Laws Of Motion Chapter-Wise Test 2

The stone retains the horizontal velocity of the truck when dropped.

Truck’s velocity at \( t = 5 \, \text{s} \): \( v = u + at = 0 + 3 \times 5 = 15 \, \text{m/s} \).

No horizontal force acts on the stone after release (air resistance neglected).

Thus, horizontal velocity remains constant at \( 15 \, \text{m/s} \) at \( t = 6 \, \text{s} \).

Truck’s velocity after \( 5 \, \text{s} \): \( v = u + at = 0 + 2 \times 5 = 10 \, \text{m/s} \).

Stone inherits this horizontal velocity upon drop.

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

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

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

Substitute: \( T - 2000 \times 10 = 2000 \times 4 \).

\( T - 20000 = 8000 \).

\( T = 20000 + 8000 = 28000 \, \text{N} \).

Initial momentum: \( p_{x1} = 0.24 \times 15 \times 0.707 = 2.5452 \, \text{kg m/s} \), \( p_{y1} = 0.24 \times 15 \times 0.707 = 2.5452 \, \text{kg m/s} \).

Final momentum: \( p_{x2} = -0.24 \times 15 \times 0.866 = -3.1176 \, \text{kg m/s} \), \( p_{y2} = 0.24 \times 15 \times 0.5 = 1.8 \, \text{kg m/s} \).

Change: \( \Delta p_x = -3.1176 - 2.5452 = -5.6628 \, \text{kg m/s} \), \( \Delta p_y = 1.8 - 2.5452 = -0.7452 \, \text{kg m/s} \).

Magnitude: \( |\Delta p| = \sqrt{(-5.6628)^2 + (-0.7452)^2} = \sqrt{32.067 + 0.555} \approx \sqrt{32.622} \approx 5.71 \, \text{N s} \).

Total mass = \( 15 \, \text{kg} \), weight = \( 15 \times 10 = 150 \, \text{N} \).

At midpoint: \( T \sin\theta = 60 \, \text{N} \), \( T \cos\theta = 150 \, \text{N} \).

Divide: \( \tan\theta = \frac{60}{150} = 0.4 \).

\( \theta = \tan^{-1}(0.4) \approx 22^\circ \) (same as Set 3 Q8, but mass change adds complexity).

Check: \( T = \sqrt{(60)^2 + (150)^2} = \sqrt{3600 + 22500} = \sqrt{26100} \approx 161.55 \, \text{N} \), consistent.

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

For \( 5 \, \text{kg} \): \( T + 10 - f_k = 5a \), \( f_k = 0.4 \times 5 \times 10 = 20 \, \text{N} \).

\( T + 10 - 20 = 5a \Rightarrow T - 10 = 5a \).

Solve: \( 30 - T = 3a \), \( T - 10 = 5a \Rightarrow 30 - (5a + 10) = 3a \).

\( 30 - 10 - 5a = 3a \Rightarrow 20 = 8a \Rightarrow a = 2.5 \, \text{m/s}^2 \).

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

For \( 4 \, \text{kg} \): \( T - mg \sin 53^\circ - f_k = 4a \).

\( N = mg \cos 53^\circ = 4 \times 10 \times 0.6 = 24 \, \text{N} \).

Friction: \( f_k = 0.15 \times 24 = 3.6 \, \text{N} \).

\( mg \sin 53^\circ = 40 \times 0.8 = 32 \, \text{N} \).

Net force: \( T - 32 - 3.6 = 4a \Rightarrow T - 35.6 = 4a \).

Solve: \( 60 - T = 6a \), \( T - 35.6 = 4a \).

Substitute: \( 60 - (4a + 35.6) = 6a \Rightarrow 60 - 35.6 - 4a = 6a \Rightarrow 24.4 = 10a \).

\( a = \frac{24.4}{10} = 2.44 \, \text{m/s}^2 \).

The \( 6.6 \, \text{kg} \) mass descends, pulling the \( 4.6 \, \text{kg} \) block with an aiding force.

For \( 6.6 \, \text{kg} \): \( 6.6g - T = 6.6a \Rightarrow 66 - T = 6.6a \).

For \( 4.6 \, \text{kg} \): \( T + 9 - f_k = 4.6a \).

Normal: \( N = mg = 4.6 \times 10 = 46 \, \text{N} \).

Friction: \( f_k = 0.2 \times 46 = 9.2 \, \text{N} \).

Net force: \( T + 9 - 9.2 = 4.6a \Rightarrow T - 0.2 = 4.6a \).

Solve: \( 66 - T = 6.6a \), \( T - 0.2 = 4.6a \).

Substitute: \( 66 - (4.6a + 0.2) = 6.6a \Rightarrow 66 - 0.2 - 4.6a = 6.6a \Rightarrow 65.8 = 11.2a \).

\( a = \frac{65.8}{11.2} \approx 5.88 \, \text{m/s}^2 \).

Angular speed: \( \omega = 40 \times \frac{2\pi}{60} = \frac{4\pi}{3} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{4\pi}{3}\right)^2 = \frac{16\pi^2}{9} \approx 17.55 \).

Centripetal force: \( F = m \omega^2 r = 0.5 \times 17.55 \times 1.5 \).

\( F \approx 0.5 \times 17.55 \times 1.5 = 13.1625 \, \text{N} \approx 13.2 \, \text{N} \).

Tension equals this force in horizontal circular motion.

The block is about to move upward, so friction acts downward along the incline.

Along incline: \( F \cos 37^\circ - mg \sin 37^\circ - f_s = 0 \) (horizontal force component opposes gravity and friction).

Normal force: \( N = mg \cos 37^\circ + F \sin 37^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 2.8 \times 10 = 28 \, \text{N} \), \( mg \sin 37^\circ = 28 \times 0.6 = 16.8 \, \text{N} \).

\( mg \cos 37^\circ = 28 \times 0.8 = 22.4 \, \text{N} \), so \( N = 22.4 + F \times 0.6 \).

Friction: \( f_s = \mu_s N = 0.5 (22.4 + 0.6F) \).

Substitute: \( F \times 0.8 - 16.8 - 0.5 (22.4 + 0.6F) = 0 \).

Expand: \( 0.8F - 16.8 - 11.2 - 0.3F = 0 \Rightarrow 0.5F - 28 = 0 \).

Solve: \( 0.5F = 28 \Rightarrow F = \frac{28}{0.5} = 56 \, \text{N} \).