Question

problem reference 7.2
at an intersection, a yellow subcompact car with mass 950 kg traveling east collides with a red truck with mass 1900 kg that is traveling north. the two vehicles stick together as a result of the of the collision, and the wreckage slides at 16.0 m/s in the direction 66.0° north of east. the collision occurs during a heavy rainstorm; ignore friction forces between the vehicles and the wet road.
what is the speed of the yellow car before the collision?
24.0 m/s
19.5 m/s
48.0 m/s
43.8 m/s

Explanation:

Step1: Apply conservation of momentum in x - direction

The initial momentum in the x - direction is \(p_{ix}=m_{car}v_{car}\) (since the truck has no initial momentum in the x - direction). The final momentum in the x - direction is \(p_{fx}=(m_{car} + m_{truck})v_f\cos\theta\). By conservation of momentum \(p_{ix}=p_{fx}\), so \(m_{car}v_{car}=(m_{car} + m_{truck})v_f\cos\theta\).

Step2: Solve for \(v_{car}\)

We know \(m_{car} = 950\space kg\), \(m_{truck}=1900\space kg\), \(v_f = 16.0\space m/s\), \(\theta = 66.0^{\circ}\).
Substitute the values into the formula \(v_{car}=\frac{(m_{car} + m_{truck})v_f\cos\theta}{m_{car}}\)

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Answer:

19.5 m/s