A 7.00 g bullet, traveling horizontally with a velocity of magnitude 380 m/s , i

Question

A 7.00 g bullet, traveling horizontally with a velocity of magnitude 380 m/s , is fired into a wooden block with mass of 0.790 kg , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 160 m/s . The block slides a distance of 45.0 cm along the surface from its initial position.
a) What is the coefficient of kinetic friction between block and surface?
b) What is the decrease in kinetic energy of the bullet?
c) What is the kinetic energy of the block at the instant after the bullet passes through it?

Explanation / Answer

m1v1+m2v2=m1u1+m2u2
v2=0
7*10^-3*380=7*10^-3*160+790*10^-3*u2
u2=1.94 m/s
so v^2-u2^2=2as
a=m2g

a=-u2^2/2s

a=-4.22 m/s^2

=0.54

b)ke=1/2*m1(u1^2-v1^2)

change ke=415.8 J

C)ke=1/2m2*u^2=1.48J

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