Question

A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block.

(a) What is their velocity just after the collision? m/s

(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity? m/s

(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping? m

Answer 1

m1 = 0.022 kg

v1 = 400 m/s

m2 = 0.5 kg

v2 = 0 m/s

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a)

Using conservation of momentum

m1 v1 + m2 v2 = ( m1 + m2) v'

0.022* 400 + 0 = ( 0.022+ 0.5) v'

v' = 16.858 m/s

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b)

Using work- energy theorem

Workdone by friciton = change in KE

u mg d = 0.5 m ( v'^2 - u^2)

0.3 * 9.8* 8 = 0.5 * ( 16.858^2 - u^2)

u = 15.4 m/s

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c)

Final velocity of after striking

(0.022+0.5)*15.4 = (0.022+0.5 + 2) V

V = 3.187 m/s

Using work energy theorem again

u mg d = 0.5 mV^2

0.3*9.8* d = 0.5 * 3.187^2

d = 1.728 m

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