Laws of Motion

Answer: (d)

1. Resultant $R = \sqrt{8^2 + 6^2} = 10\text{ N}$, so $a = \frac{10}{5} = 2\text{ m s}^{-2}$.
2. Angle with the 8 N force: $\tan\theta = \frac{6}{8} = \frac{3}{4}$.
3. So the answer is (d) $2\text{ m s}^{-2};\ \tan^{-1}(3/4)$ with 8 N force.

Answer: (d)

1. $\vec{a} = \frac{5\hat{i} + 10\hat{j}}{0.1} = 50\hat{i} + 100\hat{j}\text{ m/s}^2$
2. From rest: $\vec{r} = \frac{1}{2}\vec{a}t^2 = 2(50\hat{i} + 100\hat{j}) = 100\hat{i} + 200\hat{j}$
3. $2x = 100 \Rightarrow x = 50$; $5y = 200 \Rightarrow y = 40$
4. $x : y = 50 : 40 = 5 : 4$. So the answer is (d).

Answer: (c)

1. $\vec{a} = \frac{\vec{F}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$
2. $\vec{v} = \frac{t^4}{4}\hat{i} - \frac{3t^2}{8}\hat{j}$; at $t = 2$: $\vec{v} = 4\hat{i} - \frac{3}{2}\hat{j}$
3. $\vec{r} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$; at $t = 2$: $\vec{r} = \frac{8}{5}\hat{i} - \hat{j}$
4. So the answer is (c).

Answer: (d)

1. $m\frac{dv}{dt} = mg - kv$, with $v = 0$ at $t = 0$.
2. Solving: $v = \frac{mg}{k}\left(1 - e^{-kt/m}\right)$
3. $v$ rises from 0 with decreasing slope and levels off at terminal speed $\frac{mg}{k}$.
4. So the answer is (d).

Answer: (c)

1. $F = m_1A_1 = m_2A_2 \Rightarrow m_1 = \frac{F}{A_1},\ m_2 = \frac{F}{A_2}$
2. $a = \frac{F}{m_1 + m_2} = \frac{F}{F/A_1 + F/A_2} = \frac{A_1A_2}{A_1 + A_2}$
3. So the answer is (c) $\frac{A_1A_2}{A_1 + A_2}$.

Answer: (c)

1. Hooke's law $T = kx$: $x_1 = \frac{5}{k},\ x_2 = \frac{7}{k}$
2. $5x_1 - 2x_2 = \frac{25 - 14}{k} = \frac{11}{k}$
3. $T = k \times \frac{11}{k} = 11\text{ N}$. So the answer is (c) 11 N.

Answer: (b)

1. $a = v\frac{dv}{dx} = 4\sqrt{x} \times \frac{2}{\sqrt{x}} = 8\text{ m/s}^2$
2. $F = ma = 0.5 \times 8 = 4\text{ N}$
3. So the answer is (b) 4 N.

Answer: (d)

1. $\vec{a} = \frac{6\hat{k}}{2} = 3\hat{k}\text{ m/s}^2$; it changes only the $z$-component.
2. $\Delta\vec{v} = \vec{a}t = 3 \times \frac{5}{3}\hat{k} = 5\hat{k}$
3. $\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}$. So the answer is (d).

Answer: (c)

1. Work by retarding force: $W = -\int_{0.1}^{1.9} 10x\,dx = -5(1.9^2 - 0.1^2) = -5(3.6) = -18\text{ J}$
2. Initial KE $= \frac{1}{2}(1)(10)^2 = 50\text{ J}$, so final KE $= 50 - 18 = 32\text{ J}$
3. $\frac{1}{2}v^2 = 32 \Rightarrow v = 8\text{ m/s}$. So the answer is (c) $8\text{ m/s}$.

Answer: (c)

1. Friction, normal reaction and viscous force need physical contact between surfaces or with the fluid.
2. Gravitational force acts at a distance through space, without contact.
3. So the answer is (c) gravitational force.

Answer: (a)

1. Net force $\vec{F} = \vec{F}_1 + \vec{F}_2 = 8\hat{i} + 4\hat{j} + 4\hat{k}\text{ N}$
2. $\vec{a} = \frac{\vec{F}}{m} = \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{4} = 2\hat{i} + \hat{j} + \hat{k}$
3. So the answer is (a) $2\hat{i} + \hat{j} + \hat{k}$.

Answer: (a)

1. $\vec{F} = \frac{d\vec{p}}{dt} = -k\sin(kt)\hat{i} - k\cos(kt)\hat{j}$
2. $\vec{F} \cdot \vec{p} = -k\sin kt\cos kt + k\sin kt\cos kt = 0$
3. So $\vec{F} \perp \vec{p}$. The answer is (a) $\frac{\pi}{2}$.

Answer: (b)

1. $\vec{v} = \frac{d\vec{r}}{dt} = 10\hat{i} + 30t\hat{j}$
2. $\vec{a} = \frac{d\vec{v}}{dt} = 30\hat{j}$, so $\vec{F} = m\vec{a}$ is along $+\hat{j}$.
3. So the answer is (b) positive $y$-axis.

Answer: (d)

1. First 24 cm: $u^2 - \frac{u^2}{9} = 2a(24) \Rightarrow \frac{8u^2}{9} = 48a$
2. Remaining part: $\left(\frac{u}{3}\right)^2 = 2as \Rightarrow \frac{u^2}{9} = 2as$
3. Divide: $\frac{8}{1} = \frac{24}{s} \Rightarrow s = 3\text{ cm}$
4. Total length $= 24 + 3 = 27\text{ cm}$. So the answer is (d) $27\text{ cm}$.

Answer: (c)

1. After the force stops, speed is constant: $v = \frac{50}{10} = 5\text{ m/s}$.
2. This speed was gained in 20 s from rest: $a = \frac{5}{20} = 0.25\text{ m/s}^2$
3. $F = ma = 20 \times 0.25 = 5\text{ N}$. So the answer is (c) 5 N.

Answer: (c)

1. $a = v\frac{dv}{dx} = 10\sqrt{x} \times \frac{5}{\sqrt{x}} = 50\text{ m/s}^2$
2. $F = ma = 0.5 \times 50 = 25\text{ N}$
3. So the answer is (c) 25 N.

Answer: (d)

1. $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 6t\hat{j}$; at $t = 1$ s, $\vec{a} = 2\hat{i} + 6\hat{j}$
2. $\vec{F} = m\vec{a} = 0.5(2\hat{i} + 6\hat{j}) = \hat{i} + 3\hat{j}\text{ N}$
3. So $x = 3$. The answer is (d) 3.

Answer: 2

1. From rest, $\vec{r} = \frac{1}{2}\frac{\vec{F}}{m}t^2$, so each coordinate is proportional to the force component.
2. $\frac{a}{b} = \frac{F_x}{F_y} = \frac{10}{5} = 2$
3. So $\frac{a}{b} = 2$.

Answer: (b)

1. Up: $a_1 = g + \frac{f}{m} = 10 + 2 = 12\text{ m/s}^2$; down: $a_2 = g - \frac{f}{m} = 8\text{ m/s}^2$
2. Same height $h = \frac{1}{2}a t^2$ both ways, so $t \propto \frac{1}{\sqrt{a}}$.
3. $\frac{t_a}{t_d} = \sqrt{\frac{a_2}{a_1}} = \sqrt{\frac{8}{12}} = \frac{\sqrt{2}}{\sqrt{3}}$
4. So the answer is (b) $\sqrt{2} : \sqrt{3}$.

Answer: 10

1. $v^2 = u^2 - 2as \Rightarrow 0 = 10^2 - 2a(0.5) \Rightarrow a = 100\text{ m/s}^2$
2. $F = ma = 0.1 \times 100 = 10\text{ N}$
3. So $x = 10$ N.

Answer: (c)

1. Constant velocity on a frictionless incline: applied force balances $mg\sin\theta$, so it always points up the slope with magnitude 2 N.
2. Going up the left face (first $D/2$), the force is along the motion: $F = +2\text{ N}$.
3. Going down the right face (next $D/2$), the force still points up the slope, i.e. against the motion: $F = -2\text{ N}$.
4. So the graph is a +2 N step then a −2 N step. The answer is (c).

Answer: (c)

1. $\vec{a} = \frac{40\hat{i} + 10\hat{j}}{5} = 8\hat{i} + 2\hat{j}\text{ m/s}^2$
2. $\vec{r} = \frac{1}{2}\vec{a}t^2 = 50(8\hat{i} + 2\hat{j}) = 400\hat{i} + 100\hat{j}\text{ m}$
3. So the answer is (c).

Answer: 12

1. $a_y = \frac{3}{2} = 1.5\text{ m/s}^2$ (check: $a_x = 1$, $x = \frac{1}{2}(1)(16) = 8$)
2. $y = \frac{1}{2}a_yt^2 = \frac{1}{2}(1.5)(16) = 12\text{ m}$
3. So $b = 12$.

Answer: 500

1. $a_x = \frac{F_x}{m} = \frac{20}{2} = 10\text{ m/s}^2$
2. $x = \frac{1}{2}a_xt^2 = \frac{1}{2}(10)(10)^2 = 500\text{ m}$
3. So the displacement along $x$ is $500$ m.

Answer: (a)

1. $mv\frac{dv}{dx} = -\alpha x^2 \Rightarrow m\int_{v_0}^{0} v\,dv = -\alpha\int_0^{x} x^2dx$
2. $\frac{mv_0^2}{2} = \frac{\alpha x^3}{3}$
3. $x = \left[\frac{3mv_0^2}{2\alpha}\right]^{1/3}$. So the answer is (a).

Answer: (a)

1. Resultant $R = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{1 + 1 + 1} = \sqrt{3}\text{ N}$
2. $a = \frac{R}{m} = \frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}} = \sqrt{0.75}\text{ m/s}^2$
3. So the answer is (a) $\sqrt{0.75}$.

Answer: (b)

1. $\vec{a} = \frac{k}{m}(v_y\hat{i} + v_x\hat{j})$
2. $(\vec{v} \times \vec{a})_z = v_xa_y - v_ya_x = \frac{k}{m}(v_x^2 - v_y^2)$
3. $\frac{d}{dt}(v_x^2 - v_y^2) = 2v_xa_x - 2v_ya_y = \frac{2k}{m}(v_xv_y - v_yv_x) = 0$, so $\vec{v} \times \vec{a}$ is constant.
4. (a), (c) fail: $\vec{v} \cdot \vec{a} = \frac{2k}{m}v_xv_y \neq 0$ and changes; (d) fails: a magnetic force is always $\perp \vec{v}$.
5. So the answer is (b).

Answer: (c)

1. Going up, gravity and drag both act downward: $v\frac{dv}{dy} = -(g + kv^2)$.
2. $H = \int_0^{u} \frac{v\,dv}{g + kv^2} = \frac{1}{2k}\ln(g + kv^2)\Big|_0^{u}$
3. $H = \frac{1}{2k}\ln\left(1 + \frac{ku^2}{g}\right)$
4. So the answer is (c).

Answer: (a)

1. Balanced forces mean $\vec{F}_{net} = 0$, so $\vec{a} = 0$ (Newton's first law).
2. With zero acceleration the velocity stays constant at $150\text{ m/s}$.
3. So the answer is (a) keep moving with same velocity.

Answer: (c)

1. Retardation $a = \frac{F}{m} = \frac{2.5 \times 10^{-2}}{0.02} = 1.25\text{ m/s}^2$
2. $v^2 = u^2 - 2as = 1 - 2(1.25)(0.2) = 0.5$
3. $v = \sqrt{0.5} \approx 0.7\text{ m/s}$. So the answer is (c) 0.7 ms$^{-1}$.

Answer: (b)

1. For pieces of the same spring, $kl = \text{constant}$, so $k \propto \frac{1}{l}$.
2. $\frac{k_1}{k_2} = \frac{l_2}{l_1} = \frac{l_2}{nl_2} = \frac{1}{n}$
3. So the answer is (b) $\frac{1}{n}$.

Answer: (a)

1. While rising: $\frac{du}{dt} = -(g + \gamma u^2)$.
2. $t = \int_0^{V_0} \frac{du}{g + \gamma u^2} = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}\,u\right)\Big|_0^{V_0}$
3. $t = \frac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\frac{\gamma}{g}}V_0\right)$
4. So the answer is (a).

Answer: (d)

1. KE $= \frac{1}{2}mv^2 = \frac{1}{8}mv_0^2 \Rightarrow v = \frac{v_0}{2} = 5\text{ m/s}$.
2. $m\frac{dv}{dt} = -kv^2 \Rightarrow \frac{1}{v} - \frac{1}{v_0} = \frac{kt}{m}$
3. $\frac{1}{5} - \frac{1}{10} = \frac{k \times 10}{10^{-2}} \Rightarrow 0.1 = 1000k$
4. $k = 10^{-4}$; units of $F/v^2$ are $\text{kg m}^{-1}$.
5. So the answer is (d) $10^{-4}\text{ kg m}^{-1}$.

Answer: (d)

1. $m\frac{dv}{dt} = \frac{R}{t^2}v \Rightarrow \frac{dv}{v} = \frac{R}{m}\frac{dt}{t^2}$
2. Integrating: $\ln v = -\frac{R}{m}\cdot\frac{1}{t} + C$
3. So $\log v$ is linear in $\frac{1}{t}$; a straight line confirms the law.
4. So the answer is (d) $\log v(t)$ against $\frac{1}{t}$.

Answer: (a)

1. $\vec{a} = \frac{\Delta\vec{v}}{\Delta t} = \frac{6\hat{j} - (6\hat{i} - 2\hat{j})}{10} = \frac{-6\hat{i} + 8\hat{j}}{10}\text{ m/s}^2$
2. $\vec{F} = m\vec{a} = 5 \times (-0.6\hat{i} + 0.8\hat{j}) = -3\hat{i} + 4\hat{j}\text{ N}$
3. So the answer is (a) $(-3\hat{i} + 4\hat{j})$ N.

Answer: (b)

1. $v = \sqrt{2gh}$, so $P = m\sqrt{2gh} \propto \sqrt{h}$.
2. Height 100% more means $h' = 2h \Rightarrow P' = \sqrt{2}P \approx 1.41P$.
3. Change $= 41\%$. So the answer is (b) 41%.

Answer: (b)

1. $m\frac{dv}{dt} = F_0e^{-bt} \Rightarrow v = \int_0^t \frac{F_0}{m}e^{-bt}dt = \frac{F_0}{mb}\left(1 - e^{-bt}\right)$
2. $v$ starts at 0, always increases (force stays positive) and approaches $\frac{F_0}{mb}$ as $t \to \infty$.
3. So the answer is (b).

Answer: (c)

1. $|\vec{F}| = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\text{ N}$
2. $m = \frac{F}{a} = \frac{10\sqrt{2}}{1} = 10\sqrt{2}\text{ kg}$
3. So the answer is (c) $10\sqrt{2}$ kg.

Answer: (c)

1. At release $|F| = kx = 15 \times 0.2 = 3\text{ N}$
2. $a = \frac{F}{m} = \frac{3}{0.3} = 10\text{ m/s}^2$
3. So the answer is (c) $10\text{ m/s}^2$.

Answer: (b)

1. $\vec{a} = \frac{\vec{F}}{m} = 2t^2\hat{i} + \frac{4}{3}t\hat{j}$
2. $\vec{v} = \int_0^t \vec{a}\,dt = \frac{2}{3}t^3\hat{i} + \frac{2}{3}t^2\hat{j}$
3. At $t = 3$ s: $\vec{v} = 18\hat{i} + 6\hat{j}$. So the answer is (b).

Answer: (b)

1. $a = \frac{F}{m} = \frac{6}{1} = 6\text{ m/s}^2$
2. $v = u + at \Rightarrow 30 = 0 + 6t \Rightarrow t = 5\text{ s}$
3. So the answer is (b) 5 seconds.

Answer: (c)

1. $m = \frac{F}{a} = \frac{10}{1} = 10\text{ kg}$
2. So the answer is (c) $10\text{ kg}$.

Answer: (c)

1. $|\vec{F}| = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\text{ N}$
2. $m = \frac{F}{a} = \frac{10\sqrt{2}}{1} = 10\sqrt{2}\text{ kg}$
3. So the answer is (c) $10\sqrt{2}\text{ kg}$.