Bullet collision question

[imas_]

Homework Statement

Two bullets of mass 20 g move with same speed opposite to each other. At time, t=0 the bullets are flying towards each other with a separation of 20 m. At t=0.02 s, The bullets graze each other and both deflect through an angle of 6°. The only form of energy dissipation is in the form of heat which increases the temperature in the bullets by 50 °C. If the specific heat of each bullet is 500 J/kg K, then determine the value of coefficient of restitution.

Relevant Equations

Law Of Conservation Of Mechanical Energy
Specific Heat Capacity Equation
Newton's Law Of Collision

My attempt:

Relative velocity, v = s/t = 20/0.02 = 1000 m/s
Velocity of each bullet, u=v/2 =500 m/s

Now we have to find loss in energy,
We know, Q=mcΔT =(20/1000)(500)(50) = 500 J per bullet
Total energy loss= 2(500) = 1000 J

Now, I know that in the next step, we have to somehow correlate kinetic energy with Newton's law of collision but I don't get how to do it. Can someone please help?

 

[nasu]

Do you have a definition of the coefficient of restitution?

 

[Herman Trivilino]

Do you have a definition of the coefficient of restitution?

Usually it's the ratio of speed after to speed before.

 

[Herman Trivilino]

Relevant Equations: Law Of Conservation Of Mechanical Energy

Mechanical energy is conserved only if the collision is perfectly elastic. There is no change in temperature.

Newton's Law Of Collision

Is that the same thing as conservation of momentum?

 

[kuruman]

Is that the same thing as conservation of momentum?

Or maybe Newton's 3rd law?

 

[haruspex]

Usually it's the ratio of speed after to speed before.

Not quite, and this makes the question quite tricky. It is (minus) the ratio of the relative velocity components normal to the plane of contact.
It is possible the author intends some simplification based on the smallness of the deflection, but for a correct treatment we need to take the bullets as spheres and consider the geometry at the moment of impact.
Let the plane of contact make an angle $\alpha$ to the line of their original velocities. Resolve the incoming velocities in directions parallel to and normal to that line. Velocities parallel to it do not change. Write the equation for how the velocities normal to it change, given the coefficient of restitution. Then write the equation for the angle of departure.
You should get a quadratic in $\tan(\alpha)$.
Since $\alpha<6°$, maybe we can ignore the quadratic term.

 

[nasu]

Usually it's the ratio of speed after to speed before.

The question was for the OP.