A 15g bullet strikes and becomes embedded in a 1.24kg 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 0.28, and impact drives the block a distance of 11.0m before it comes to rest.

a) What was the muzzle speed of the bullet?

b) What is the deceleration of the block?

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In the situation described in the problem a considerable amount of kinetic energy of the bullet emerging out of the gun will be used up for the bullet to get embedded in the block of wood. In all probability, in a real situation, most of the energy of the bullet will be used up this way. However, the the question provides no data on the energy used up this way. Therefore to solve the problem we assume that the entire kinetic energy of the bullet has been used up in moving the block of wood horizontally against force of friction.

Further it is assumed that the block starts moving with the speed of muzzle speed of bullet just as it is hit by it and then decelerates at uniform rate till it comes to a stop.

Given:

Mass of bullet = m1 = 15 g = 0,015 kg

Mass of wood block = m2 = 1.24 kg

Coefficient of friction = M = 0.28

Distance moved by block = d = 11.0 m

We know acceleration due to gravity = g = 9.81 m/s^2

Calculating muzzle speed of bullet:

The normal force (fn) exerted by the block on horizontal surface is given by:

fn = m2*g

And frictional force of block (f) against the horizontal surface is given by:

f = fn*M = m2*g*M

Work done or energy expended (e) in moving the block is given by:

e = f*d = m2*g*M*d = 1.24*9.81*0.28*11 = 37.466352 J

This energy is supplied by the kinetic energy of bullet which is given by the formula

Kinetic energy of bullet = e = 1/2*m1*v^2 = 1/2*0.015*v^2 = 0.0075v^2

Where v = muzzle speed of the bullet

Therefore:

0.0075v^2 = 37.466352

Or: v = (37.466352/0.0075)^1/2= 70.6789 m/s

Calculating declaration of block:

deceleration = a = (v^2)/(2*d) = 70.6789^2/(2*11) = 227.0688 m/s^2

Answer:

Muzzle speed of bullet = 70.6789 m/s

Deceleration of block = 227.0688 m/s^2

We pressume that the kinetic energy partly dissipated in work done in pushing the block after the bullet hitting and also partly used up in getting ebedded in the block. Mind that getting embedded , the sound of hit heat or deshape also require energy.So the impact is not elastic and under this inelastic collision the solution is as below.

Le v1 be the velocity of the bullet of mass m1 and v2 be the velocity of the block of mass m2 after hit.

Therefore, the kinetic energy of of the block and bullet together =(1/2)(m1+m2)v2^2 which is used up in pushing the block with aforce of (m1+m2)g* coefficient of Kinetic friction, M* times the displacement, s of the block after hit by bullet = (m1+m2)g*M*s. Therefore,

(1/2)(m1+m2) v2^2 = (m1+m2)g*M*s

So, v2^2 = 2g*M*s .......................(1)

So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/s

Therefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M

=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.

Since the momentums are conserved,

m1v1 = (m1+m2)v2 or

v1 = (m1+m2)v2 / m1 , substituting m1= 15 g or 15/1000 kg = 0.015 kg, m2 = 1.24 kg, v2 = sqrt(60.4296) , we get,

v1=(0.015+1.24)sqrt(60.4296)/(0.015)

= 650.3952 m/s is the speed of the bullet.

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