Question

A 0.143 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s. It has a head-on collision with a 0.298 kg glider that is moving to the left with a speed of 2.25 m/s. Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.143kg glider.Find the magnitude of the final velocity of the 0.298kg glider.

Answer 1

`v_(2f) = -0.207 m/s`

`v_(1f) = -3.36 m/s`

Step-by-step explanation:

As we know that system of two gliders are moving on frictionless track

so momentum conservation is applied on it

So we will have

`m_1v_(1i) + m_2v_(2i) = m_1 v_(1f) + m_2v_(2f)`

so we will have

`0.143(0.900) + 0.298(-2.25) = 0.143 v_(1f) + 0.298v_(2f)`

`-0.5418 = 0.143 v_(1f) + 0.298 v_(2f)`

now we know that collision is elastic so we have

`e = (v_(2f) - v_(1f))/(0.900 + 2.25)`

`v_(2f) - v_(1f) = 3.15`

so by solving above equations we have

`v_(2f) = -0.207 m/s`

`v_(1f) = -3.36 m/s`