Question

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the first car moves in the same direction as before with a speed v/3. Find the final speed of the second car.

Answer 1

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$

$m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}$

Divide both side by $m_1$, then multiply by 6 we have

$6v = 2v + 3v_2$

$3v_2 = 4v$

$v_2 = \frac{4v}{3}$

So the final speed of the second car is 4/3 of the first car original speed