Question

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

Answer 1

Answer:

here is your answer

Explanation:

mass of bullet is M1=10g

mass of block is m2=900g

velocity of bullet is v1=400m/s

velocity of block is v2

m1v1=m2v2

putting value answer will be 4m/s

Answer 2

Explanation:

let final velocity of bag along with the bullet embedded in it be v

for bullet m1 = 10g=0.01kg

v1=400m/s;=vm/s

for bag m2=900g=0.09kg;v2=0

v2=vm/s

now according to law of conversion of momentum

m1u1 + m2u2 = m1v1 + m2v2

m1u1 + m2v2 = ( m1 + m2 ) v

v = m1v1 + m2v2 / m1 + m2

=0.01 × 400 + 0.9 × 0 / 0.01 + 0.9

= 4/0.91

= 400/ 90

= 4.4 m / sec