Laws of Motion Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 1300 \, \text{kg} \) car turns on a banked road (\( \theta = 15^\circ \)) with radius \( 50 \, \text{m} \) at the optimum speed to avoid friction. What is the speed? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Optimum speed: \( v_0 = \sqrt{rg \tan\theta} \) (no friction contribution).

Substitute: \( r = 50 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ = 0.268 \).

\( v_0^2 = 50 \times 10 \times 0.268 = 500 \times 0.268 = 134 \).

\( v_0 = \sqrt{134} \approx 11.58 \, \text{m/s} \).

10 m/s
11.6 m/s
12.5 m/s
13 m/s
2

A \( 5.4 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.25 \)) is connected to a \( 7.4 \, \text{kg} \) mass over a pulley. A \( 17 \, \text{N} \) force opposes the \( 5.4 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

The \( 7.4 \, \text{kg} \) mass descends, pulling the \( 5.4 \, \text{kg} \) block against friction and the opposing force.

For \( 7.4 \, \text{kg} \): \( 7.4g - T = 7.4a \Rightarrow 74 - T = 7.4a \).

For \( 5.4 \, \text{kg} \): \( T - f_k - 17 = 5.4a \).

Normal: \( N = mg = 5.4 \times 10 = 54 \, \text{N} \).

Friction: \( f_k = 0.25 \times 54 = 13.5 \, \text{N} \).

Net force: \( T - 13.5 - 17 = 5.4a \Rightarrow T - 30.5 = 5.4a \).

Solve: \( 74 - T = 7.4a \), \( T - 30.5 = 5.4a \).

Substitute: \( 74 - (5.4a + 30.5) = 7.4a \Rightarrow 74 - 30.5 - 5.4a = 7.4a \Rightarrow 43.5 = 12.8a \).

\( a = \frac{43.5}{12.8} \approx 3.4 \, \text{m/s}^2 \).

3 m/s²
3.4 m/s²
3.8 m/s²
4.2 m/s²
2

A \( 7.2 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.3 \)) is pulled by a \( 5.2 \, \text{kg} \) mass over a pulley. A \( 18 \, \text{N} \) force opposes the \( 7.2 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

For \( 5.2 \, \text{kg} \): \( 5.2g - T = 5.2a \Rightarrow 52 - T = 5.2a \).

For \( 7.2 \, \text{kg} \): \( T - f_k - 18 = 7.2a \).

Normal: \( N = mg = 7.2 \times 10 = 72 \, \text{N} \).

Friction: \( f_k = 0.3 \times 72 = 21.6 \, \text{N} \).

Net force: \( T - 21.6 - 18 = 7.2a \Rightarrow T - 39.6 = 7.2a \).

Solve: \( 52 - T = 5.2a \), \( T - 39.6 = 7.2a \).

Substitute: \( 52 - (7.2a + 39.6) = 5.2a \Rightarrow 52 - 39.6 - 7.2a = 5.2a \Rightarrow 12.4 = 12.4a \).

\( a = 1 \, \text{m/s}^2 \), \( T - 39.6 = 7.2 \times 1 \Rightarrow T - 39.6 = 7.2 \Rightarrow T = 46.8 \, \text{N} \).

42 N
46.8 N
50 N
54 N
2

A wooden block of mass \( 3 \, \text{kg} \) on a soft floor sinks with an acceleration of \( 0.2 \, \text{m/s}^2 \) when a \( 27 \, \text{kg} \) mass is placed on it. What is the normal force on the floor? (Take \( g = 10 \, \text{m/s}^2 \))

Total mass \( m = 3 + 27 = 30 \, \text{kg} \).

Net force \( F_{\text{net}} = ma = 30 \times 0.2 = 6 \, \text{N} \).

Weight \( mg = 30 \times 10 = 300 \, \text{N} \).

Normal force \( N \) satisfies: \( mg - N = ma \).

\( 300 - N = 6 \Rightarrow N = 300 - 6 = 294 \, \text{N} \).

290 N
294 N
300 N
306 N
2

A \( 50 \, \text{kg} \) man in a lift accelerating upward at \( 2 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

The scale reads the normal force, which is the apparent weight.

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

Substitute: \( N - 50 \times 10 = 50 \times 2 \).

\( N - 500 = 100 \Rightarrow N = 600 \, \text{N} \).

The reading is \( 600 \, \text{N} \).

500 N
550 N
600 N
650 N
3

A \( 60 \, \text{kg} \) man in a lift accelerating upward at \( 1.5 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

Scale reads normal force.

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

\( N - 60 \times 10 = 60 \times 1.5 \).

\( N - 600 = 90 \Rightarrow N = 690 \, \text{N} \).

650 N
690 N
720 N
750 N
2

A \( 75 \, \text{kg} \) man in a lift accelerating downward at \( 2 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

Scale reads normal force.

Net force: \( N - mg = ma \), where \( a = -2 \, \text{m/s}^2 \).

\( N - 75 \times 10 = 75 \times (-2) \).

\( N - 750 = -150 \Rightarrow N = 600 \, \text{N} \).

550 N
600 N
650 N
700 N
2

A \( 2500 \, \text{kg} \) rocket accelerates at \( 8 \, \text{m/s}^2 \) upward while ejecting gas at \( 50 \, \text{kg/s} \) with speed \( 100 \, \text{m/s} \) relative to the rocket. What is the total thrust? (Take \( g = 10 \, \text{m/s}^2 \))

Thrust from gas: \( T_{\text{gas}} = \frac{dm}{dt} \times v_{\text{rel}} = 50 \times 100 = 5000 \, \text{N} \).

Net force: \( T_{\text{total}} - mg = ma \).

\( T_{\text{total}} - 2500 \times 10 = 2500 \times 8 \Rightarrow T_{\text{total}} - 25000 = 20000 \).

\( T_{\text{total}} = 25000 + 20000 = 45000 \, \text{N} \).

40000 N
45000 N
50000 N
55000 N
2

A \( 0.3 \, \text{kg} \) ball moving at \( 20 \, \text{m/s} \) at \( 30^\circ \) to a wall rebounds at the same speed and angle. What is the impulse magnitude? (Take \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Impulse is the total change in momentum.

Initial: \( p_{x1} = 0.3 \times 20 \times 0.866 = 5.196 \, \text{kg m/s} \), \( p_{y1} = 0.3 \times 20 \times 0.5 = 3 \, \text{kg m/s} \).

Final: \( p_{x2} = -5.196 \, \text{kg m/s} \), \( p_{y2} = 3 \, \text{kg m/s} \).

\( \Delta p_x = -5.196 - 5.196 = -10.392 \, \text{kg m/s} \), \( \Delta p_y = 0 \).

Magnitude: \( |\Delta p| = \sqrt{(-10.392)^2} \approx 10.4 \, \text{N s} \).

8 N s
10.4 N s
12 N s
14 N s
2

Two billiard balls, each of mass \( 0.1 \, \text{kg} \), moving in opposite directions at \( 5 \, \text{m/s} \), collide and rebound with the same speed. What is the impulse on one ball due to the other?

Impulse = change in momentum of one ball.

For one ball: initial momentum \( p_i = 0.1 \times 5 = 0.5 \, \text{kg m/s} \).

After collision (rebounds): \( p_f = 0.1 \times (-5) = -0.5 \, \text{kg m/s} \).

Change \( \Delta p = p_f - p_i = -0.5 - 0.5 = -1 \, \text{kg m/s} \).

Magnitude of impulse = \( 1 \, \text{N s} \).

0.5 N s
1 N s
1.5 N s
2 N s
2

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