In a free space, a rifle of mass M shoots a bullet of mass m at a stationary block of mass M at a distance D away from it. When the bullet has moved through a distance d towards the block, the centre of mass of the bullet-block system is at a distance of

In a free space, a rifle of mass M shoots a bullet of mass m at a stationary block of mass M at a distance D away from it. When the bullet has moved through a distance d towards the block, the centre of mass of the bullet-block system is at a distance of

A
$\frac{(D - d) m}{M + m} from the bullet$
B
$\frac{md + MD}{M + m} from the block$
C
$\frac{2 md + MD}{M + m} from the block$
D
$\frac{(D - d) M}{M + m} from the bullet$

Solution:

Distance between bullet and block at this instant is $(D - d)$ .
Distance of CM from bullet

$= \frac{M (D - d) + m \times 0}{M + m} = \frac{M}{M + m} (D - d)$

Similarly, distance of CM from block $= \frac{m}{M + m} (D - d)$

Hence, option (4) is correct.

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