Question

A bullet of mass 2.0×10−3 kg embeds itself in a wooden block with mass 0.991 kg , which then compresses a spring (k = 200 N/m ) by a distance 3.5×10−2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.52.

a)What is the initial speed of the bullet?

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?

Answer #1

Given,

m = 2 x 10^-3 kg ; m = .991 kg ; k = 200 N/m ; d = 3.5 x 10^-2 ; uk = 0.52

a)PEspring = 1/2 k x^2

PEspring = 0.5 x 200 x 0.035^2 = 0.1225 J

work done against friction

Wf = uk (m + M) g x

Wf = 0.52 (0.002 + 0.991) x 9.81 x 0.035 = 0.1773 J

Intial KE of the system

KE = PEsping + Wf

KE = 0.1225 + 0.1773 = 0.2998 J

1/2 (m + M) v^2 = 0.2998 J

v = sqrt (2 x 0.2998/0.993) = 0.78 m/s

from conservation of momentum

m vi = (m + M)v

vi = (m + M) v/mi

vi = 0.993 x 0.78/0.002 = 387.27 m/s

**Hence, vi = 387.27 m/s**

b)KEi = 1/2 m vi^2 = 0.5 x 0.002 x 387.27^2 = 150 J

W = 0.2998

fraction = W/KEi = (150 - 0.2998)/150 = 0.998

**fraction loss W/KEi = 0.998**

%loss = (150 - 0.2998)/150 = 99.8 %

**Hence, %age loss = 99.8 %**

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