Laws of Motion Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A body of mass \( 10 \, \text{kg} \) is on a horizontal surface with a coefficient of static friction \( 0.2 \). What is the maximum static frictional force? (Take \( g = 10 \, \text{m/s}^2 \))

Maximum static friction is given by \( f_s = \mu_s N \).

The normal force \( N \) equals the weight: \( N = mg = 10 \times 10 = 100 \, \text{N} \).

Coefficient of static friction \( \mu_s = 0.2 \).

So, \( f_s = 0.2 \times 100 = 20 \, \text{N} \).

10 N
20 N
30 N
40 N
2

A \( 0.35 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.2 \, \text{m} \) at \( 50 \, \text{rev/min} \). What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Angular speed: \( \omega = 50 \times \frac{2\pi}{60} = \frac{5\pi}{3} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{5\pi}{3}\right)^2 = \frac{25\pi^2}{9} \approx 27.42 \).

Tension: \( T = m \omega^2 r = 0.35 \times 27.42 \times 1.2 \).

\( T \approx 0.35 \times 27.42 \times 1.2 \approx 11.52 \, \text{N} \).

10 N
11.5 N
13 N
15 N
2

A \( 0.25 \, \text{kg} \) ball moving at \( 18 \, \text{m/s} \) at \( 53^\circ \) to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

Perpendicular component (x-axis) reverses upon rebound.

Initial: \( p_{x1} = m v \cos 53^\circ = 0.25 \times 18 \times 0.6 = 2.7 \, \text{kg m/s} \).

Final: \( p_{x2} = -m v \cos 53^\circ = -0.25 \times 18 \times 0.6 = -2.7 \, \text{kg m/s} \).

Impulse: \( \Delta p_x = -2.7 - 2.7 = -5.4 \, \text{N s} \).

Magnitude: \( |\Delta p_x| = 5.4 \, \text{N s} \).

4 N s
5.4 N s
6.5 N s
8 N s
2

A \( 1 \, \text{kg} \) stone in a vertical circle of radius \( 4 \, \text{m} \) has a speed of \( 20 \, \text{m/s} \) at the bottom. A tangential force of \( 4 \, \text{N} \) acts downward from bottom to top. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 1 \times 10 = 1 \times (20)^2 / 4 \).

\( T_b - 10 = 100 \Rightarrow T_b = 110 \, \text{N} \).

Work by tangential force increases KE: Arc length \( \pi r = 4\pi \approx 12.56 \, \text{m} \), \( W = 4 \times 12.56 = 50.24 \, \text{J} \).

Initial KE: \( \frac{1}{2} m v_b^2 = 0.5 \times 1 \times 400 = 200 \, \text{J} \).

PE gain: \( 2mg = 2 \times 1 \times 10 \times 4 = 40 \, \text{J} \).

Final KE: \( 200 + 50.24 - 40 = 210.24 \, \text{J} \).

\( \frac{1}{2} m v_t^2 = 210.24 \Rightarrow 0.5 v_t^2 = 210.24 \Rightarrow v_t^2 = 420.48 \Rightarrow v_t \approx 20.5 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 10 = 1 \times 420.48 / 4 \Rightarrow T_t + 10 = 105.12 \).

\( T_t = 105.12 - 10 = 95.12 \, \text{N} \approx 95.1 \, \text{N} \).

90 N
95.1 N
100 N
105 N
2

A \( 2000 \, \text{kg} \) car on a banked road (\( \theta = 25^\circ \), \( \mu_s = 0.4 \)) turns at radius \( 80 \, \text{m} \) while accelerating tangentially at \( 1.5 \, \text{m/s}^2 \). What is the maximum speed before slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 25^\circ \approx 0.466 \))

The maximum speed depends on radial forces (centripetal), not tangential acceleration, which affects speed change over time.

Formula for max speed on a banked road with friction: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \).

Calculate the numerator: \( \mu_s + \tan\theta = 0.4 + 0.466 = 0.866 \).

Calculate the denominator: \( 1 - \mu_s \tan\theta = 1 - 0.4 \times 0.466 = 1 - 0.1864 = 0.8136 \).

Substitute: \( v_{\text{max}}^2 = 80 \times 10 \times \frac{0.866}{0.8136} \).

\( \frac{0.866}{0.8136} \approx 1.0645 \), so \( v_{\text{max}}^2 = 800 \times 1.0645 \approx 851.6 \).

\( v_{\text{max}} = \sqrt{851.6} \approx 29.18 \, \text{m/s} \).

The tangential acceleration doesn’t alter the radial friction limit, so this is the maximum speed before slipping.

27 m/s
29.2 m/s
31 m/s
33 m/s
2

A \( 0.4 \, \text{kg} \) stone in a vertical circle of radius \( 2 \, \text{m} \) has a tension of \( 20 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow 20 - 0.4 \times 10 = 0.4 \times v_b^2 / 2 \).

\( 20 - 4 = 0.2 v_b^2 \Rightarrow 16 = 0.2 v_b^2 \Rightarrow v_b^2 = 80 \Rightarrow v_b \approx 8.94 \, \text{m/s} \).

Energy conservation: \( \frac{1}{2} m v_b^2 = \frac{1}{2} m v_t^2 + 2mg \) (no tangential force).

\( 0.5 \times 0.4 \times 80 = 0.5 \times 0.4 \times v_t^2 + 2 \times 0.4 \times 10 \).

\( 16 = 0.2 v_t^2 + 8 \Rightarrow 0.2 v_t^2 = 8 \Rightarrow v_t^2 = 40 \Rightarrow v_t \approx 6.32 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 4 = 0.4 \times 40 / 2 \Rightarrow T_t + 4 = 8 \Rightarrow T_t = 4 \, \text{N} \).

2 N
4 N
6 N
8 N
2

A \( 1600 \, \text{kg} \) truck accelerates at \( 3 \, \text{m/s}^2 \) for \( 3 \, \text{s} \), then a \( 1.5 \, \text{kg} \) stone is dropped. What is the horizontal distance traveled by the stone in \( 2 \, \text{s} \)? (Neglect air resistance)

Truck’s velocity after \( 3 \, \text{s} \): \( v = u + at = 0 + 3 \times 3 = 9 \, \text{m/s} \).

Stone inherits this horizontal velocity upon drop.

No horizontal forces act after drop (air resistance neglected).

Distance: \( s = v \times t = 9 \times 2 = 18 \, \text{m} \).

15 m
18 m
20 m
22 m
2

A \( 10 \, \text{kg} \) mass is suspended by a rope with a \( 40 \, \text{N} \) horizontal force at its midpoint. What is the angle the rope makes with the vertical? (Take \( g = 10 \, \text{m/s}^2 \))

At midpoint: horizontal \( T \sin\theta = 40 \, \text{N} \), vertical \( T \cos\theta = mg = 10 \times 10 = 100 \, \text{N} \).

Divide: \( \tan\theta = \frac{40}{100} = 0.4 \).

\( \theta = \tan^{-1}(0.4) \).

Using \( \tan 22^\circ \approx 0.404 \), \( \theta \approx 22^\circ \).

15°
22°
30°
45°
2

A \( 8.4 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.25 \)) is pulled by a \( 6.4 \, \text{kg} \) mass over a pulley. A \( 19 \, \text{N} \) force opposes the \( 8.4 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

For \( 6.4 \, \text{kg} \): \( 6.4g - T = 6.4a \Rightarrow 64 - T = 6.4a \).

For \( 8.4 \, \text{kg} \): \( T - f_k - 19 = 8.4a \).

Normal: \( N = mg = 8.4 \times 10 = 84 \, \text{N} \).

Friction: \( f_k = 0.25 \times 84 = 21 \, \text{N} \).

Net force: \( T - 21 - 19 = 8.4a \Rightarrow T - 40 = 8.4a \).

Solve: \( 64 - T = 6.4a \), \( T - 40 = 8.4a \).

Substitute: \( 64 - (8.4a + 40) = 6.4a \Rightarrow 64 - 40 - 8.4a = 6.4a \Rightarrow 24 = 14.8a \).

\( a \approx 1.62 \, \text{m/s}^2 \), \( T - 40 = 8.4 \times 1.62 \Rightarrow T - 40 \approx 13.61 \Rightarrow T \approx 53.61 \, \text{N} \).

50 N
53.6 N
57 N
60 N
2

A \( 0.5 \, \text{kg} \) stone tied to a string of length \( 1 \, \text{m} \) is whirled at \( 60 \, \text{rev/min} \). What is the maximum tension the string can withstand if it breaks at \( 50 \, \text{N} \)?

Centripetal force \( F = m \omega^2 r \).

\( \omega = 60 \, \text{rev/min} = \frac{60 \times 2\pi}{60} = 2\pi \, \text{rad/s} \).

\( \omega^2 = (2\pi)^2 = 4\pi^2 \).

\( F = 0.5 \times 4\pi^2 \times 1 \approx 19.74 \, \text{N} \).

Since \( 19.74 \, \text{N} < 50 \, \text{N} \), maximum tension possible here is \( 50 \, \text{N} \) (breaking limit).

20 N
30 N
40 N
50 N
4

A \( 0.38 \, \text{kg} \) stone in a vertical circle of radius \( 1.9 \, \text{m} \) has a tension of \( 28 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow 28 - 0.38 \times 10 = 0.38 \times v_b^2 / 1.9 \).

\( 28 - 3.8 = 0.2 v_b^2 \Rightarrow 24.2 = 0.2 v_b^2 \Rightarrow v_b^2 = 121 \Rightarrow v_b = 11 \, \text{m/s} \).

Energy: \( \frac{1}{2} m v_b^2 = \frac{1}{2} m v_t^2 + 2mg \).

\( 0.5 \times 0.38 \times 121 = 0.5 \times 0.38 \times v_t^2 + 2 \times 0.38 \times 10 \).

\( 22.99 = 0.19 v_t^2 + 7.6 \Rightarrow 0.19 v_t^2 = 15.39 \Rightarrow v_t^2 \approx 81 \Rightarrow v_t \approx 9 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 3.8 = 0.38 \times 81 / 1.9 \Rightarrow T_t + 3.8 \approx 16.2 \Rightarrow T_t \approx 12.4 \, \text{N} \).

10 N
12.4 N
14 N
16 N
2

A \( 85 \, \text{kg} \) man in a lift accelerating upward at \( 1.8 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

Scale reads normal force.

Net force: \( N - mg = ma \).

\( N - 85 \times 10 = 85 \times 1.8 \).

\( N - 850 = 153 \Rightarrow N = 1003 \, \text{N} \).

950 N
1003 N
1050 N
1100 N
2

A \( 60 \, \text{kg} \) man stands in a lift that accelerates downward at \( 3 \, \text{m/s}^2 \) and then jumps with an upward acceleration of \( 7 \, \text{m/s}^2 \) relative to the lift. What is the normal force during the jump? (Take \( g = 10 \, \text{m/s}^2 \))

Relative to ground: \( a_{\text{net}} = a_{\text{jump}} + a_{\text{lift}} = 7 + (-3) = 4 \, \text{m/s}^2 \) upward.

\( N - mg = ma_{\text{net}} \).

\( N - 60 \times 10 = 60 \times 4 \).

\( N - 600 = 240 \Rightarrow N = 840 \, \text{N} \).

600 N
720 N
840 N
960 N
3

A \( 0.2 \, \text{kg} \) ball moving at \( 25 \, \text{m/s} \) at \( 53^\circ \) to a wall rebounds elastically at \( 37^\circ \) with speed \( 20 \, \text{m/s} \). What is the impulse magnitude? (Take \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

Initial: \( p_{x1} = 0.2 \times 25 \times 0.6 = 3 \, \text{kg m/s} \), \( p_{y1} = 0.2 \times 25 \times 0.8 = 4 \, \text{kg m/s} \).

Final: \( p_{x2} = -0.2 \times 20 \times 0.8 = -3.2 \, \text{kg m/s} \), \( p_{y2} = 0.2 \times 20 \times 0.6 = 2.4 \, \text{kg m/s} \).

\( \Delta p_x = -3.2 - 3 = -6.2 \, \text{kg m/s} \), \( \Delta p_y = 2.4 - 4 = -1.6 \, \text{kg m/s} \).

Impulse \( |\Delta p| = \sqrt{(-6.2)^2 + (-1.6)^2} = \sqrt{38.44 + 2.56} = \sqrt{41} \approx 6.4 \, \text{N s} \).

5 N s
6.4 N s
7 N s
8 N s
2

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