Question

Ball 1 with an initial speed of 14 m/s has a perfectly elastic collision with ball 2 that is initially at rest. After, the speed of ball 2 is 21 m/s. What will be the speed of ball 2 if the initial speed of ball 1 is doubled

Answer 1

"42 m/s" is the appropriate answer.

Step-by-step explanation:

Given:

Initial speed of Ball 1,

= 14 m/s

Now,

If the initial speed of ball 1 is doubled, then the speed of ball 2 will be:

⇒
`(V_(fx))_2=(2m_1)/(m_1+m_2)(v_1x_1)`

hence,

Doubling (
`v_1x_1`) will double
`(V_(fx))_2`,

⇒
`(V_(fx))_2=2* 21`

`=42 \ m/s`