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

Answer:

"42 m/s" is the appropriate answer.

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:

⇒ $(V_{fx})_2=\frac{2m_1}{m_1+m_2}(v_1x_1)$

hence,

Doubling ($v_1x_1$) will double $(V_{fx})_2$,

⇒ $(V_{fx})_2=2\times 21$

              $=42 \ m/s$