# A 15g bullet strikes and becomes embedded in a 1.25kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction (µ) between the block and the surface is 0.3, and impact drives the block a distance of 10.0m before it comes to rest. What was the muzzle speed of the bullet?

Using the law of conservation of momentum, total momentum, before and after the impact stays the same. That is

MbVb + MwVw = (Mb + Mw)V

where, Mb and Mw are the masses of the bullet and wood block; Vb and Vw are velocities of bullet and wood block and V is the velocity of combined system (after bullet gets embedded in the block).

The friction force can be calculated as:

F = `mu` (Mb + Mw)g

and the work done by friction force is calculated as:

W = `mu` (Mb + Mw)g x distance

This work done by friction force opposes the motion of bullet + block system (and is balanced by its kinetic energy),

Hence, `1/2` (Mb + Mw) `V^2`  = `mu` (Mb + Mw) g x distance

or, `V^2`  = 2`mu` g x distance = 2 x 0.3 x 9.81 x 10

or, Initial velocity of the wood + bullet system, V = 7.67 m/s

and using this value in the first equation, we get

0.015 x Vb + 1.25 x 0 = (0.015 + 1.25) x 7.67

thus, Vb = 646.84 m/s

Hence the muzzle velocity of bullet is 646.84 m/s.

Hope this helps.

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