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$$^{-1}$$ and makes an elastic collision with it. Thereafter $$B$$ makes a completely inelastic collision with $$C$$. All motions occur along the same straight line. The final speed of $$C$$ is:

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We need to find the final speed of object $$C$$ after a series of two successive collisions on a frictionless horizontal surface.

1. Identify the Given Parameters

From the problem statement  page:

• Mass of object $$A$$ ($$m_A$$) = $$m$$
• Mass of object $$B$$ ($$m_B$$) = $$2m$$
• Mass of object $$C$$ ($$m_C$$) = $$2m$$
• Initial velocity of $$A$$ ($$u_A$$) = $$9\text{ m s}^{-1}$$
• Objects $$B$$ and $$C$$ are initially at rest ($$u_B = 0$$, $$u_C = 0$$).

2. First Collision: Elastic Collision Between A and B

Since the collision between $$A$$ and $$B$$ is perfectly elastic ($$e = 1$$), we can find the final velocity of $$B$$ ($$v_B$$) immediately after this impact using the standard 1D elastic collision formula:

$$v_B = \frac{2m_A u_A}{m_A + m_B} + \frac{(m_B - m_A)u_B}{m_A + m_B}$$

Since $$u_B = 0$$, the second term drops out. Substituting the mass values:

$$v_B = \frac{2(m)(9)}{m + 2m} = \frac{18m}{3m} = 6\text{ m s}^{-1}$$

3. Second Collision: Completely Inelastic Collision Between B and C

After the first collision, $$B$$ moves toward $$C$$ with a speed of $$6\text{ m s}^{-1}$$. When $$B$$ collides with $$C$$, they stick together because the collision is completely inelastic.

Applying the principle of conservation of linear momentum for this pair:

$$m_B v_B + m_C u_C = (m_B + m_C) v_f$$

Since $$C$$ was at rest ($$u_C = 0$$), the equation becomes:

$$(2m)(6) + 0 = (2m + 2m) v_f$$

$$12m = 4m \cdot v_f$$

Solving for the combined final velocity ($$v_f$$):

$$v_f = \frac{12m}{4m} = 3\text{ m s}^{-1}$$

Conclusion

The final speed of object $$C$$ is 3 $$\text{m s}^{-1}$$.