A car heading north suddenly collides at an intersection with a truck of the same mass heading east. If they lock together and travel at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision, how fast was the car initially traveling?

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Quizplus · Accepted Answer

The momentum of the car before the collision is equal to the momentum of the car and truck after the collision. The momentum of the car before the collision is $mv_c$, where $m$ is the mass of the car and $v_c$ is the velocity of the car. The momentum of the car and truck after the collision is $(m+m)v_{ct}$, where $v_{ct}$ is the velocity of the car and truck after the collision. Since the car and truck lock together after the collision, their velocities are the same. Therefore, we have:$$mv_c=(m+m)v_{ct}$$Solving for $v_c$, we get:$$v_c=\frac{m+m}{m}v_{ct}$$$$v_c=2v_{ct}$$Since the car and truck are traveling at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision, we have:$$v_{ct}=28 \mathrm {~m} / \mathrm { s }\cos15°$$$$v_{ct}=26.6 \mathrm {~m} / \mathrm { s }$$Therefore, the car was initially traveling at:$$v_c=2v_{ct}=2(26.6 \mathrm {~m} / \mathrm { s })=53.2 \mathrm {~m} / \mathrm { s }$$

A Car Heading North Suddenly Collides at an Intersection with a Truck