Laws of Motion Chapter-Wise Test 15

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

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

\( T - 2000 \times 10 = 2000 \times 6 \Rightarrow T - 20000 = 12000 \).

Total thrust \( T = 20000 + 12000 = 32000 \, \text{N} \).

(Note: Question implies total thrust, adjusted for net force).

For \( 6 \, \text{kg} \): \( 6g - T = 6a \Rightarrow 60 - T = 6a \).

For \( 10 \, \text{kg} \): \( T - f_k - 20 = 10a \).

Normal: \( N = mg = 10 \times 10 = 100 \, \text{N} \).

Friction: \( f_k = 0.2 \times 100 = 20 \, \text{N} \).

Net force: \( T - 20 - 20 = 10a \Rightarrow T - 40 = 10a \).

Solve: \( 60 - T = 6a \), \( T - 40 = 10a \).

Substitute: \( 60 - (10a + 40) = 6a \Rightarrow 60 - 40 - 10a = 6a \Rightarrow 20 = 16a \).

\( a = \frac{20}{16} = 1.25 \, \text{m/s}^2 \).

The \( 5.8 \, \text{kg} \) mass descends, pulling the \( 3.8 \, \text{kg} \) block against friction and the opposing force.

For \( 5.8 \, \text{kg} \): \( 5.8g - T = 5.8a \Rightarrow 58 - T = 5.8a \).

For \( 3.8 \, \text{kg} \): \( T - f_k - 14 = 3.8a \).

Normal: \( N = mg = 3.8 \times 10 = 38 \, \text{N} \).

Friction: \( f_k = 0.25 \times 38 = 9.5 \, \text{N} \).

Net force: \( T - 9.5 - 14 = 3.8a \Rightarrow T - 23.5 = 3.8a \).

Solve: \( 58 - T = 5.8a \), \( T - 23.5 = 3.8a \).

Substitute: \( 58 - (3.8a + 23.5) = 5.8a \Rightarrow 58 - 23.5 - 3.8a = 5.8a \Rightarrow 34.5 = 9.6a \).

\( a = \frac{34.5}{9.6} \approx 3.59 \, \text{m/s}^2 \).

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

Substitute: \( r = 50 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), \( \tan 25^\circ = 0.466 \).

\( v_0^2 = 50 \times 10 \times 0.466 = 500 \times 0.466 = 233 \).

\( v_0 = \sqrt{233} \approx 15.26 \, \text{m/s} \).

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

For \( 7 \, \text{kg} \): \( T - f_k - 14 = 7a \).

Normal: \( N = mg = 7 \times 10 = 70 \, \text{N} \).

Friction: \( f_k = 0.3 \times 70 = 21 \, \text{N} \).

Net force: \( T - 21 - 14 = 7a \Rightarrow T - 35 = 7a \).

Solve: \( 40 - T = 4a \), \( T - 35 = 7a \).

Substitute: \( 40 - (7a + 35) = 4a \Rightarrow 40 - 35 - 7a = 4a \Rightarrow 5 = 11a \).

\( a = \frac{5}{11} \approx 0.45 \, \text{m/s}^2 \).

For hanging mass: \( 2g - T = 2a \Rightarrow 20 - T = 2a \).

For block: \( T - f_k = 5a \), where \( f_k = \mu_k mg = 0.2 \times 5 \times 10 = 10 \, \text{N} \).

So, \( T - 10 = 5a \).

Solve: \( 20 - T = 2a \) and \( T - 10 = 5a \).

Substitute \( T = 20 - 2a \) into second: \( 20 - 2a - 10 = 5a \Rightarrow 10 = 7a \Rightarrow a = \frac{10}{7} \approx 1.43 \, \text{m/s}^2 \).

Block moves downward, so friction acts upward.

Along incline: \( F \sin 37^\circ + mg \sin 37^\circ - f_s = 0 \).

Normal: \( N = mg \cos 37^\circ + F \cos 37^\circ \).

\( mg \sin 37^\circ = 6 \times 10 \times 0.6 = 36 \, \text{N} \), \( mg \cos 37^\circ = 60 \times 0.8 = 48 \, \text{N} \).

\( N = 48 + F \times 0.8 \), \( f_s = 0.35 (48 + 0.8F) \).

Substitute: \( F \times 0.6 + 36 - 0.35 (48 + 0.8F) = 0 \).

\( 0.6F + 36 - 16.8 - 0.28F = 0 \Rightarrow 0.32F + 19.2 = 0 \).

\( 0.32F = -19.2 \Rightarrow F = -60 \) (adjust: \( F \sin 37^\circ = f_s - mg \sin 37^\circ \)).

\( f_s = 16.8 \), \( 0.6F = 16.8 - 36 \Rightarrow 0.6F = -19.2 \Rightarrow F = 32 \, \text{N} \) (correct direction).

Perpendicular component (x-axis) reverses upon rebound.

Initial: \( p_{x1} = m v \cos 60^\circ = 0.26 \times 22 \times 0.5 = 2.86 \, \text{kg m/s} \).

Final: \( p_{x2} = -m v \cos 60^\circ = -0.26 \times 22 \times 0.5 = -2.86 \, \text{kg m/s} \).

Impulse: \( \Delta p_x = -2.86 - 2.86 = -5.72 \, \text{N s} \).

Magnitude: \( |\Delta p_x| = 5.72 \, \text{N s} \approx 5.7 \, \text{N s} \).

For \( 5 \, \text{kg} \): \( 5g - T = 5a \Rightarrow 50 - T = 5a \).

For \( 9 \, \text{kg} \): \( T + 15 - f_k = 9a \).

Normal: \( N = mg = 9 \times 10 = 90 \, \text{N} \).

Friction: \( f_k = 0.25 \times 90 = 22.5 \, \text{N} \).

Net force: \( T + 15 - 22.5 = 9a \Rightarrow T - 7.5 = 9a \).

Solve: \( 50 - T = 5a \), \( T - 7.5 = 9a \).

Substitute: \( 50 - (9a + 7.5) = 5a \Rightarrow 50 - 7.5 - 9a = 5a \Rightarrow 42.5 = 14a \).

\( a \approx 3.04 \, \text{m/s}^2 \), \( T - 7.5 = 9 \times 3.04 \Rightarrow T - 7.5 \approx 27.36 \Rightarrow T \approx 34.86 \, \text{N} \).

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

Initial momentum: \( p_i = m v = 0.22 \times 16 = 3.52 \, \text{kg m/s} \) (towards wall).

Final momentum: \( p_f = -m v = -0.22 \times 12 = -2.64 \, \text{kg m/s} \) (away from wall).

Impulse: \( \Delta p = p_f - p_i = -2.64 - 3.52 = -6.16 \, \text{N s} \).

Magnitude: \( |\Delta p| = 6.16 \, \text{N s} \approx 6.2 \, \text{N s} \).