A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thik wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.
Answers
Answer:
mass of bullet=10g=0.01kg
velocity of bullet=1.5
s
m
mass of plank=90g=0.09kg
Total momentum before collision=mass of plank × velocity
=0.01×1.5+0.09×0=0.015kg
s
m
Total momentum after collision=(mass of bullet +mass of plank )× velocity
=0.1kg×V=0.1Vkg
s
m
According to law of conservation of momentum,
Total momentum before collision=Total momentum after collision
0.015kg
s
m
=0.1vkg
s
m
velocity acquired by the plank and the bullet=0.15
s
m
Answer:
Mass of bullet = 10g = 0.01kg
velocity of bullet = 1.5 m/s
Mass of plank = 90g = 0.09kg
Total momentum before collision = ( mass of plank ) × ( velocity)
= (0.01× 1.5) +(0.09×0) = 0.015kg m/s
Total momentum after collision= ( mass of bullet + mass of plank ) × ( velocity)
= (0.1 kg × v) = 0.1v kg m/s
According to law of conservation of momentum,
Total momentum before collision = Total momentum after collision.
M1 V1= M2 V2
0.015 kg m/s = 0.1 v kg m/s
Thus, velocity acquired by the plank and the bullet = 0.15 m/ s.
Explanation:
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