A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at $13.0 \mathrm {~m} / \mathrm { s }$ in the same direction.If the second car has a mass of $790 \mathrm {~kg}$ and a speed of $15.1 \mathrm {~m} / \mathrm { s }$ right after the collision,what is the velocity of the first car after this sudden collision?
A)14.0 m/s
B)18.2 m/s
C)24.2 m/s
D)-14.0 m/s

Question

Quizplus · Accepted Answer

Using the conservation of momentum, the velocity of the first car after the collision can be calculated as follows: $ (328 \, \text{kg} \times 19.1 \, \text{m/s}) + (790 \, \text{kg} \times 13.0 \, \text{m/s}) = (328 \, \text{kg} \times v) + (790 \, \text{kg} \times 15.1 \, \text{m/s}) $. Solving for $ v $, we find $ v = 14.0 \, \text{m/s} $.

A 328-Kg Car Moving at 19 13.0 m/s13.0 \mathrm {~m} / \mathrm { s }13.0 m/s

A 328-Kg Car Moving at 19 $13.0 \mathrm {~m} / \mathrm { s }$