Laws of Motion Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 0.29 \, \text{kg} \) ball moving at \( 18 \, \text{m/s} \) strikes a wall perpendicularly and rebounds at \( 14 \, \text{m/s} \). What is the impulse? (Take \( g = 10 \, \text{m/s}^2 \))

Impulse is the change in momentum, with direction reversing upon rebound.

Initial momentum: \( p_i = m v = 0.29 \times 18 = 5.22 \, \text{kg m/s} \) (towards wall).

Final momentum: \( p_f = -m v = -0.29 \times 14 = -4.06 \, \text{kg m/s} \) (away from wall).

Impulse: \( \Delta p = p_f - p_i = -4.06 - 5.22 = -9.28 \, \text{N s} \).

Magnitude: \( |\Delta p| = 9.28 \, \text{N s} \approx 9.3 \, \text{N s} \).

8 N s
9.3 N s
10.5 N s
12 N s
2

A \( 1150 \, \text{kg} \) car turns on a banked road (\( \theta = 30^\circ \), \( \mu_s = 0.25 \)) with radius \( 35 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 30^\circ = 0.577 \))

Maximum speed: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \).

Numerator: \( \mu_s + \tan\theta = 0.25 + 0.577 = 0.827 \).

Denominator: \( 1 - 0.25 \times 0.577 = 1 - 0.14425 = 0.85575 \).

\( v_{\text{max}}^2 = 35 \times 10 \times \frac{0.827}{0.85575} \approx 350 \times 0.966 \approx 338.1 \).

\( v_{\text{max}} = \sqrt{338.1} \approx 18.39 \, \text{m/s} \).

16 m/s
18.4 m/s
20 m/s
22 m/s
2

A car of mass \( 1000 \, \text{kg} \) moves with a constant velocity of \( 20 \, \text{m/s} \) on a rough road. What is the net force acting on the car?

The car moves with constant velocity, meaning its acceleration is zero.

By Newton’s First Law, if \( a = 0 \), the net force \( F = ma = 0 \).

The driving force and frictional force cancel each other out.

Thus, the net force acting on the car is zero.

0 N
500 N
1000 N
2000 N
1

A \( 0.2 tare \text{kg} \) stone is whirled in a circle of radius \( 1.5 \, \text{m} \) with a string that breaks at \( 100 \, \text{N} \). What is the maximum angular speed? (Take \( g = 10 \, \text{m/s}^2 \))

Centripetal force \( F = m \omega^2 r \), max \( F = 100 \, \text{N} \).

Substitute: \( 100 = 0.2 \times \omega^2 \times 1.5 \).

\( 100 = 0.3 \omega^2 \).

\( \omega^2 = \frac{100}{0.3} = \frac{1000}{3} \approx 333.33 \).

\( \omega = \sqrt{333.33} \approx 18.26 \, \text{rad/s} \).

15 rad/s
18.3 rad/s
20 rad/s
22 rad/s
2

A \( 3.2 \, \text{kg} \) block on a \( 45^\circ \) incline (\( \mu_s = 0.45 \)) is pulled upward by a horizontal force just sufficient to start motion. What is the force? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 45^\circ = 0.707 \), \( \cos 45^\circ = 0.707 \))

The block is on the verge of moving upward, so friction acts downward along the incline.

Along incline: \( F \cos 45^\circ - mg \sin 45^\circ - f_s = 0 \) (horizontal force component opposes gravity and friction).

Normal force: \( N = mg \cos 45^\circ + F \sin 45^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 3.2 \times 10 = 32 \, \text{N} \), \( mg \sin 45^\circ = 32 \times 0.707 = 22.624 \, \text{N} \).

\( mg \cos 45^\circ = 32 \times 0.707 = 22.624 \, \text{N} \), so \( N = 22.624 + F \times 0.707 \).

Friction: \( f_s = \mu_s N = 0.45 (22.624 + 0.707F) \).

Substitute: \( F \times 0.707 - 22.624 - 0.45 (22.624 + 0.707F) = 0 \).

Expand: \( 0.707F - 22.624 - 10.1808 - 0.31815F = 0 \Rightarrow 0.38885F - 32.8048 = 0 \).

Solve: \( 0.38885F = 32.8048 \Rightarrow F \approx \frac{32.8048}{0.38885} \approx 84.34 \, \text{N} \).

80 N
84.3 N
88 N
92 N
2

A \( 900 \, \text{kg} \) car turns on a banked road (\( \theta = 30^\circ \), \( \mu_s = 0.3 \)) with radius \( 45 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 30^\circ = 0.577 \))

Maximum speed: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \).

\( \mu_s + \tan\theta = 0.3 + 0.577 = 0.877 \).

\( 1 - 0.3 \times 0.577 = 1 - 0.1731 = 0.8269 \).

\( v_{\text{max}}^2 = 45 \times 10 \times \frac{0.877}{0.8269} \approx 450 \times 1.0606 \approx 477.27 \).

\( v_{\text{max}} \approx \sqrt{477.27} \approx 21.85 \, \text{m/s} \).

20 m/s
21.9 m/s
23 m/s
25 m/s
2

A \( 3 \, \text{kg} \) block on a \( 60^\circ \) incline (\( \mu_s = 0.6 \)) is pulled upward by a force at \( 30^\circ \) to the incline with minimum force to start motion. What is the force? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Along incline: \( F \cos 30^\circ - mg \sin 60^\circ - f_s = 0 \).

Normal: \( N = mg \cos 60^\circ + F \sin 30^\circ \).

\( mg \sin 60^\circ = 3 \times 10 \times 0.866 = 25.98 \, \text{N} \), \( mg \cos 60^\circ = 15 \, \text{N} \).

\( N = 15 + F \times 0.5 \), \( f_s = 0.6 (15 + 0.5F) \).

\( F \times 0.866 - 25.98 - 0.6 (15 + 0.5F) = 0 \Rightarrow 0.866F - 25.98 - 9 - 0.3F = 0 \).

\( 0.566F = 34.98 \Rightarrow F \approx 61.8 \, \text{N} \).

50 N
61.8 N
70 N
80 N
2

A \( 0.46 \, \text{kg} \) stone in a vertical circle of radius \( 1.7 \, \text{m} \) has a speed of \( 10 \, \text{m/s} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.46 \times 10 = 0.46 \times (10)^2 / 1.7 \).

\( T_b - 4.6 = 0.46 \times 58.82 \Rightarrow T_b - 4.6 \approx 27.06 \Rightarrow T_b \approx 31.66 \, \text{N} \).

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

\( 0.5 \times 0.46 \times 100 = 0.5 \times 0.46 \times v_t^2 + 2 \times 0.46 \times 10 \).

\( 23 = 0.23 v_t^2 + 9.2 \Rightarrow 0.23 v_t^2 = 13.8 \Rightarrow v_t^2 \approx 60 \Rightarrow v_t \approx 7.75 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 4.6 = 0.46 \times 60 / 1.7 \Rightarrow T_t + 4.6 \approx 16.24 \Rightarrow T_t \approx 11.64 \, \text{N} \).

9 N
11.6 N
13 N
15 N
2

A \( 0.44 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.6 \, \text{m} \) with a tension of \( 15 \, \text{N} \). What is the speed? (Take \( g = 10 \, \text{m/s}^2 \))

Tension provides centripetal force: \( T = m v^2 / r \).

Substitute: \( 15 = 0.44 \times v^2 / 1.6 \).

\( 15 = 0.275 v^2 \Rightarrow v^2 = \frac{15}{0.275} \approx 54.55 \).

\( v = \sqrt{54.55} \approx 7.39 \, \text{m/s} \).

6.5 m/s
7.4 m/s
8 m/s
8.5 m/s
2

A \( 1000 \, \text{kg} \) car turns on a banked road (\( \theta = 15^\circ \), \( \mu_s = 0.2 \)) with radius \( 50 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Maximum speed: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \).

Numerator: \( \mu_s + \tan\theta = 0.2 + 0.268 = 0.468 \).

Denominator: \( 1 - 0.2 \times 0.268 = 1 - 0.0536 = 0.9464 \).

\( v_{\text{max}}^2 = 50 \times 10 \times \frac{0.468}{0.9464} \approx 500 \times 0.4947 \approx 247.35 \).

\( v_{\text{max}} = \sqrt{247.35} \approx 15.73 \, \text{m/s} \).

14 m/s
15.7 m/s
17 m/s
18 m/s
2

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