Question

At a crossing a truck travelling towards the north collides with a car travelling towards
the east. After the collision the car and the truck stick together and move off at an angle
of 30 º east of north. If the speed of the car before the collision was 20 ms−1, and the
mass of the truck is twice the mass of the car, calculate the speed of the truck before
and after the collision.

Answer 1

Agree that car’s speed before/after collision is v_1/u, mass m, speed of the truck before/after collision is u_1/u, mass 2m. Place 0X axis eastward, 0Y northward, then apply momentum conservation principle for 0X and 0Y axes:
0X:mv_1=(m+2m)u cos⁡〖30°,〗
0Y:2mu_1=(m+2m)u sin⁡〖30°,〗
therefore,
u_1=v_1/2 tan⁡〖30°〗=20/2·1/√3=5.77 m/s,
u=2/(6 sin⁡〖30°〗 ) v_1 tan⁡〖30°〗=2/(6·0.5)·20·1/√3=7.70 m/s.