Question

Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2m and 2m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is : 

 

• A. 6 m/s
• B. 3 m/s
• C. 4 m/s
• D. 9 m/s

Hint:

For a 1D elastic collision of mass $m_1$ (velocity $u_1$) with a stationary mass $m_2$, the final velocity of $m_2$ is $v_2 = \frac{2m_1}{m_1+m_2}u_1$. This shortcut could be used for the first step.

Answer 1

The Correct Option is B

To solve this problem, we will break it down into two parts: the elastic collision between objects A and B, and the completely inelastic collision between objects B and C.

1. Elastic Collision Between A and B:
   • Initial velocity of A, \(u_A = 9 \, \text{m/s}\), and initial velocity of B, \(u_B = 0\).
   • Mass of A, \(m_A = m\), and mass of B, \(m_B = 2m\).
   • Using the formula for elastic collisions, the final velocities \(v_A\) and \(v_B\) are given by: v_A = \frac{(m_A - m_B)u_A + 2m_Bu_B}{m_A + m_B} v_B = \frac{(m_B - m_A)u_B + 2m_Au_A}{m_A + m_B}
   • Substituting the values: v_A = \frac{(m - 2m) \times 9 + 2 \times 2m \times 0}{m + 2m} = \frac{-m \times 9}{3m} = -3 \, \text{m/s} v_B = \frac{(2m - m) \times 0 + 2 \times m \times 9}{m + 2m} = \frac{18m}{3m} = 6 \, \text{m/s}
2. Completely Inelastic Collision Between B and C:
   • Now, B moves with \(v_B = 6 \, \text{m/s}\) and collides with C, which is initially at rest.
   • Mass of C, \(m_C = 2m\).
   • In a completely inelastic collision, both objects move together with a common velocity \(v\).
   • Using the conservation of momentum: (2m \times 6) + (2m \times 0) = (2m + 2m) \times v 12m = 4m \times v v = \frac{12m}{4m} = 3 \, \text{m/s}

The final speed of C after both collisions is 3 m/s.