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?  

Expert Answers

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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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