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

Step 1: Elastic collision between A and B.
Let initial velocity of A be $u_A = 9$ m/s, and initial velocity of B be $u_B = 0$.
Masses are $m_A = m$ and $m_B = 2m$.
Let final velocities be $v_A$ and $v_B$.
By conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
$m(9) + 2m(0) = m v_A + 2m v_B \implies 9 = v_A + 2v_B$ (Eq. 1)
For an elastic collision, the coefficient of restitution e = 1.
$e = \frac{v_B - v_A}{u_A - u_B} \implies 1 = \frac{v_B - v_A}{9 - 0} \implies 9 = v_B - v_A$ (Eq. 2)
Adding (Eq. 1) and (Eq. 2):
$(v_A + 2v_B) + (v_B - v_A) = 9 + 9$
$3v_B = 18 \implies v_B = 6$ m/s.
So, after the first collision, B moves with a speed of 6 m/s.
Step 2: Completely inelastic collision between B and C.
Now B (mass 2m, velocity 6 m/s) collides with C (mass 2m, velocity 0).
In a completely inelastic collision, the objects stick together. Let their final common velocity be $v_f$.
By conservation of momentum: $m_B v_B + m_C u_C = (m_B + m_C)v_f$
$(2m)(6) + (2m)(0) = (2m + 2m)v_f$
$12m = (4m)v_f$
$v_f = \frac{12m}{4m} = 3$ m/s.
The final speed of the combined mass (B+C) is 3 m/s. Therefore, the final speed of C is 3 m/s.