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# A bullet of mass 1.3103 {\rm kg} embeds itself in a wooden block with mass 0.984 {\rm kg}, which then compresses a spring (k = 140 {\rm N/m}) by a...

A bullet of mass 1.3×10−3 {rm kg} embeds itself in a wooden block with mass 0.984 {rm kg}, which then compresses a spring (k = 140 {rm N/m}) by a distance 5.0×10−2 {rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.43. What is the initial speed of the bullet? 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?

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Initial velocity of bullet is u
After collision its velocity is v
m= mass of bullet
M= mass of box
So, mu =(M+m)v [conservation of momentum]
Or, v = F = frictional force =
From conservation of...

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