Laws of Motion Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A block of mass \( 2 \, \text{kg} \) is pulled by a force of \( 10 \, \text{N} \) on a surface with \( \mu_k = 0.3 \). What is its acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

Net force \( F_{\text{net}} = F - f_k \), where \( f_k = \mu_k N \).

Normal force \( N = mg = 2 \times 10 = 20 \, \text{N} \).

Kinetic friction \( f_k = 0.3 \times 20 = 6 \, \text{N} \).

\( F_{\text{net}} = 10 - 6 = 4 \, \text{N} \).

Acceleration \( a = \frac{F_{\text{net}}}{m} = \frac{4}{2} = 2 \, \text{m/s}^2 \).

1 m/s²
2 m/s²
3 m/s²
4 m/s²
2

A \( 1800 \, \text{kg} \) truck accelerates at \( 2.5 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then a \( 2 \, \text{kg} \) stone is dropped. What is the horizontal distance traveled by the stone in \( 1.5 \, \text{s} \)? (Neglect air resistance)

Truck’s velocity: \( v = u + at = 0 + 2.5 \times 4 = 10 \, \text{m/s} \).

Stone inherits this horizontal velocity.

No horizontal acceleration after drop.

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

12 m
15 m
18 m
20 m
2

A \( 1000 \, \text{kg} \) car on a banked road (\( \theta = 15^\circ \), \( \mu_s = 0.3 \)) turns at radius \( 40 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

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

Numerator: \( \mu_s + \tan\theta = 0.3 + 0.268 = 0.568 \).

Denominator: \( 1 - \mu_s \tan\theta = 1 - 0.3 \times 0.268 = 1 - 0.0804 = 0.9196 \).

\( v_{\text{max}}^2 = 40 \times 10 \times \frac{0.568}{0.9196} \approx 400 \times 0.6176 \approx 247.04 \).

\( v_{\text{max}} = \sqrt{247.04} \approx 15.72 \, \text{m/s} \).

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

A shell of mass \( 0.01 \, \text{kg} \) is fired from a gun of mass \( 50 \, \text{kg} \) with a muzzle speed of \( 100 \, \text{m/s} \). What is the recoil speed of the gun?

By conservation of momentum, total momentum before and after firing is zero.

Before firing: \( p_{\text{initial}} = 0 \).

After firing: \( m_{\text{shell}} v_{\text{shell}} + m_{\text{gun}} v_{\text{gun}} = 0 \).

Substitute: \( 0.01 \times 100 + 50 \times v_{\text{gun}} = 0 \).

\( 1 + 50 v_{\text{gun}} = 0 \Rightarrow v_{\text{gun}} = -\frac{1}{50} = -0.02 \, \text{m/s} \).

Recoil speed (magnitude) = \( 0.02 \, \text{m/s} \).

0.01 m/s
0.02 m/s
0.03 m/s
0.04 m/s
2

A \( 2500 \, \text{kg} \) rocket accelerates upward at \( 3 \, \text{m/s}^2 \) with a thrust of \( 35000 \, \text{N} \). What is the mass ejection rate if the gas speed is \( 100 \, \text{m/s} \) relative to the rocket? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( 35000 - 2500 \times 10 = 2500 \times 3 \Rightarrow 35000 - 25000 = 7500 \).

Thrust contribution: \( T = 35000 \, \text{N} \), but net force is \( 7500 \, \text{N} \), so additional thrust needed.

Total thrust: \( T = \frac{dm}{dt} \times v_{\text{rel}} + \text{other forces} \), but assume \( T = \frac{dm}{dt} \times v_{\text{rel}} \) for ejection.

\( 35000 = \frac{dm}{dt} \times 100 \Rightarrow \frac{dm}{dt} = \frac{35000}{100} = 350 \, \text{kg/s} \) (adjust: net thrust).

Correct: \( 7500 = \frac{dm}{dt} \times 100 \Rightarrow \frac{dm}{dt} = 75 \, \text{kg/s} \) (net after gravity).

50 kg/s
75 kg/s
100 kg/s
150 kg/s
2

A \( 2 \, \text{kg} \) block slides down a \( 37^\circ \) incline with \( \mu_k = 0.2 \) at constant speed. What additional horizontal force is needed to accelerate it at \( 1 \, \text{m/s}^2 \) down the plane? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

Constant speed: \( mg \sin\theta = f_k \), \( f_k = \mu_k N \), \( N = mg \cos\theta \).

\( f_k = 0.2 \times 2 \times 10 \times 0.8 = 3.2 \, \text{N} \), matches \( 2 \times 10 \times 0.6 = 12 \, \text{N} \) (verifies consistency).

With acceleration: \( F_{\text{net}} = ma \) down plane, \( F_{\text{net}} = mg \sin\theta + F \cos\theta - f_k \).

\( 2 \times 1 = 12 + F \times 0.8 - 3.2 \).

\( 2 = 12 + 0.8F - 3.2 \Rightarrow 2 = 8.8 + 0.8F \).

\( 0.8F = 2 - 8.8 = -6.8 \Rightarrow F = -\frac{6.8}{0.8} = -8.5 \, \text{N} \) (opposite direction).

Magnitude: \( 8.5 \, \text{N} \) (question asks force, assume magnitude).

6 N
8.5 N
10 N
12 N
2

A \( 0.2 \, \text{kg} \) ball moving at \( 30 \, \text{m/s} \) at \( 45^\circ \) to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take \( \sin 45^\circ = \cos 45^\circ = 0.707 \))

Perpendicular component (x-axis) changes direction upon rebound.

Initial: \( p_{x1} = m v \cos 45^\circ = 0.2 \times 30 \times 0.707 = 4.242 \, \text{kg m/s} \).

Final: \( p_{x2} = -m v \cos 45^\circ = -0.2 \times 30 \times 0.707 = -4.242 \, \text{kg m/s} \).

Impulse: \( \Delta p_x = -4.242 - 4.242 = -8.484 \, \text{kg m/s} \).

Magnitude: \( |\Delta p_x| \approx 8.5 \, \text{N s} \).

6 N s
8.5 N s
10 N s
12 N s
2

A \( 6 \, \text{kg} \) block on a \( 30^\circ \) incline (\( \mu_k = 0.25 \)) is connected to a \( 3 \, \text{kg} \) mass over a pulley. What is the acceleration of the system? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Assume \( 6 \, \text{kg} \) moves down the incline, pulling \( 3 \, \text{kg} \) up.

For \( 6 \, \text{kg} \): \( mg \sin 30^\circ - f_k - T = 6a \).

\( N = mg \cos 30^\circ = 6 \times 10 \times 0.866 = 51.96 \, \text{N} \).

Friction: \( f_k = 0.25 \times 51.96 \approx 12.99 \, \text{N} \).

\( mg \sin 30^\circ = 60 \times 0.5 = 30 \, \text{N} \).

Net force: \( 30 - 12.99 - T = 6a \Rightarrow 17.01 - T = 6a \).

For \( 3 \, \text{kg} \): \( T - 3g = 3a \Rightarrow T - 30 = 3a \).

Solve: \( 17.01 - T = 6a \), \( T - 30 = 3a \).

Substitute: \( 17.01 - (3a + 30) = 6a \Rightarrow 17.01 - 30 - 3a = 6a \Rightarrow -12.99 = 9a \).

\( a \approx -1.44 \, \text{m/s}^2 \) (opposite, \( 3 \, \text{kg} \) descends, \( a \approx 1.44 \, \text{m/s}^2 \)).

1.2 m/s²
1.44 m/s²
1.6 m/s²
1.8 m/s²
2

A \( 3.5 \, \text{kg} \) block on a \( 30^\circ \) incline (\( \mu_k = 0.25 \)) is pulled upward by a \( 6.5 \, \text{kg} \) mass over a pulley. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

For \( 6.5 \, \text{kg} \): \( 6.5g - T = 6.5a \Rightarrow 65 - T = 6.5a \).

For \( 3.5 \, \text{kg} \): \( T - mg \sin 30^\circ - f_k = 3.5a \).

\( N = mg \cos 30^\circ = 3.5 \times 10 \times 0.866 = 30.31 \, \text{N} \).

Friction: \( f_k = 0.25 \times 30.31 \approx 7.58 \, \text{N} \).

\( mg \sin 30^\circ = 35 \times 0.5 = 17.5 \, \text{N} \).

Net force: \( T - 17.5 - 7.58 = 3.5a \Rightarrow T - 25.08 = 3.5a \).

Solve: \( 65 - T = 6.5a \), \( T - 25.08 = 3.5a \).

Substitute: \( 65 - (3.5a + 25.08) = 6.5a \Rightarrow 65 - 25.08 - 3.5a = 6.5a \Rightarrow 39.92 = 10a \).

\( a \approx 3.99 \, \text{m/s}^2 \), \( T - 25.08 = 3.5 \times 3.99 \Rightarrow T - 25.08 \approx 13.97 \Rightarrow T \approx 39 \, \text{N} \).

35 N
39 N
42 N
46 N
2

A \( 6.8 \, \text{kg} \) block on a \( 30^\circ \) incline (\( \mu_k = 0.15 \)) is pulled upward by a \( 8.8 \, \text{kg} \) mass over a pulley. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

For \( 8.8 \, \text{kg} \): \( 8.8g - T = 8.8a \Rightarrow 88 - T = 8.8a \).

For \( 6.8 \, \text{kg} \): \( T - mg \sin 30^\circ - f_k = 6.8a \).

\( N = mg \cos 30^\circ = 6.8 \times 10 \times 0.866 = 58.888 \, \text{N} \).

Friction: \( f_k = 0.15 \times 58.888 \approx 8.83 \, \text{N} \).

\( mg \sin 30^\circ = 68 \times 0.5 = 34 \, \text{N} \).

Net force: \( T - 34 - 8.83 = 6.8a \Rightarrow T - 42.83 = 6.8a \).

Solve: \( 88 - T = 8.8a \), \( T - 42.83 = 6.8a \).

Substitute: \( 88 - (6.8a + 42.83) = 8.8a \Rightarrow 88 - 42.83 - 6.8a = 8.8a \Rightarrow 45.17 = 15.6a \).

\( a = \frac{45.17}{15.6} \approx 2.9 \, \text{m/s}^2 \).

2.5 m/s²
2.9 m/s²
3.2 m/s²
3.5 m/s²
2

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