A bullet of mass 1.2 times 10^-3 kg embeds itself in a wooden block with mass 0.

Question

A bullet of mass 1.2 times 10^-3 kg embeds itself in a wooden block with mass 0.985 kg, which then compresses a spring (k = 200 N/m) by a distance 6.0 times 10^-2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.45. What is the initial speed of the bullet? Express your answer using two significant figures. What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) In the bullet and the block? Express your answer using two significant figures.

Explanation / Answer

Given M = 1.2 * 10^-3 + 0.985

M = 0.9862 kg

distance x = 0.06 m

mue = 0.45

let v be the speed of bullet and wooden block

so from conservation of energy

1/2 * M * v^2 - (f * x) = 1/2 * k * x^2

(1/2 * M * v^2) - ( mue * M * g * x) = 1/2 * k * x^2

(1/2 * 0.9862 * v^2) - (0.45 * 0.9862 * 9.8 * 0.06) = 1/2 * 200 * 0.06^2

v = 1.12 m/s

now let v' be the velocity of bullet

from conservation of momentum

m * v' = M * V

1.2 * 10^-3 * v' = 0.9862 * 1.12

v' = 920.45 m/s

b)

fractional loss of kinetic energy is

KE(final) / KE(initial)

1/2 * M * v^2 / (1/2 * m * v'^2)

(1/2 * 0.9862 * 1.12^2) / (1/2 * 1.2 * 10^-3 * 920.45^2)

1.2 * 10^-3

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