Question

Two blocks sliding on a frictionless surface with the less massive block moving twice as fast as the more massive block. Eventually, the smaller block catches up and collides with it, and sticks. What is the speed of the two blocks after the collision given the speed of the more massive block is ?????????

Answer 1

Explanation:

For the given figure, we will apply the law of conservation of momentum as follows.

          $m(2v_{1}) + 3m(v_{1}) = (m + 3m)v_{f}$

                  $5mv_{1} = 4mv_{f}$

                  $v_{f} = \frac{5}{4}v_{1}$

Therefore, we can conclude that speed of the two blocks after the collision is $\frac{5}{4}v_{1}$.

Answer 2

Answer:

Explanation:

Let the mass of lighter block is m and then the mass of massive block is M.

velocity of massive block is v and the velocity of lighter block is 2v.

Let the speed of the combined block is v'

Use conservation of momentum

m x 2v + M x v = ( M + m) x v'

v ( M + 2m) = ( M + m) v'

$v' = \frac{M+2m}{M+m}v$