}{m} \).

Substitute: \( F = 10 \, \text{N} \), \( m = 2 \, \text{kg} \).

So, \( a = \frac{10}{2} = 5 \, \text{m/s}^2 \).

Use the equation \( v = u + at \), where initial velocity \( u = 0 \), \( t = 5 \, \text{s} \).

Substitute: \( v = 0 + 5 \times 5 = 25 \, \text{m/s} \).

10 m/s
15 m/s
20 m/s
25 m/s
4

A \( 2000 \, \text{kg} \) truck accelerates at \( 2 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then a \( 1 \, \text{kg} \) ball is dropped. What is the horizontal distance from the drop point after \( 1.5 \, \text{s} \)? (Neglect air resistance)

Truck’s speed at \( t = 5 \, \text{s} \): \( v = 0 + 2 \times 5 = 10 \, \text{m/s} \).

Ball retains horizontal velocity of \( 10 \, \text{m/s} \).

After drop, no horizontal acceleration.

Distance = \( v \times t = 10 \times 1.5 = 15 \, \text{m} \).

12 m
15 m
18 m
20 m
2

A \( 0.6 \, \text{kg} \) stone in a vertical circle of radius \( 1.5 \, \text{m} \) has a tension of \( 20 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

Bottom: \( T_b - mg = m \omega^2 r \Rightarrow 20 - 0.6 \times 10 = 0.6 \times \omega^2 \times 1.5 \).

\( 20 - 6 = 0.9 \omega^2 \Rightarrow 14 = 0.9 \omega^2 \Rightarrow \omega^2 = \frac{14}{0.9} \approx 15.56 \).

Top: \( T_t + mg = m \omega^2 r \Rightarrow T_t + 0.6 \times 10 = 0.6 \times 15.56 \times 1.5 \).

\( T_t + 6 = 14 \Rightarrow T_t = 14 - 6 = 8 \, \text{N} \).

6 N
8 N
10 N
12 N
2

A \( 0.15 \, \text{kg} \) ball moving at \( 25 \, \text{m/s} \) at \( 60^\circ \) to a wall rebounds at \( 30^\circ \) with the same speed. What is the impulse magnitude? (Take \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Initial momentum: \( p_{x1} = 0.15 \times 25 \times 0.5 = 1.875 \, \text{kg m/s} \), \( p_{y1} = 0.15 \times 25 \times 0.866 = 3.2475 \, \text{kg m/s} \).

Final momentum: \( p_{x2} = -0.15 \times 25 \times 0.866 = -3.2475 \, \text{kg m/s} \), \( p_{y2} = 0.15 \times 25 \times 0.5 = 1.875 \, \text{kg m/s} \).

Change: \( \Delta p_x = -3.2475 - 1.875 = -5.1225 \, \text{kg m/s} \), \( \Delta p_y = 1.875 - 3.2475 = -1.3725 \, \text{kg m/s} \).

Magnitude: \( |\Delta p| = \sqrt{(-5.1225)^2 + (-1.3725)^2} = \sqrt{26.24 + 1.883} \approx \sqrt{28.123} \approx 5.3 \, \text{N s} \).

4.5 N s
5.3 N s
6 N s
7 N s
2

A \( 3000 \, \text{kg} \) truck accelerates at \( 3 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then a \( 2 \, \text{kg} \) ball is dropped. What is the horizontal distance traveled by the ball in \( 2 \, \text{s} \) relative to the ground? (Neglect air resistance)

The truck’s initial velocity is zero, and it accelerates at \( 3 \, \text{m/s}^2 \) for \( 4 \, \text{s} \).

Velocity at \( t = 4 \, \text{s} \): \( v = u + at = 0 + 3 \times 4 = 12 \, \text{m/s} \).

The ball is dropped with this horizontal velocity relative to the ground.

After dropping, no horizontal forces act (air resistance neglected), so the ball’s horizontal velocity remains \( 12 \, \text{m/s} \).

Distance traveled in \( 2 \, \text{s} \): \( s = v \times t = 12 \times 2 = 24 \, \text{m} \).

Relative to the ground, this is the total horizontal displacement from the drop point.

20 m
24 m
28 m
32 m
2

A batsman hits a ball of mass \( 0.2 \, \text{kg} \) coming at \( 15 \, \text{m/s} \) straight back with the same speed. What is the impulse imparted to the ball?

Impulse equals the change in momentum of the ball.

Initial momentum \( p_i = m \times v = 0.2 \times 15 = 3 \, \text{kg m/s} \) (towards batsman).

Final momentum \( p_f = 0.2 \times (-15) = -3 \, \text{kg m/s} \) (opposite direction).

Change in momentum \( = p_f - p_i = -3 - 3 = -6 \, \text{N s} \).

The magnitude of impulse is \( 6 \, \text{N s} \).

3 N s
6 N s
9 N s
12 N s
2

A book of mass \( 2 \, \text{kg} \) is at rest on a horizontal table. What is the magnitude of the normal force exerted by the table on the book? (Take \( g = 10 \, \text{m/s}^2 \))

The book is at rest, so the net force on it must be zero by Newton’s First Law.

The weight of the book acts downward and is given by \( mg = 2 \times 10 = 20 \, \text{N} \).

The normal force \( R \) acts upward and balances the weight.

Thus, \( R = mg = 20 \, \text{N} \).

10 N
20 N
30 N
40 N
2

A \( 4 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.25 \)) is pulled by a \( 6 \, \text{kg} \) mass over a pulley. A \( 15 \, \text{N} \) force at \( 45^\circ \) upward aids the \( 4 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 45^\circ = \cos 45^\circ = 0.707 \))

For \( 6 \, \text{kg} \): \( 6g - T = 6a \Rightarrow 60 - T = 6a \).

For \( 4 \, \text{kg} \): \( T + F \cos 45^\circ - f_k = 4a \), \( N = mg - F \sin 45^\circ \).

\( N = 4 \times 10 - 15 \times 0.707 = 40 - 10.605 = 29.395 \, \text{N} \).

\( f_k = 0.25 \times 29.395 \approx 7.35 \, \text{N} \), \( F \cos 45^\circ = 15 \times 0.707 \approx 10.605 \, \text{N} \).

\( T + 10.605 - 7.35 = 4a \Rightarrow T + 3.255 = 4a \).

Solve: \( 60 - T = 6a \), \( T + 3.255 = 4a \Rightarrow 60 - (4a - 3.255) = 6a \Rightarrow 63.255 = 10a \Rightarrow a \approx 6.33 \, \text{m/s}^2 \), \( T = 60 - 6 \times 6.33 \approx 22 \, \text{N} \).

20 N
22 N
25 N
28 N
2

A \( 5 \, \text{kg} \) block on a \( 37^\circ \) incline (\( \mu_k = 0.3 \)) is pulled up by a \( 7 \, \text{kg} \) mass over a pulley with a \( 20 \, \text{N} \) force opposing the \( 5 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

For \( 5 \, \text{kg} \): \( T - mg \sin\theta - f_k - 20 = 5a \), \( f_k = 0.3 \times 5 \times 10 \times 0.8 = 12 \, \text{N} \).

\( T - 5 \times 10 \times 0.6 - 12 - 20 = 5a \Rightarrow T - 30 - 12 - 20 = 5a \Rightarrow T - 62 = 5a \).

For \( 7 \, \text{kg} \): \( 7g - T = 7a \Rightarrow 70 - T = 7a \).

Solve: \( 70 - T = 7a \), \( T - 62 = 5a \Rightarrow 70 - (5a + 62) = 7a \Rightarrow 8 = 12a \Rightarrow a = 0.67 \, \text{m/s}^2 \).

0.5 m/s²
0.67 m/s²
0.8 m/s²
1 m/s²
2

A \( 10 \, \text{kg} \) mass in a lift accelerating upward at \( 3 \, \text{m/s}^2 \) is pushed downward with a force of \( 50 \, \text{N} \) by a spring. What is the normal force on the floor? (Take \( g = 10 \, \text{m/s}^2 \))

Net force on mass: \( N - mg - F_{\text{spring}} = ma \).

\( N - 10 \times 10 - 50 = 10 \times 3 \).

\( N - 100 - 50 = 30 \).

\( N - 150 = 30 \Rightarrow N = 180 \, \text{N} \).

130 N
150 N
180 N
200 N
3

A man of mass \( 60 \, \text{kg} \) stands on a weighing scale in a lift moving upwards with an acceleration of \( 2 \, \text{m/s}^2 \). What is the reading on the scale? (Take \( g = 10 \, \text{m/s}^2 \))

The scale reads the normal force \( N \), which is the apparent weight.

Net force \( F_{\text{net}} = ma \), and \( N - mg = ma \).

So, \( N = mg + ma \).

Substitute: \( m = 60 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), \( a = 2 \, \text{m/s}^2 \).

\( N = 60 \times 10 + 60 \times 2 = 600 + 120 = 720 \, \text{N} \).

600 N
660 N
720 N
780 N
3

A \( 4 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.3 \)) is connected to a \( 6 \, \text{kg} \) mass over a pulley. A \( 12 \, \text{N} \) force opposes the \( 4 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

The \( 6 \, \text{kg} \) mass descends, pulling the \( 4 \, \text{kg} \) block horizontally.

For \( 6 \, \text{kg} \): \( 6g - T = 6a \Rightarrow 60 - T = 6a \).

For \( 4 \, \text{kg} \): \( T - f_k - 12 = 4a \).

Normal force: \( N = mg = 4 \times 10 = 40 \, \text{N} \).

Friction: \( f_k = 0.3 \times 40 = 12 \, \text{N} \).

Net force: \( T - 12 - 12 = 4a \Rightarrow T - 24 = 4a \).

Solve: \( 60 - T = 6a \), \( T - 24 = 4a \).

Substitute: \( 60 - (4a + 24) = 6a \Rightarrow 60 - 24 - 4a = 6a \Rightarrow 36 = 10a \).

\( a = \frac{36}{10} = 3.6 \, \text{m/s}^2 \).

3 m/s²
3.6 m/s²
4 m/s²
4.5 m/s²
2

A \( 7.4 \, \text{kg} \) block on a \( 53^\circ \) incline (\( \mu_k = 0.15 \)) is pulled upward by a \( 9.4 \, \text{kg} \) mass over a pulley. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

For \( 9.4 \, \text{kg} \): \( 9.4g - T = 9.4a \Rightarrow 94 - T = 9.4a \).

For \( 7.4 \, \text{kg} \): \( T - mg \sin 53^\circ - f_k = 7.4a \).

\( N = mg \cos 53^\circ = 7.4 \times 10 \times 0.6 = 44.4 \, \text{N} \).

Friction: \( f_k = 0.15 \times 44.4 = 6.66 \, \text{N} \).

\( mg \sin 53^\circ = 74 \times 0.8 = 59.2 \, \text{N} \).

Net force: \( T - 59.2 - 6.66 = 7.4a \Rightarrow T - 65.86 = 7.4a \).

Solve: \( 94 - T = 9.4a \), \( T - 65.86 = 7.4a \).

Substitute: \( 94 - (7.4a + 65.86) = 9.4a \Rightarrow 94 - 65.86 - 7.4a = 9.4a \Rightarrow 28.14 = 16.8a \).

\( a = \frac{28.14}{16.8} \approx 1.68 \, \text{m/s}^2 \).

1.4 m/s²
1.68 m/s²
1.9 m/s²
2.1 m/s²
2

A \( 0.32 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.5 \, \text{m} \) at \( 48 \, \text{rev/min} \). What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Angular speed: \( \omega = 48 \times \frac{2\pi}{60} = \frac{8\pi}{5} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{8\pi}{5}\right)^2 = \frac{64\pi^2}{25} \approx 25.27 \).

Tension: \( T = m \omega^2 r = 0.32 \times 25.27 \times 1.5 \).

\( T \approx 0.32 \times 25.27 \times 1.5 \approx 12.13 \, \text{N} \).

10 N
12.1 N
14 N
16 N
2

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