Past exam of the physics 1 paper of the University of Oxford / 2016 resit / 9 / Solution

Let $\lambda =1{\mathrm{s}}^{-1}$ be the launch rate and $\mu =\lambda m$. Treat many small impacts as a continuous momentum flux, justified by $m\ll M$. Write $x(t)$ for truck displacement from its initial position and $U=\dot{x}$, with $t=0$ at the first impact. First neglect vertical motion, so every projectile has horizontal velocity $v$.

A fixed launch rate is different from a fixed impact rate on a receding truck. Projectiles reaching the truck at time $t$ were launched at retarded time $s=t-x(t)/v$, after an irrelevant constant initial-flight offset. The moving-target interception rate is thereforeFor an elastic collision with a much heavier truck, each ball transfers momentum approximately $2m(v-U)$. The elastic-ball truck keeps mass $M$, soSolving from rest and integrating once more givesThe rebound reverses the ball's motion relative to the truck; its laboratory velocity $2U-v$ can still be forwards once $U>v/2$. The truck approaches $v$ asymptotically, while its decreasing impact rate prevents exponential approach in laboratory time. The approximation omits the velocity jumps of individual finite-mass impacts.

For sticking snowballs, let $A(t)$ be the instantaneous truck-plus-snow mass, with $A(0)=M$. The variable-mass system obeysIntegrating the second equation gives $AU=v(A-M)$. Therefore $A\dot{A}=\mu M$, and with $z=\sqrt{1+2\mu t/M}$,Retaining the accumulated mass is necessary even though each individual snowball is small.

For the range calculation, take the intended idealization of horizontal launch from height $h$, no air resistance, and interception possible until the projectile reaches the ground. Its flight time is $\tau =\sqrt{2h/g}$ and horizontal range is $D=v\tau$. If the initial truck distance from the boy is $X_0<D$, substitute $x=D-X_0$ in the last expression. The speed at the range limit isFor an initially nearby truck, $X_0=0$ in this point-target model and $a=\sqrt{2\mu \tau /M}$. A different launch angle or finite interception height changes the available range and must be specified separately.

If instead one idealizes one impact each second, regardless of the truck's changing position, the answers are different: elastic balls give $U=v(1-e^{-2\mu t/M})$ and $x=v[t-M(1-e^{-2\mu t/M})/(2\mu )]$; sticking balls give $U=v\mu t/(M+\mu t)$ and $x=vt-(Mv/\mu )\ln (1+\mu t/M)$. Those expressions assume controlled arrival times, whereas the boxed results respect the stated fixed launch rate.