A 3.62 g bullet moving at 270 m/s enters and stops in an initially stationary 2.30 kg wooden block on a horizontal frictionless surface.

Question

Physics

Melvin Young

13 September, 20:20

A 3.62 g bullet moving at 270 m/s enters and stops in an initially stationary 2.30 kg wooden block on a horizontal frictionless surface.

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Answers (1)

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Izabelle Stein · Answer 1

Momentum before collision must be equal to momentum after collision

(m1 + m 2) v = m1v1 + m 2v 2

3.62g*270 m/s=2.30kg * (x)

convert the units to be the same that is convert the kg mass to grams

1kg=1000g

2.30kg=y

230/100*1000=2300g

3.62*270=2300X

977.4g/m/s=2300X

977.4/2300=0.4250M/S

Finding a velocity after the gun is embedded on the block of wood

3.62*270+2300g*0.425 = (3.62+2300) * V

977.5+977.5=2303V

1955=2303V

V=1955/2303

V=0.8489M/S