A bullet of mass 1.9×10 3 kg embeds itself in a wooden block with mass 0.981 kg

Question

A bullet of mass 1.9×103 kg embeds itself in a wooden block with mass 0.981 kg , which then compresses a spring (k = 180 N/m ) by a distance 6.0×102 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.45.

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

Energy from block and bullet - Frictional energy = Energy absorbed by spring
Es = Energy absorbed by spring = (1/2)kx^2
where
k = spring constant = 180 N/m
x = length at which spring was compressed = 0.06 m
Es = (1/2)(180)(0.06)2
Es = 0.324 joules
Frictional energy = Wf = 0.45(0.0019 + 0.981)(9.8)(0.06)
Wf = 0.260
KE = Energy from block/bullet
KE - 0.324 = 0.260
KE = 0.584
Let V = velocity of the block/bullet after bullet is embedded in the block
0.584 = (1/2)(0.0019 + 0.981)V^2
0.584 = (1/2)(0.9829)V^2
V = 1.09 m/sec.
Using the law of conservation of momentum,
0.0019*Vi = (0.0019 + 0.981)(1.09)
where
Vi = initial speed of the bullet
0.0019Vi = 1.07
Vi = 563.15 m/sec.

b)

Initial KE of bullet = (1/2)(0.0019)(563.15)^2 = 301.28 joules

Energy to overcome friction = Wf = 0.260 J

0.260 * 100/301.28 = 0.086 %

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