First slide

Collisions

Question

A wooden block of mass M rests on a horizontal surface. A bullet of mass m moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance x on the surface. If the coefficient of friction between wood and the surface is $μ$ , the speed of the bullet at the time of striking the block is (where m is mass of the bullet)

Moderate

A

$\sqrt{\frac{2 Mg}{μm}}$
B

$\sqrt{\frac{2 μmg}{Mx}}$
C

$\sqrt{2 μgx} (\frac{M + m}{m})$
D

$\sqrt{\frac{2 μmx}{M + m}}$

Solution

Let speed of the bullet = v
Speed of the system after the collision = V
By conservation of momentum $mv = (m + M) V$
$\Rightarrow V = \frac{mv}{M + m}$
So the initial K.E. acquired by the system
$= \frac{1}{2} (M + m) V^{2} = \frac{1}{2} (m + M) {(\frac{mv}{M + m})}^{2} = \frac{1}{2} \frac{m^{2} v^{2}}{(m + M)}$
This kinetic energy goes against friction work done by friction = $μR \times x = μ (m + M) g \times x$
By the law of conservation of energy
$\frac{1}{2} \frac{m^{2} v^{2}}{(m + M)} = μ (m + M) g \times x \Rightarrow v^{2} = 2 μgx {(\frac{m + M}{m})}^{2}$
$∴ v = \sqrt{2 μgx} (\frac{M + m}{m})$

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