Mechanics

(a) 14%

For a product of powers, maximum percentage errors add after multiplying by the powers: $\frac{\Delta P}{P}\times100 = 3(1)+2(2)+3+4 = 14\%$. Divisors add their errors too (errors never cancel), which is why 10% (from subtracting) is wrong.

(c) $2\hat i-3\hat j$ m/s

Acceleration is only vertical, so the horizontal component stays $2$ m/s. At the same level the vertical speed is equal in magnitude and reversed in direction, giving $-3$ m/s. So $\vec v_B = 2\hat i-3\hat j$.

(b) $h_1=\frac{h_2}{3}=\frac{h_3}{5}$

For free fall from rest, distances in equal successive time intervals are in the ratio $1:3:5$. Hence $h_2=3h_1$ and $h_3=5h_1$, i.e. $h_1=h_2/3=h_3/5$. Option (c) would need ratio $1:3:9$, which is wrong.

(a) zero

The blocks move upward at constant speed, so acceleration is zero for every block. By Newton's first law the net force on the $2m$ block (tension above, tension below and weight) is zero, even though its weight is $2mg$.

(c) $\mu=2\tan\theta$

On the smooth half the block gains speed with $a=g\sin\theta$. On the rough half it must lose the same speed over the same distance, so the retardation has the same magnitude: $\mu g\cos\theta-g\sin\theta=g\sin\theta$. This gives $\mu=2\tan\theta$. Using $\mu=\tan\theta$ would only keep the speed constant, not stop the block.

(a) 9 J

Displacement $\vec d=(4\hat i+3\hat j-\hat k)-(2\hat i+\hat k)=2\hat i+3\hat j-2\hat k$. Work $W=\vec F\cdot\vec d=(3)(2)+(1)(3)=9$ J. The mass of the particle is not needed.

(b) 5 kg

Momentum is conserved and the initial momentum is zero. The first two momenta are $12$ and $2\times8=16$ kg m/s at right angles, so their resultant is $\sqrt{12^2+16^2}=20$ kg m/s. The third part must carry equal and opposite momentum: $m(4)=20$, so $m=5$ kg.

(a) $\frac{3g}{2L}$

Just after cutting, torque about the hinge is due to weight acting at the centre: $\tau=Mg\frac L2$. Moment of inertia about the end is $\frac{ML^2}{3}$, so $\alpha=\frac{\tau}{I}=\frac{MgL/2}{ML^2/3}=\frac{3g}{2L}$.

(d) Disc

Energy conservation for rolling: $\frac12 mv^2\left(1+\frac{k^2}{R^2}\right)=mgh$, so $h=\frac{v^2}{2g}\left(1+\frac{k^2}{R^2}\right)$. Setting $h=\frac{3v^2}{4g}$ gives $\frac{k^2}{R^2}=\frac12$, which is the disc (ring is 1, hollow sphere $2/3$, solid sphere $2/5$).

(b) $\frac23mgR$

Potential energy is $-\frac{GMm}{r}$. At the surface it is $-\frac{GMm}{R}$; at height $2R$ ($r=3R$) it is $-\frac{GMm}{3R}$. Change $=\frac{GMm}{R}\left(1-\frac13\right)=\frac23\frac{GMm}{R}=\frac23mgR$ using $GM=gR^2$. The formula $mgh=2mgR$ is wrong because $g$ is not constant at such heights.

(d) $-4G$

Potential is a scalar and adds: $V=-G\cdot2\left(\frac11+\frac12+\frac14+\frac18+\dots\right)$. The geometric series sums to $\frac{1}{1-1/2}=2$, so $V=-G(2)(2)=-4G$.

(a) length = 50 cm, diameter = 0.5 mm

For the same material and tension, extension $\Delta l=\frac{Fl}{AY}\propto\frac{l}{d^2}$. Values of $l/d^2$ (cm, mm units): (a) $50/0.25=200$, (b) $100/1=100$, (c) $200/4=50$, (d) $300/9\approx33$. The first wire is largest.

(d) angle of contact between the surface and the liquid

Wettability is decided by the angle of contact: a small angle (acute) means the liquid wets the surface, a large (obtuse) angle means it does not. Surface tension and viscosity affect other behaviour but do not by themselves define wetting.

(a) $\Delta l$ versus $l^2$

Volume is fixed, so area $A=\frac Vl$. Then $\Delta l=\frac{Fl}{AY}=\frac{Fl^2}{VY}$. Since $F$, $V$, $Y$ are constants, $\Delta l\propto l^2$, and only this plot is a straight line (through the origin).

(c) $\frac{11}{5}(\hat i+\hat j)$

Average velocity is total displacement divided by total time, using only the end points. $\Delta\vec r=(13-2)\hat i+(14-3)\hat j=11(\hat i+\hat j)$ m over $5$ s, so $\vec V_{av}=\frac{11}{5}(\hat i+\hat j)$. Option (d) divides the final position (not the displacement) by $5$.

(b) $\frac{g(1-2\mu)}{3}$

The whole system of three masses moves with the same acceleration $a$. The driving force is $m_1g=mg$, and friction acts on both table masses: $\mu(m_2+m_3)g=2\mu mg$. $mg-2\mu mg=(3m)a$, so $a=\frac{g(1-2\mu)}{3}$. Option (c) wrongly divides by $2m$, counting only two masses in the system.

(c) $[F\,V^{-1}\,T]$

Force is $F=ma$, with $a=\frac{V}{T}$, so $m=\frac{F}{a}=\frac{FT}{V}$. Hence $[m]=[F\,V^{-1}\,T]$. The other options place $V$ or $T$ with the wrong power, which would not give the dimensions of $F/a$.

(d) 5 : 7

Rolling without slipping: $a=\frac{g\sin\theta}{1+I/mR^2}=\frac{g\sin\theta}{1+2/5}=\frac57g\sin\theta$. Sliding without friction: $a=g\sin\theta$. Ratio $=\frac{5}{7}:1=5:7$. Option (c) inverts the ratio.

(d) $\frac{2ma}{g+a}$

Let the upthrust be $U$ (constant). Descending: $mg-U=ma$, so $U=m(g-a)$. After removing $\Delta m$, it rises with acceleration $a$: $U-(m-\Delta m)g=(m-\Delta m)a$, so $m-\Delta m=\frac{U}{g+a}=\frac{m(g-a)}{g+a}$. Thus $\Delta m=m-\frac{m(g-a)}{g+a}=\frac{2ma}{g+a}$.

(b) energy $=3VT\left(\frac1r-\frac1R\right)$ is released.

With $n$ small drops, $nr^3=R^3$ and $V=\frac43\pi R^3$. Initial area $=n\cdot4\pi r^2=\frac{3V}{r}$, final area $=4\pi R^2=\frac{3V}{R}$. The area falls, so energy $=T\,\Delta A=3VT\left(\frac1r-\frac1R\right)$ is released. Option (a) is wrong as surface energy decreases on coalescence; (d) has the wrong factor $4$ instead of $3$.

(a) $\frac32mv^2$

Momentum conservation: the two pieces of mass $m$ move at right angles, giving a resultant $\sqrt2\,mv$. The third piece (mass $2m$) must have equal and opposite momentum, so its speed is $u=\frac{\sqrt2\,mv}{2m}=\frac{v}{\sqrt2}$. $KE=2\times\frac12mv^2+\frac12(2m)\frac{v^2}{2}=mv^2+\frac{mv^2}{2}=\frac32mv^2$. Option (d) forgets the kinetic energy of the third piece.

(b) $10^{-2}$ m

For a black hole the escape velocity must equal $c$: $\sqrt{\frac{2GM}{R}}=c$, so $R=\frac{2GM}{c^2}$. $R=\frac{2\times6.67\times10^{-11}\times5.98\times10^{24}}{(3\times10^8)^2}\approx8.9\times10^{-3}$ m $\approx10^{-2}$ m. The other options are orders of magnitude off the calculated value.

(d) 3.5

The trajectory $y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$ is identical for the same $\theta$ only if $\frac{g}{u^2}$ is the same. $\frac{g'}{3^2}=\frac{9.8}{5^2}$, so $g'=9.8\times\frac{9}{25}=3.53\approx3.5\ \mathrm{m\,s^{-2}}$. Option (b) comes from using the inverse ratio $\frac{25}{9}$ instead of $\frac{9}{25}$.

(c) 157 N

For a solid cylinder, $I=\frac12MR^2=\frac12\times50\times0.25=6.25\ \mathrm{kg\,m^2}$. Angular acceleration $\alpha=2\ \mathrm{rev\,s^{-2}}=4\pi\ \mathrm{rad\,s^{-2}}$. Torque $TR=I\alpha$, so $T=\frac{6.25\times4\pi}{0.5}=50\pi\approx157$ N. Option (b), $78.5$ N, results from using $\alpha=2\pi$ rad/s$^2$, forgetting that one revolution is $2\pi$ rad and the value is $2$ rev/s$^2$.

(b) $x=\frac{m_2L}{m_1+m_2}$

Work needed is $\frac12I\omega_0^2$, so it is least when the moment of inertia $I=m_1x^2+m_2(L-x)^2$ is least. $\frac{dI}{dx}=2m_1x-2m_2(L-x)=0$ gives $x=\frac{m_2L}{m_1+m_2}$, which is the centre of mass (second derivative $2(m_1+m_2)>0$, a minimum). Option (c) is the distance of the centre of mass from the other end.

(c) $7.76\ \mathrm{km\,s^{-1}}$

Orbital speed $v=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{gR^2}{R+h}}$ with $R+h=6.63\times10^6$ m. $v=\sqrt{\frac{9.8\times(6.38\times10^6)^2}{6.63\times10^6}}\approx7.76\times10^3\ \mathrm{m\,s^{-1}}$. Option (d) would be the answer if $g$ were wrongly taken at the satellite height without scaling.

(a) $t=\frac{\pi}{\omega}$

Orthogonal vectors have zero dot product: $\vec A\cdot\vec B=\cos\omega t\cos\frac{\omega t}{2}+\sin\omega t\sin\frac{\omega t}{2}=\cos\frac{\omega t}{2}$. Setting $\cos\frac{\omega t}{2}=0$ gives $\frac{\omega t}{2}=\frac{\pi}{2}$, so $t=\frac{\pi}{\omega}$. At $t=0$ (option b) both vectors equal $\hat i$, so they are parallel, not orthogonal.

(c) $6.66\ \mathrm{kg\,m^2\,s^{-2}}$

Initial speed $v = 54\ \mathrm{km\,h^{-1}} = 15\ \mathrm{m\,s^{-1}}$, so $\omega_0 = v/r = 15/0.45 = 33.3\ \mathrm{rad\,s^{-1}}$. Angular deceleration $\alpha = \omega_0/t = 33.3/15 = 2.22\ \mathrm{rad\,s^{-2}}$. Braking torque $\tau = I\alpha = 3\times2.22 = 6.66$ N m.

(c) $\dfrac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\dfrac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}$

For a collision, A must reach B, so the relative velocity of A with respect to B, $\vec v_1-\vec v_2$, must point along the line from A to B, i.e. along $\vec r_2-\vec r_1$. Hence the unit vector of $\vec r_1-\vec r_2$ equals the unit vector of $\vec v_2-\vec v_1$. Options (a) and (d) involve angular momentum or dot products and say nothing about meeting, and (b) is not required.

(c) $-1$

Angular momentum about the origin is conserved only if the torque $\vec r\times\vec F$ is zero, i.e. $\vec F$ is parallel to $\vec r$. Since $\vec r = 2\hat i-6\hat j-12\hat k$ and $\vec F=\alpha\hat i+3\hat j+6\hat k$, the components must be in the same ratio: $\alpha/2 = 3/(-6) = 6/(-12) = -1/2$, giving $\alpha=-1$.

(c) 1.70

Power = pressure $\times$ volume flow rate. $P = \rho g h = 13.6\times10^3\times10\times0.15 = 2.04\times10^4$ Pa. Flow rate $Q = 5\times10^{-3}\ \mathrm{m^3}/60\ \mathrm{s} = 8.33\times10^{-5}\ \mathrm{m^3/s}$. Power $= PQ \approx 1.70$ W.

(d) $20\ \mathrm{m\,s^{-1}}$

After the collision half the energy is lost and the ball rises 20 m, so the kinetic energy just after impact is $mg(20)$, and just before impact it is twice that: $\tfrac12 mv^2 = 2mg(20)$, giving $v^2 = 800$. Before that, falling 20 m from $v_0$: $v^2 = v_0^2+2gh$, so $v_0^2 = 800-400=400$ and $v_0=20\ \mathrm{m\,s^{-1}}$. Forgetting the factor 2 would give a wrong answer such as 10 m/s.

(d) $\dfrac{VR^2}{nr^2}$

By the equation of continuity, the flow rate in the tube equals the total flow rate through the holes: $\pi R^2 V = n\,\pi r^2\,v$. So the ejection speed is $v = \dfrac{VR^2}{nr^2}$.

(c) $1,-1,-1$

Take $v_c = \eta^x\rho^y r^z$. Dimensions: $[\eta]=ML^{-1}T^{-1}$, $[\rho]=ML^{-3}$, $[r]=L$, $[v]=LT^{-1}$. Equating powers of $T$ gives $x=1$; of $M$ gives $x+y=0$ so $y=-1$; of $L$ gives $-x-3y+z=1$, so $z=-1$. Thus $v_c = \eta/(\rho r)$.

(d) water rises upto the top of capillary tube and stays there without overflowing.

The height $h=\dfrac{2T\cos\theta}{\rho g r}$ is fixed by the radius of curvature of the meniscus. If the tube is shorter than $h$, the meniscus simply adjusts: its radius of curvature increases so that $R=r/\cos\theta$ becomes larger and the pressure balance still holds at the lower height. So the water stays at the top without overflowing, since overflowing would need a continuous energy source.

(b) the acceleration of S is always directed towards the centre of the earth.

Gravity is a central force, so the acceleration always points to the earth's centre. Linear momentum changes (speed varies, being larger near perigee), so (a) is wrong. Angular momentum is conserved in both magnitude and direction because the torque is zero, so (c) is wrong. The total mechanical energy is constant in a conservative field, so (d) is wrong.

(d) 2 : 1

Both wires must stretch equally for the lower ends to be level. Extension $\Delta l = \dfrac{Fl}{AY}$, with $l$ and $A$ the same, so $F\propto Y$. Hence $F_{steel}:F_{brass}=Y_{steel}:Y_{brass}=2:1$.

(a) Magnitude of the velocity of particle is 8 meter/second

The particle moves on a circle of radius $R=4$ m with $\omega=2\pi$ rad/s, so the speed is $v=R\omega=4\times2\pi=8\pi\approx25.1\ \mathrm{m\,s^{-1}}$, not 8 m/s. The other statements are true: the path is a circle, the acceleration $\vec a=-\omega^2\vec R$ points toward the centre, and its magnitude is $v^2/R$.

(d) 0.6 and 0.5

The box begins to slip at $30^\circ$, so $\mu_s=\tan30^\circ\approx0.58\approx0.6$. Sliding 4.0 m in 4.0 s from rest: $s=\tfrac12at^2$ gives $a=0.5\ \mathrm{m\,s^{-2}}$. Then $a=g\sin30^\circ-\mu_k g\cos30^\circ$, i.e. $0.5=5-8.66\mu_k$, so $\mu_k\approx0.52\approx0.5$.

(a) $\sqrt{5gR}$

At the top of the loop the minimum condition is $mg=\dfrac{mv_{top}^2}{R}$, so $v_{top}^2=gR$. Energy conservation from bottom to top (height $2R$): $\tfrac12 mv^2=\tfrac12 mv_{top}^2+mg(2R)$, giving $v^2=gR+4gR=5gR$, so $v=\sqrt{5gR}$.

(c) $90^\circ$

$|\vec A+\vec B|^2=A^2+B^2+2AB\cos\theta$ and $|\vec A-\vec B|^2=A^2+B^2-2AB\cos\theta$. Setting them equal gives $4AB\cos\theta=0$, so $\cos\theta=0$ and $\theta=90^\circ$.

(b) 2600 km

With $r$ measured from the earth's centre: potential $V=-\dfrac{GM}{r}$ and $g=\dfrac{GM}{r^2}$, so $\dfrac{|V|}{g}=r=\dfrac{5.4\times10^7}{6.0}=9\times10^6$ m $=9000$ km. Height $h=r-R_e=9000-6400=2600$ km.

(d) Velocity is perpendicular to $\vec r$ and acceleration is directed towards the origin.

$\vec v=\dfrac{d\vec r}{dt}=\omega(-\sin\omega t\,\hat x+\cos\omega t\,\hat y)$, and $\vec v\cdot\vec r=0$, so $\vec v\perp\vec r$. $\vec a=-\omega^2(\cos\omega t\,\hat x+\sin\omega t\,\hat y)=-\omega^2\vec r$, which points toward the origin (uniform circular motion).