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Q.

A block of mass m rests on a rough horizontal surface (Coefficient of friction is $μ$ ). When a bullet of mass $\frac{m}{2}$ strikes horizontally, and get embedded in it, the block moves a distance ‘d’ before coming to rest. The initial velocity of the bullet is $k \sqrt{2 μgd}$ , then the value of k is

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a

2

b

3

c

4

d

5

answer is B.

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Detailed Solution

Let initial velocity of the bullet be v.
By linear momentum conservation
$\frac{m}{2} v = (\frac{m}{2} + m) v_{1}$
$(v_{1} = combined velocity)$
$v_{1} = \frac{v}{3} . ..................... (1)$
Retardation $= μg$
$0 = {(\frac{v}{3})}^{2} - 2 μgd \Rightarrow v = 3 \sqrt{2 μgd}$

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A block of mass m rests on a rough horizontal surface (Coefficient of friction is μ). When a bullet of mass m2 strikes horizontally, and get embedded in it, the block moves a distance ‘d’ before coming to rest. The initial velocity of the bullet is k2μgd, then the value of k is