Given. Car A: weight 2000 lb, travelling north (velocity $v_A$). Car B: weight 3600 lb, travelling east (velocity $v_B$). After a perfectly plastic (stick-together) collision, the wreckage skids at 40° north of east. One car was at exactly 35 mi/hr; the other was faster.
Given data
Quantity
Value
Weight of car A (north)
2000 lb
Weight of car B (east)
3600 lb
Skid direction after impact
40° north of east
Claimed speed limit
35 mi/hr
Find. a) Which car was actually at 35 mi/hr; b) the speed of the other car.
Figure 4 — top view: momentum components of A (north) and B (east) combine into the observed post-impact skid direction.
Approach. Momentum is conserved in both x and y during the (perfectly plastic) collision. Since car A's momentum is entirely in the y-direction and car B's is entirely in the x-direction, the ratio of the combined momentum components fixes the ratio $v_A/v_B$ via the observed 40° skid angle. Test each car at 35 mi/hr and keep the case where the OTHER car comes out faster (consistent with both drivers blaming the other for speeding).
Momentum-ratio relation. With $m=W/g$ (same g for both, so weights may be used directly), momentum conservation in x and y gives
$$\tan(40^\circ) = \dfrac{p_y}{p_x} = \dfrac{W_A v_A}{W_B v_B}$$
Test: assume car A was at 35 mi/hr.
$$v_B = \dfrac{W_A v_A}{W_B\tan(40^\circ)} = \dfrac{2000(35)}{3600\tan(40^\circ)} = 23.17\ \text{mi/hr}$$
This makes B SLOWER than the limit — inconsistent with "each claimed the other was faster." Reject.
Test: assume car B was at 35 mi/hr.
$$v_A = \dfrac{W_B v_B\tan(40^\circ)}{W_A} = \dfrac{3600(35)\tan(40^\circ)}{2000} = \boxed{52.86\ \text{mi/hr}}$$
This makes A faster than the limit — consistent with the scenario. Accept: car B was at the 35 mi/hr limit; car A was speeding.
Check. With $v_B=35$, $v_A=52.86$: $\theta=\arctan\!\big(W_Av_A/(W_Bv_B)\big)=\arctan(2000\cdot52.86/(3600\cdot35))=40.0^\circ$ — matches the given skid direction exactly.
