Vardaan Institute - Detailed Solutions: Master Sheet (Laws of Motion)

Section A: Newton's Laws & Impulse Solutions

A force $F = (6\hat{i} - 8\hat{j} + 10\hat{k})\text{ N}$ produces an acceleration of $1\text{ m/s}^2$ in a body. Calculate mass of the body.

Sol: $|F| = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt{2}\text{ N}$. Mass $m = \frac{|F|}{a} = \frac{10\sqrt{2}}{1} = 10\sqrt{2} \approx 14.14\text{ kg}$.

A cricket ball of mass $150\text{ g}$ moving at $20\text{ m/s}$ is caught by a player in $0.1\text{ s}$. Find impulse and average force.

Sol: Impulse $I = \Delta p = m v = 0.15 \times 20 = 3\text{ N}\cdot\text{s}$. Average force $F = \frac{I}{\Delta t} = \frac{3}{0.1} = 30\text{ N}$.

Section B: Connected Bodies Solutions

Two masses $m_1 = 3\text{ kg}$ and $m_2 = 5\text{ kg}$ are connected by a light string passing over a smooth pulley. Find $a$ and $T$.

Sol: $a = \frac{m_2 - m_1}{m_1 + m_2} g = \frac{5 - 3}{8} g = \frac{g}{4} = 2.45\text{ m/s}^2$. $T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2(3)(5)}{8} g = \frac{30}{8} g = 36.75\text{ N}$.

A block of mass $m=2\text{ kg}$ rests on an inclined plane of angle $\theta = 30^\circ$. Find normal reaction and friction force.

Sol: Normal $N = m g \cos 30^\circ = 2 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 16.97\text{ N}$. Friction $f = m g \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8\text{ N}$.

Section C: Friction & Banking Solutions

A car negotiates a banked curve of radius $R = 50\text{ m}$ at angle $\theta = 45^\circ$. Find optimum speed for zero friction.

Sol: $v = \sqrt{R g \tan 45^\circ} = \sqrt{50 \times 9.8 \times 1} = \sqrt{490} = 7\sqrt{10} \approx 22.14\text{ m/s}$.

Define angle of friction $\lambda$ and angle of repose $\theta_{repose}$, and prove $\mu_s = \tan\lambda = \tan\theta_{repose}$.

Sol: Limiting friction $f_s = \mu_s N$. Angle between resultant contact force and normal is $\tan\lambda = \frac{f_s}{N} = \mu_s$. On incline at impending slip $m g \sin\theta = f_s = \mu_s m g \cos\theta \implies \tan\theta = \mu_s$.

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