Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 6

After the push, the forces are weight $mg$ downward, normal reaction $N$ upward and sliding friction $F=\mu N$ opposite the velocity. There is no continuing applied push. The table has constant slopeVertical balance gives $N=mg$, and horizontal motion gives $ma=-\mu mg$. ThusThe friction magnitude is $0.0300\mathrm{N}$.

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.

Let $m_p=0.250\mathrm{kg}$ and let $m_w$ be the hanging mass. Taking up-slope and downward respectively as positive,Eliminate the common tension to obtainHere $\theta =10^{\circ }$, $\mu =0.1$ and $a=1.60\mathrm{m}{\mathrm{s}}^{-2}$. The pulley changes the direction of the string force; in the ideal model it does not change its magnitude.

The required constant deceleration follows from $v^2=u^2+2as$:Both gravity and sliding friction oppose ascent, so $g(\sin \theta +0.1\cos \theta )=5.76$. WriteThe physical incline solution is $\theta =30.0^{\circ }$ more precisely $30.0389^{\circ }$. The combined mass cancels from the equation. The top of the track is consequently $H=2\sin \theta =1.00117\mathrm{m}$ above its base.

Let $x=0.400\mathrm{m}$ be the compression and $m_t=0.550\mathrm{kg}$ the combined mass. Starting from rest at the top, it falls through $H+x$ before maximum compression. HenceThe gravitational drop during compression must be included. Equating $kx$ to the weight would instead describe a static equilibrium, not the turning point after a fall.

Immediately after the parcel leaves, the spring still stores $\frac{1}{2}k{x}^2$, but only the $m_s=0.300\mathrm{kg}$ sledge is launched. Choose gravitational potential zero at the ramp base. Its initial position is $-x$, so conservation of energy to the apex givesThereforeabove the ramp base. Its upward travel from the compressed position is $H_s+x=2.56882\mathrm{m}$, a different distance from the height above the base.

For actual complete bounce distances $D_1=4\mathrm{m}$ and $D_{n+1}=r{D}_n$, the geometric series givesA natural intended estimate compares the observed first round trip with the loss-free round trip from the initially compressed position,If the reduction factor is taken to be $r=D_1/D_{\mathrm{ideal}}=0.77857$, the answer is $D_{\mathrm{total}}\simeq 18.1\mathrm{m}$. The first term is the actual 4 m, not the hypothetical loss-free distance.

This numerical inference needs the stated comparison assumption. Strictly, knowing the first actual bounce distance and saying subsequent distances share a ratio does not independently specify that ratio. A change in spring compression also means that an energy-loss fraction is not automatically the same as a vertical-distance ratio.