A bullet of mass 1.8×10 3 kg embeds itself in a wooden block with mass 0.980 kg

Question

A bullet of mass 1.8×103 kg embeds itself in a wooden block with mass 0.980 kg , which then compresses a spring (k = 140 N/m ) by a distance 4.0×102 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.46.

Part A

What is the initial speed of the bullet?

Express your answer using two significant figures.

Part 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?

Express your answer using two significant figures.

Explanation / Answer

step;1

Given that

mass of bullet m=1.8*10^-3 kg

mass of the wooden block M=0.98 kg

spring constant k=140 n/m

distance x=4*10^-2 m

coefficient of kinetic friction =0.46

step;2

now we find the energy absorbed by spring

the energy absorbed by spring Es=1/2kx^2=1/2*140*0.04^2=0.112 J

fractional energy Wf=0.46(0.98+0.0018)*9.8*0.04=0.18 J

K.E=0.112+0.18=0.292 J

now we find the finail velocity after bullet embedded by block

0.292=1/2(0.98+0.0018)V^2

speed V=0.8 m/s

now we find the initial velocity of bullet

0.0018u=(0.98+0.0018)0.8

initial velocity u=436.4 m/s

initial kinetic energy of the bullet K.Ei=1/2*0.0018*436.4^2=171.4 J

fraction =Wf/K.Ei*100

=0.18*100/171.4

=0.11 %

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