Question

A truck with a mass of 1360 kg and moving with a speed of 16.5 m/s rear-ends a 805 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answer 1

Explanation:

Applying the law of conservation of momentum,

Momentum before collision = Momentum after collision.

$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$

$1810 \mathrm{~kg} \times 16 \frac{\mathrm{m}}{\mathrm{s}}+0=1810$ $\mathrm{~kg} \times \mathrm{v}_{1}+673 \times v_{2}$

$28960=1810 \mathrm{~kg} \times \mathrm{v}_{1}+673 \times v_{2}$

For an elastic collision,

Relative velocity of approach = Relative velocity of separation.

$16 \frac{\mathrm{m}}{\mathrm{s}}=\mathrm{v}_{2}-\mathrm{v}_{1}$

$\mathrm{v}_{2}=v_{1}+16 \quad \ldots \ldots \ldots \ldots(\mathrm{b})$

Using the value of $v_{2}$ in equation (a)

$28960 &=1810 \mathrm{v}_{1}+673\left(\mathrm{v}_{1}+16\right)$

$=1810 \mathrm{y}+673 \mathrm{v}_{1}+10768$

$28960-10768 &=2483 \mathrm{v}_{1}$

$18192 &=2483 \mathrm{v}_{1}$

$v_{1} &=7.33 \frac{\mathrm{m}}{\mathrm{s}}$

Using the value of $v_{1}$ in equation (b)

$v_{2} &=7.33+16$

$=23.33 \frac{\mathrm{m}}{\mathrm{s}}$

Therefore the speed of truck after collision was found to be $7.33 \frac{\mathrm{m}}{\mathrm{s}}$

And the speed of car after collision was found to be $23.33 \frac{\mathrm{m}}{\mathrm{s}}$