Question

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with 5m/s in the same direction. They suffer an elastic collision and block B is now moving with 8m/s. What is the speed of block A after collision? 10m/s
8m/s
5m/s
13m/s
18m/s

Answer 1

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is $\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}$

Substituting values

    $3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s$

Speed of block A after collision = 10 m/s

Option A is the correct answer.