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

Answer on Question #72637 Physics / Mechanics | Relativity

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 $\alpha = 30{}^{\circ}$ east of north. If the speed of the car before the collision was $v = 20 \, \mathrm{ms}^{-1}$, and the mass of the truck is twice the mass of the car $m$, calculate the speed of the truck before and after the collision.

Solution:

Let us apply momentum conservation law

$$2mv_1 = 3mv \sin 30{}^{\circ}$$ $$m \times 20 = 3mv \cos 30{}^{\circ}$$

Thus

$$v_1 = 10 \tan 30{}^{\circ} = 5.77 \, \frac{\mathrm{m}}{\mathrm{s}}$$ $$v = \frac{2}{3} \frac{v_1}{\sin 30{}^{\circ}} = 7.7 \, \frac{\mathrm{m}}{\mathrm{s}}$$

Answers: $7.7 \frac{\mathrm{m}}{\mathrm{s}}$, $5.77 \frac{\mathrm{m}}{\mathrm{s}}$.

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