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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.

Respuesta :

maria254
momentum before collision must be equal to momentum after collision

(m1 + m2)v = m1v1 + m2v2



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

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