Question

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car B at −10 m/s when they collide head–on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision? −8 m/s −10 m/s 0 m/s 10 m/s

Answer 1

-10 m/s

Explanation:

We have the following information:

`\begin{gathered} m_A=m_B=100\text{ kg} \\ v_(A1)=8\text{ m/s} \\ v_(B1)=-10\text{ m/s} \\ v_(B2)=8\text{ m/s} \end{gathered}`

Now by momentum conservation equation, we have:

`\begin{gathered} m_A\cdot v_(A1)+m_2\cdot v_(B1)=m_A\cdot v_(A2)+m_2\cdot v_(B2) \\ \text{replacing and solving for }v_(A2)\colon \\ 100\cdot8+100\cdot-10=100\cdot v_(A2)+100\cdot8 \\ 8-10=v_(A2)+8 \\ v_(A2)=8-10-8 \\ v_(A2)=-10 \end{gathered}`

so car A will move with a velocity of -10 m/s