Question

A bullet having a mass of 10 g and

moving with a speed of 1.5 m/s,

penetrates a thick 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.
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Answer 1

Explanation:

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

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Answer 2

0.15 m/s

Explanation:

m1=10g = 10 x 10-3 (raised to minus 3)

u1=1.5 m/s

m2=90g=90 x 10-3(raised to minus 3)

u2=0 m/s

m1u1+m2u2=m1v1+m2v2 [law of conservation of momentum]

m1u1=(m1+m2)v----u2=0, v1=v2=v

v=m1u1

m1+m2

=10 x 10-3(raised to) x 1.5

10 x 10-3(raised to) + 90 x 10-3(raised to)

v=10 x 1.5

100

=0.15 m/s