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 # 73682, Physics- Mechanics -Relativity:

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 $30{}^{\circ}$ 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.

Solution: Let us consider mass of car is m and $V_{i}$ be the velocity before collision. And after collision together velocity is v. So, as per question given mass of the truck is 2m and $V_{t}$ be its velocity before collision.

Let us apply momentum conservation law,

$2\mathrm{m}\times \mathrm{V}_{\mathrm{t}} = (2\mathrm{m} + \mathrm{m})\mathrm{v}\cos 30{}^{\circ}$ and $\mathrm{m}\times 20 = (2\mathrm{m} + \mathrm{m})\mathrm{v}\sin 30{}^{\circ}$

Or, $V_{t} = 1.5$ v (0.866) ...(1) or, $v = 13.33$ m/s (as $\sin 30{}^{\circ} = 0.5$ )

(2)

Put the value of $\mathbf{v}$ from equation (2) to equation (1), we get,

$V_{t} = 17.32 \, \text{m/s}$ .

Answer: Speed of the truck before and after collisions are $17.32 \, \text{m/s}$ and $13.33 \, \text{m/s}$ respectively.

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