Question

On a vehicle crash simulation, a car of mass 1200 kg is travelling at a velocity of 40 mph in the easterly direction when it is hit by a truck of mass 3300 kg travelling at a velocity of 25 mph from the westerly direction. Assuming that the two vehicles become entangled calculate in m/s their combined velocity after the crash.
Take 1.61 km/hr = 1 mph

Answer 1

Answer on Question#53380 - Physics - Mechanics - Kinematics - Dynamics

On a vehicle crash simulation, a car of mass $m = 1200 \, \mathrm{kg}$ is travelling at a velocity of $v_{m} = 40 \, \mathrm{mph}$ in the easterly direction when it is hit by a truck of mass $M = 3300 \, \mathrm{kg}$ travelling at a velocity of $v_{M} = 25 \, \mathrm{mph}$ from the westerly direction. Assuming that the two vehicles become entangled calculate in $m/s$ their combined velocity after the crash.

Take $1.61 \, \mathrm{km/hr} = 1 \, \mathrm{mph}$

Solution:

Let's choose the east direction as the positive one. According to the law of conservation of momentum we obtain the resultant momentum of entangled vehicles

$$(M + m) \cdot v = M \cdot v_{M} - m \cdot v_{m},$$

where $v$ – is the speed of the vehicles after collision. Therefore

$$v = \frac{M \cdot v_{M} - m \cdot v_{m}}{(M + m)} = \frac{3300 \, \mathrm{kg} \cdot 25 \, \mathrm{mph} - 1200 \, \mathrm{kg} \cdot 40 \, \mathrm{mph}}{3300 \, \mathrm{kg} + 1200 \, \mathrm{kg}} = 7.67 \, \mathrm{mph} = 3.43 \, \frac{\mathrm{m}}{\mathrm{s}}$$

**Answer**: $3.43 \, \frac{\mathrm{m}}{\mathrm{s}}$ in the easterly direction.

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