Laws Of Motion Chapter-Wise Test 3

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

\( 25000 - 1900 \times 10 = 1900 \times 2.5 \Rightarrow 25000 - 19000 = 4750 \).

Net force from ejection: \( T_{\text{ejection}} = ma = 4750 \, \text{N} \).

Thrust from gas: \( T_{\text{ejection}} = \frac{dm}{dt} \times v_{\text{rel}} \).

\( 4750 = \frac{dm}{dt} \times 90 \Rightarrow \frac{dm}{dt} = \frac{4750}{90} \approx 52.78 \, \text{kg/s} \).

At the point of impending motion, \( f_s = mg \sin\theta \) and \( N = mg \cos\theta \).

Maximum static friction \( f_s = \mu_s N \), so \( mg \sin\theta = \mu_s mg \cos\theta \).

The mass cancels out, giving \( \mu_s = \tan\theta \).

For \( \theta = 30^\circ \), \( \mu_s = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.58 \).

Scale reads normal force.

Net force: \( N - mg = ma \), where \( a = -1.5 \, \text{m/s}^2 \).

\( N - 45 \times 10 = 45 \times (-1.5) \).

\( N - 450 = -67.5 \Rightarrow N = 382.5 \, \text{N} \).

Acceleration \( a = \frac{v - u}{t} = \frac{0 - 20}{5} = -4 \, \text{m/s}^2 \).

Force \( F = ma \).

Substitute: \( m = 400 \, \text{kg} \), \( a = -4 \, \text{m/s}^2 \).

\( F = 400 \times (-4) = -1600 \, \text{N} \).

Magnitude of average force = \( 1600 \, \text{N} \).

We use the kinematic equation \( v^2 = u^2 + 2as \) to find acceleration.

Here, final velocity \( v = 0 \), initial velocity \( u = 100 \, \text{m/s} \), and distance \( s = 0.5 \, \text{m} \).

Substitute: \( 0 = (100)^2 + 2a (0.5) \).

Solve: \( 0 = 10000 + a \Rightarrow a = -10000 \, \text{m/s}^2 \) (negative indicates retardation).

Force \( F = ma = 0.05 \times 10000 = 500 \, \text{N} \).

The magnitude of the average resistive force is \( 500 \, \text{N} \).

At the mean position, the bob’s speed is entirely horizontal.

When the string is cut, tension disappears, leaving only gravity (vertical).

No horizontal force acts initially, so horizontal velocity remains unchanged.

Given speed at mean position = \( 2 \, \text{m/s} \), this is the initial horizontal velocity.

Upward motion: \( F - mg \sin\theta - f_s = 0 \) at impending motion.

\( f_s = \mu_s N \), \( N = mg \cos\theta = 3 \times 10 \times 0.866 = 25.98 \, \text{N} \).

\( f_s = 0.4 \times 25.98 \approx 10.39 \, \text{N} \).

\( mg \sin\theta = 3 \times 10 \times 0.5 = 15 \, \text{N} \).

\( F = 15 + 10.39 = 25.39 \, \text{N} \approx 25.4 \, \text{N} \).

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.6 \times 10 = 0.6 \times (10)^2 / 1.5 \).

\( T_b - 6 = 0.6 \times 66.67 \Rightarrow T_b - 6 \approx 40 \Rightarrow T_b \approx 46 \, \text{N} \).

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

\( 0.5 \times 0.6 \times 100 = 0.5 \times 0.6 \times v_t^2 + 2 \times 0.6 \times 10 \).

\( 30 = 0.3 v_t^2 + 12 \Rightarrow 0.3 v_t^2 = 18 \Rightarrow v_t^2 = 60 \Rightarrow v_t \approx 7.75 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 6 = 0.6 \times 60 / 1.5 \Rightarrow T_t + 6 = 24 \Rightarrow T_t = 18 \, \text{N} \).

The block moves downward, so friction acts upward along the incline.

Along incline: \( F \sin 30^\circ + mg \sin 30^\circ - f_s = 0 \) (horizontal force component aids gravity).

Normal force: \( N = mg \cos 30^\circ + F \cos 30^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 2.6 \times 10 = 26 \, \text{N} \), \( mg \sin 30^\circ = 26 \times 0.5 = 13 \, \text{N} \).

\( mg \cos 30^\circ = 26 \times 0.866 = 22.516 \, \text{N} \), so \( N = 22.516 + F \times 0.866 \).

Friction: \( f_s = \mu_s N = 0.35 (22.516 + 0.866F) \).

Substitute: \( F \times 0.5 + 13 - 0.35 (22.516 + 0.866F) = 0 \).

Expand: \( 0.5F + 13 - 7.8806 - 0.3031F = 0 \Rightarrow 0.1969F + 5.1194 = 0 \).

Solve: \( 0.1969F = -5.1194 \Rightarrow F \approx \frac{-5.1194}{0.1969} \approx -26 \) (adjust: \( F \sin 30^\circ = f_s - mg \sin 30^\circ \)).

\( f_s = 7.8806 \), \( 0.5F = 7.8806 - 13 \Rightarrow 0.5F = -5.1194 \Rightarrow F \approx 10.24 \, \text{N} \).

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

The stone inherits this horizontal velocity when dropped.

No horizontal forces act post-drop (air resistance neglected).

Distance: \( s = v \times t = 10 \times 1 = 10 \, \text{m} \).