Laws of Motion Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 0.34 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.8 \, \text{m} \) at \( 42 \, \text{rev/min} \). What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Angular speed: \( \omega = 42 \times \frac{2\pi}{60} = \frac{7\pi}{5} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{7\pi}{5}\right)^2 = \frac{49\pi^2}{25} \approx 19.36 \).

Tension: \( T = m \omega^2 r = 0.34 \times 19.36 \times 1.8 \).

\( T \approx 0.34 \times 19.36 \times 1.8 \approx 11.85 \, \text{N} \).

10 N
11.9 N
13 N
15 N
2

A car of mass \( 1200 \, \text{kg} \) takes a turn of radius \( 50 \, \text{m} \) on a banked road at \( 15^\circ \). What is the optimum speed to avoid friction? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Optimum speed \( v_0 = \sqrt{rg \tan\theta} \).

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

\( v_0 = \sqrt{50 \times 10 \times 0.268} \).

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

10 m/s
11.6 m/s
13 m/s
15 m/s
2

A \( 65 \, \text{kg} \) man in a lift accelerating upward at \( 2.5 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

Scale reads normal force.

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

\( N - 65 \times 10 = 65 \times 2.5 \).

\( N - 650 = 162.5 \Rightarrow N = 812.5 \, \text{N} \).

750 N
812.5 N
850 N
900 N
2

A \( 6 \, \text{kg} \) block on a \( 53^\circ \) incline (\( \mu_k = 0.2 \)) is pulled down by a \( 4 \, \text{kg} \) mass over a pulley with a \( 15 \, \text{N} \) force opposing the \( 6 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

For \( 6 \, \text{kg} \): \( mg \sin\theta - f_k - T - 15 = 6a \), \( f_k = 0.2 \times 6 \times 10 \times 0.6 = 7.2 \, \text{N} \).

\( 6 \times 10 \times 0.8 - 7.2 - T - 15 = 6a \Rightarrow 48 - 7.2 - 15 - T = 6a \Rightarrow 25.8 - T = 6a \).

For \( 4 \, \text{kg} \): \( T - 4g = 4a \Rightarrow T - 40 = 4a \).

Solve: \( 25.8 - (4a + 40) = 6a \Rightarrow 25.8 - 40 - 4a = 6a \Rightarrow -14.2 = 10a \).

\( a = -1.42 \, \text{m/s}^2 \) (up incline), magnitude \( 1.42 \, \text{m/s}^2 \).

1 m/s²
1.42 m/s²
1.8 m/s²
2 m/s²
2

A stone of mass \( 0.2 \, \text{kg} \) is dropped from a height. What is the net force acting on it during its downward motion? (Take \( g = 10 \, \text{m/s}^2 \), ignore air resistance)

The stone is in free fall, and the only force acting on it is gravity.

The weight of the stone is \( F = mg \).

Substitute: \( m = 0.2 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \).

So, \( F = 0.2 \times 10 = 2 \, \text{N} \), directed downward.

Since air resistance is ignored, this is the net force.

1 N
2 N
3 N
4 N
2

A \( 6.2 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.2 \)) is pulled by a \( 4.2 \, \text{kg} \) mass over a pulley. A \( 13 \, \text{N} \) force aids the \( 6.2 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

For \( 4.2 \, \text{kg} \): \( 4.2g - T = 4.2a \Rightarrow 42 - T = 4.2a \).

For \( 6.2 \, \text{kg} \): \( T + 13 - f_k = 6.2a \).

Normal: \( N = mg = 6.2 \times 10 = 62 \, \text{N} \).

Friction: \( f_k = 0.2 \times 62 = 12.4 \, \text{N} \).

Net force: \( T + 13 - 12.4 = 6.2a \Rightarrow T + 0.6 = 6.2a \).

Solve: \( 42 - T = 4.2a \), \( T + 0.6 = 6.2a \).

Substitute: \( 42 - (6.2a - 0.6) = 4.2a \Rightarrow 42 + 0.6 - 6.2a = 4.2a \Rightarrow 42.6 = 10.4a \).

\( a \approx 4.1 \, \text{m/s}^2 \), \( T + 0.6 = 6.2 \times 4.1 \Rightarrow T + 0.6 \approx 25.42 \Rightarrow T \approx 24.82 \, \text{N} \).

22 N
24.8 N
27 N
30 N
2

A \( 0.3 \, \text{kg} \) ball moving at \( 40 \, \text{m/s} \) at \( 60^\circ \) to a wall rebounds at \( 53^\circ \) with speed \( 32 \, \text{m/s} \). What is the impulse magnitude? (Take \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

Impulse is the change in momentum, a vector quantity, so we compute components along and perpendicular to the wall.

Initial momentum: \( p_{x1} = m v \cos 60^\circ = 0.3 \times 40 \times 0.5 = 6 \, \text{kg m/s} \) (towards wall).

\( p_{y1} = m v \sin 60^\circ = 0.3 \times 40 \times 0.866 = 10.392 \, \text{kg m/s} \) (along wall).

Final momentum: \( p_{x2} = -m v \cos 53^\circ = -0.3 \times 32 \times 0.6 = -5.76 \, \text{kg m/s} \) (away from wall).

\( p_{y2} = m v \sin 53^\circ = 0.3 \times 32 \times 0.8 = 7.68 \, \text{kg m/s} \) (along wall).

Change: \( \Delta p_x = -5.76 - 6 = -11.76 \, \text{kg m/s} \), \( \Delta p_y = 7.68 - 10.392 = -2.712 \, \text{kg m/s} \).

Magnitude: \( |\Delta p| = \sqrt{(-11.76)^2 + (-2.712)^2} = \sqrt{138.2976 + 7.354944} \).

\( |\Delta p| = \sqrt{145.652544} \approx 12.07 \, \text{N s} \).

10 N s
12.1 N s
14 N s
16 N s
2

A \( 50 \, \text{kg} \) man in a lift descending at \( 5 \, \text{m/s}^2 \) pushes a \( 10 \, \text{kg} \) box downward with \( 20 \, \text{N} \). What is the normal force on the floor? (Take \( g = 10 \, \text{m/s}^2 \))

For box: \( N_b - mg - F = ma \Rightarrow N_b - 10 \times 10 - 20 = 10 \times (-5) \).

\( N_b - 100 - 20 = -50 \Rightarrow N_b - 120 = -50 \Rightarrow N_b = 70 \, \text{N} \).

For man: \( N_m - mg - N_b = ma \Rightarrow N_m - 50 \times 10 - 70 = 50 \times (-5) \).

\( N_m - 500 - 70 = -250 \Rightarrow N_m - 570 = -250 \Rightarrow N_m = 320 \, \text{N} \).

Total \( N = N_m + N_b = 320 + 70 = 390 \, \text{N} \).

350 N
390 N
420 N
450 N
2

A \( 0.25 \, \text{kg} \) ball at \( 36 \, \text{m/s} \) hits a wall at \( 60^\circ \) and rebounds at \( 45^\circ \) with speed \( 30 \, \text{m/s} \). What is the impulse perpendicular to the wall? (Take \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \), \( \sin 45^\circ = \cos 45^\circ = 0.707 \))

Perpendicular (x-axis): Initial \( p_x = 0.25 \times 36 \times 0.5 = 4.5 \, \text{kg m/s} \).

Final \( p_x = -0.25 \times 30 \times 0.707 = -5.3025 \, \text{kg m/s} \).

Impulse \( \Delta p_x = -5.3025 - 4.5 = -9.8025 \, \text{N s} \).

Magnitude = \( 9.8 \, \text{N s} \).

8 N s
9.8 N s
11 N s
12 N s
2

A \( 70 \, \text{kg} \) man in a lift accelerating downward at \( 3 \, \text{m/s}^2 \) pushes a \( 15 \, \text{kg} \) box upward with \( 50 \, \text{N} \). What is the normal force on the floor? (Take \( g = 10 \, \text{m/s}^2 \))

For box: \( F - mg - N_b = ma \Rightarrow 50 - 15 \times 10 - N_b = 15 \times (-3) \).

\( 50 - 150 - N_b = -45 \Rightarrow -100 - N_b = -45 \Rightarrow N_b = -55 \) (impossible, adjust: \( N_b \) on man).

Correct: Box’s \( N_b \) on man, man’s \( N_m \) on floor: \( N_b - 150 = -45 \Rightarrow N_b = 105 \, \text{N} \) (down on man).

For man: \( N_m - mg - N_b = ma \Rightarrow N_m - 70 \times 10 - 105 = 70 \times (-3) \).

\( N_m - 700 - 105 = -210 \Rightarrow N_m - 805 = -210 \Rightarrow N_m = 595 \, \text{N} \).

Total on floor: \( N = N_m = 595 \, \text{N} \) (box’s normal is on man, not floor directly).

550 N
595 N
620 N
650 N
2

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