Laws of Motion Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle of mass \( 1 \, \text{kg} \) moves with a velocity described by \( y = 2t + 5t^2 \). What is the force acting on it at \( t = 2 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Position \( y = 2t + 5t^2 \).

Velocity \( v = \frac{dy}{dt} = 2 + 10t \).

Acceleration \( a = \frac{dv}{dt} = 10 \, \text{m/s}^2 \).

Force \( F = ma = 1 \times 10 = 10 \, \text{N} \) (constant, matches gravity example in PDF).

At \( t = 2 \, \text{s} \), \( F = 10 \, \text{N} \).

5 N
10 N
15 N
20 N
2

A \( 70 \, \text{kg} \) man in a lift accelerating downward at \( 1.2 \, \text{m/s}^2 \) 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 \( a = -1.2 \, \text{m/s}^2 \).

\( N - 70 \times 10 = 70 \times (-1.2) \).

\( N - 700 = -84 \Rightarrow N = 616 \, \text{N} \).

580 N
616 N
650 N
680 N
2

A \( 0.2 \, \text{kg} \) ball moving at \( 15 \, \text{m/s} \) strikes a wall perpendicularly and rebounds at \( 10 \, \text{m/s} \). What is the impulse? (Take \( g = 10 \, \text{m/s}^2 \))

Impulse is the change in momentum, direction reverses.

Initial momentum: \( p_i = m v = 0.2 \times 15 = 3 \, \text{kg m/s} \) (towards wall).

Final momentum: \( p_f = -m v = -0.2 \times 10 = -2 \, \text{kg m/s} \) (away from wall).

Impulse: \( \Delta p = p_f - p_i = -2 - 3 = -5 \, \text{N s} \).

Magnitude: \( |\Delta p| = 5 \, \text{N s} \).

3 N s
5 N s
7 N s
9 N s
2

A \( 8 \, \text{kg} \) mass and a \( 12 \, \text{kg} \) mass are connected over a pulley. A \( 20 \, \text{N} \) force pulls the \( 8 \, \text{kg} \) mass upward. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

For \( 12 \, \text{kg} \): \( 12g - T = 12a \Rightarrow 120 - T = 12a \).

For \( 8 \, \text{kg} \): \( T + 20 - 8g = 8a \Rightarrow T + 20 - 80 = 8a \Rightarrow T - 60 = 8a \).

Solve: \( 120 - T = 12a \), \( T - 60 = 8a \).

\( 120 - (8a + 60) = 12a \Rightarrow 120 - 60 - 8a = 12a \Rightarrow 60 = 20a \).

\( a = \frac{60}{20} = 3 \, \text{m/s}^2 \).

2 m/s²
3 m/s²
4 m/s²
5 m/s²
2

A \( 0.8 \, \text{kg} \) stone in a vertical circle of radius \( 2 \, \text{m} \) has a speed of \( 12 \, \text{m/s} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.8 \times 10 = 0.8 \times (12)^2 / 2 \).

\( T_b - 8 = 0.8 \times 72 \Rightarrow T_b - 8 = 57.6 \Rightarrow T_b = 65.6 \, \text{N} \).

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

\( 0.5 \times 0.8 \times 144 = 0.5 \times 0.8 \times v_t^2 + 2 \times 0.8 \times 10 \).

\( 57.6 = 0.4 v_t^2 + 16 \Rightarrow 0.4 v_t^2 = 41.6 \Rightarrow v_t^2 = 104 \Rightarrow v_t \approx 10.2 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 8 = 0.8 \times 104 / 2 \Rightarrow T_t + 8 = 41.6 \Rightarrow T_t = 33.6 \, \text{N} \).

30 N
33.6 N
36 N
40 N
2

A \( 50 \, \text{kg} \) man in a lift descending at \( 4 \, \text{m/s}^2 \) jumps upward with a speed of \( 5 \, \text{m/s} \) relative to the lift. What is the normal force on the floor during the jump? (Take \( g = 10 \, \text{m/s}^2 \))

Man’s acceleration relative to ground = lift’s (\( -4 \, \text{m/s}^2 \)) + jump (\( a_{\text{jump}} \)).

During jump, assume instant force \( N \) accelerates him: \( N - mg = ma_{\text{jump}} \).

Lift frame: \( v_{\text{rel}} = 5 \, \text{m/s} \), ground speed = \( -4 + 5 = 1 \, \text{m/s} \) upward.

Net acceleration relative to ground = \( 1 - (-4) = 5 \, \text{m/s}^2 \) upward initially.

\( N - 50 \times 10 = 50 \times 5 \Rightarrow N - 500 = 250 \Rightarrow N = 750 \, \text{N} \).

500 N
600 N
750 N
1000 N
3

A \( 9.2 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.2 \)) is pulled by a \( 7.2 \, \text{kg} \) mass over a pulley. A \( 20 \, \text{N} \) force opposes the \( 9.2 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

For \( 7.2 \, \text{kg} \): \( 7.2g - T = 7.2a \Rightarrow 72 - T = 7.2a \).

For \( 9.2 \, \text{kg} \): \( T - f_k - 20 = 9.2a \).

Normal: \( N = mg = 9.2 \times 10 = 92 \, \text{N} \).

Friction: \( f_k = 0.2 \times 92 = 18.4 \, \text{N} \).

Net force: \( T - 18.4 - 20 = 9.2a \Rightarrow T - 38.4 = 9.2a \).

Solve: \( 72 - T = 7.2a \), \( T - 38.4 = 9.2a \).

Substitute: \( 72 - (9.2a + 38.4) = 7.2a \Rightarrow 72 - 38.4 - 9.2a = 7.2a \Rightarrow 33.6 = 16.4a \).

\( a \approx 2.05 \, \text{m/s}^2 \), \( T - 38.4 = 9.2 \times 2.05 \Rightarrow T - 38.4 \approx 18.86 \Rightarrow T \approx 57.26 \, \text{N} \).

53 N
57.3 N
60 N
63 N
2

A body of mass \( 6 \, \text{kg} \) is acted upon by two perpendicular forces of \( 8 \, \text{N} \) and \( 6 \, \text{N} \). What is the magnitude of its acceleration?

Net force is the vector sum of the two perpendicular forces.

Magnitude \( F = \sqrt{(8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{N} \).

Acceleration \( a = \frac{F}{m} \).

Substitute: \( m = 6 \, \text{kg} \), \( F = 10 \, \text{N} \).

\( a = \frac{10}{6} \approx 1.67 \, \text{m/s}^2 \).

1 m/s²
1.67 m/s²
2 m/s²
2.5 m/s²
2

A \( 0.48 \, \text{kg} \) stone in a vertical circle of radius \( 1.5 \, \text{m} \) has a tension of \( 32 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow 32 - 0.48 \times 10 = 0.48 \times v_b^2 / 1.5 \).

\( 32 - 4.8 = 0.32 v_b^2 \Rightarrow 27.2 = 0.32 v_b^2 \Rightarrow v_b^2 = 85 \Rightarrow v_b \approx 9.22 \, \text{m/s} \).

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

\( 0.5 \times 0.48 \times 85 = 0.5 \times 0.48 \times v_t^2 + 2 \times 0.48 \times 10 \).

\( 20.4 = 0.24 v_t^2 + 9.6 \Rightarrow 0.24 v_t^2 = 10.8 \Rightarrow v_t^2 = 45 \Rightarrow v_t \approx 6.71 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 4.8 = 0.48 \times 45 / 1.5 \Rightarrow T_t + 4.8 = 14.4 \Rightarrow T_t = 9.6 \, \text{N} \).

8 N
9.6 N
11 N
12.5 N
2

A \( 1700 \, \text{kg} \) truck accelerates at \( 1.5 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then a \( 1.8 \, \text{kg} \) stone is dropped. What is the horizontal distance traveled by the stone in \( 2.5 \, \text{s} \)? (Neglect air resistance)

Truck’s velocity after \( 4 \, \text{s} \): \( v = u + at = 0 + 1.5 \times 4 = 6 \, \text{m/s} \).

Stone inherits this horizontal velocity upon drop.

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

Distance: \( s = v \times t = 6 \times 2.5 = 15 \, \text{m} \).

12 m
15 m
18 m
20 m
2

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