A bullet of mass m = 0.0010 kg embeds itself in a wooden block with mass M = 0.9

Question

A bullet of mass m = 0.0010 kg embeds itself in a wooden block with mass M = 0.999 kg, which then compresses a spring (k = 150 N/m ) by a distance x = 0.045 m before coming to rest. The coefficient of kinetic friction between the block and table is = 0.50.

(a)

What is the initial velocity (assumed horizontal) of the bullet?

(b)

What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

Explanation / Answer

m = 0.0010 kg
M = 0.999 kg
k = 150 N/m
x = 0.045 m
= 0.50

(a)
Using Energy conservation,
Initial Kinetic Energy = Final Spring potential Energy + Energy lost in friction
1/2 * (m+M)*vf^2 = 1/2 * k * x^2 + *(m+M)*g*x
1/2 * (0.0010+0.999) * vf^2 = 1/2 * 150 * 0.045^2 + 0.5*(0.0010+0.999)*9.8*0.045
vf = 0.863 m/s

Using Momentum conservation,
Initial Momentum = Final Momentum
m*vi = (m+M)*vf
0.0010 * vi = (0.0010 + 0.999) * 0.863
vi = 863 m/s

(b)
initial kinetic energy = 1/2 * 0.0010 * 863^2 = 372.38 J
Final kinetic energy = 1/2 * (0.0010 + 0.999) * 0.863^2 = 0.372 J
= (372.38 - 0.372) / 372.38
= 0.999

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