1.

A bullet of mass 10 g moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass 900 g. what is the velocity acquired by the wooden block

Answer»

Hi,

Here is your answer

GIVEN:

Mass of BULLET(m1)=10g=0.01kg

Mass of wooden block(m2)=900g=0.9kg

Initial velocity of bullet(u1)=400m/s

Initial velocity of wooden block(u2)=0m/s

To Find:

Final velocity of the system(v)

Solution:

By the law of CONSERVATION of momentum

m1u1+m2u2=(m1+m2)×v

Substituting the given datas, we get

0.01×400+0.9×0=(0.01+0.9)v

4/0.91=v

v=4.4m/s

Notes:

1.Momentum before collision is equal to the momentum after collision in an isolated system

2. Momentum is the PRODUCT of mass and velocity. It's unit is kg m/s

Hope this helps you.


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