Final Speed of Object After Successive Elastic and Inelastic Collisions

To find the final speed of object C, we analyze the problem in two stages of collisions occurring along the straight line on the frictionless horizontal surface.

Stage 1: Elastic collision between object A and object B

• Mass of object A: $m_A = m$
• Mass of object B: $m_B = 2m$
• Initial velocity of object A: $u_A = 9\text{ m/s}$
• Initial velocity of object B: $u_B = 0\text{ m/s}$

Since the collision is perfectly elastic ($e = 1$), linear momentum and kinetic energy are conserved. The velocity of object B immediately after the collision, $v_B$, is given by the standard elastic collision formula:
$v_B = \frac{2m_A}{m_A + m_B}u_A + \frac{m_B - m_A}{m_A + m_B}u_B$

Substituting the given values into the formula:
$v_B = \frac{2m}{m + 2m}(9) + 0 = \frac{2}{3} \times 9 = 6\text{ m/s}$

Stage 2: Completely inelastic collision between object B and object C

• Mass of object B: $m_B = 2m$
• Velocity of object B before collision: $v_B = 6\text{ m/s}$
• Mass of object C: $m_C = m$
• Velocity of object C before collision: $u_C = 0\text{ m/s}$

Since the collision between B and C is completely inelastic, the two objects stick together and move with a common final velocity $v_f$.

Applying the law of conservation of linear momentum for the system consisting of B and C:
$m_B v_B + m_C u_C = (m_B + m_C) v_f$

Substitute the given values:
$(2m)(6) + 0 = (2m + m) v_f$
$12m = 3m v_f$
$v_f = \frac{12m}{3m} = 4\text{ m/s}$

Since object C moves together with object B after the collision, the final speed of object C is $4\text{ m/s}$.