Collisions: notes and previous year questions

Answer: Option B

Two stages, each with its own law. The impact is over in an instant, so momentum carries you across it: $0.1 \times 400 = (0.1 + 3.9)v$, and the block and bullet move off together at $v = 10\ \mathrm{m/s}$. Then friction stops them over $20\ \mathrm{m}$, and the work–energy theorem gives $\mu Mg \times d = \tfrac{1}{2}Mv^{2}$ with $M = 4\ \mathrm{kg}$; the mass cancels, leaving $\mu = \dfrac{v^{2}}{2gd} = \dfrac{100}{2 \times 10 \times 20} = 0.25$. What you must not do is feed the bullet's $\tfrac{1}{2} \times 0.1 \times 400^{2} = 8000\ \mathrm{J}$ into the friction equation — the sticking collision throws away all but $200\ \mathrm{J}$ of it before the sliding even starts.

Answer: Option C

The blocks fly off at angles, but the question needs none of them. The collision is elastic, so the total kinetic energy is the same before and after — and kinetic energy is a plain number, with no direction to keep track of. $\tfrac{1}{2}Mv^{2} = \tfrac{1}{2}M\left(\dfrac{v}{3}\right)^{2} + \tfrac{1}{2}Mv'^{2}$ gives $v'^{2} = v^{2} - \dfrac{v^{2}}{9} = \dfrac{8}{9}v^{2}$, so $v' = \dfrac{2\sqrt{2}}{3}v$. Reaching for momentum is the trap here: it is a vector, and with angles unknown it would give two equations in too many unknowns. Energy does it in one line.

Answer: 5

Follow the first bob round. The minimum speed at the bottom for a full circle on a string is $\sqrt{5gl_{1}}$, and climbing through $2l_{1}$ leaves it $v^{2} = 5gl_{1} - 4gl_{1} = gl_{1}$ at the top — the bare minimum there. At the top it strikes an identical bob at rest, elastically, so they swap: the first bob stops and the second leaves with $\sqrt{gl_{1}}$. The second bob is at the lowest point of its own circle, where the minimum it needs is $\sqrt{5gl_{2}}$. So $gl_{1} = 5gl_{2}$ and $\dfrac{l_{1}}{l_{2}} = 5$. The trap is handing the second bob the first bob's launch speed $\sqrt{5gl_{1}}$, which gives a ratio of $1$ — it arrives at the top with far less.

Answer: Options A, C, E

In A, C and E every large force is internal — powder on bullet and bullet on rifle, piece on piece, trolley on trolley — so the total is kept. C is the subtle one: gravity is external and does act on the shell, but over $0.001\ \mathrm{s}$ its impulse $mg\Delta t$ is tiny next to the momentum the shell already carries, so the momentum just after the burst equals the momentum just before. B fails for exactly that reason stretched out: over $2\ \mathrm{s}$ gravity hands the stone $2mg$ worth of momentum. D fails because the system is the ball alone; the wall is outside it, and the wall's large push is what reverses the ball.

Answer: Option A

The block hit by the bouncing bullet. Call the bullet's mass $m$ and speed $u$, the block's mass $M$, and the bounce-back speed $v$. The lodged bullet hands over only its own momentum: $(M + m)V = mu$. The bouncing bullet also has to be turned round, and the block, pushing it back, takes an equal push forwards: $MV' = mu + mv$. So the block gets more momentum, and it is lighter than block-plus-bullet, so $V' > V$ and it swings higher. It is the same reason a ball bouncing off a wall gives it twice the impulse of one that stops dead. The trap is B, thinking the lodged bullet "gives all it has" and so gives the most; a bounce gives more than all. C forgets that the momentum the bullet carries away backwards has to be made up by the block. And D is not needed: the bounce speed sets how much higher, but any bounce at all, $v > 0$, makes $mu + mv$ larger than $mu$.

Answer: Option A

The motion has two stages, and each has its own law. The impact is a sticking collision: over that instant momentum is conserved and kinetic energy is not — most of it goes into splintered wood and heat. The swing is the reverse: the string pulls at right angles to the path and does no work, gravity is conservative, so mechanical energy is kept — but momentum is not, since gravity and the string are external forces acting for the whole swing. C mixes that up. B is the student's mistake, and it gets the bullet's speed far too low.