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

Options

A. 44 m/s
B. 4.4 m/s
C. 0.44 m/s
D. 440 m/s

Answer

B

Explanation

Here, Mass of the bullet,`m_(1)=10g`
`=10/1000`kg
=0.01 kg
And, Velocity of the bullet, `v_(1)`=400 m/s
So, Momentum of bullet=`m_(1)xxv_(1)`
`=0.01xx400kg.m//s`…….(1)
Now, this bullet of mass 10 g gets embedded into a wooden block of mass 900 g . So ,the mass of wooden blockk alongwith the embedded bullet will become 900+10=910 g. Thus,
Mass of wooden block+Bullet, `m_(2)=900+10`
=910 g
`=910/1000` kg
=0.91 kg
And, Velocity of wooden block+bullet,`v_(2)=?` (To be calculated)
So, Momentum of wooden block+bullet=`m_(2)xxv_(2)`
`=0.91xxv_(2)`kg.m/s...(2)
Now, according to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal.
So, `m_(1)xxv_(1)=m_(2)xxv_(2)`
or 0.01`xx400=0.91xxv_(2)`
And, `v_(2)=(0.01xx400)/0.91`
=4.4 m/s
Thus, the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second.