Question

A bullet of mass 20 grams moving with a velocity of 300m/s gets embedded in a freely suspended wooden block of mass 880 gram. What is the velocity acquired by the block ​

Answer 1

Hi friends here your answer

given data :-

mass of bullet = 20÷ 1000

= 2× 10-^2

velocity u1 = 300 m/s

mass of wooden block = 880÷1000

= 0.88 kg

now

law of momentum conservation

momentum before collision = momentum after collision

mu1 + mu2 = mv1 +mv2

0.02 × 300 + 0 = v ( m1 + m2)

6 = v ( 0.02 + 0.88)

6 = v × 0.9

v = 60 /9

v = 6.66 m/s

Answer 2

Explanation:

This is completely inelastic collision. The linear momentum is conserved.

If a body of mass, m collides with a body of mass M and gets embedded in it, according to law of conservation of momentum,

mv =( m+M)V.

Here,

m is mass of colliding body. m= 20 g.

M is mass of target body. M=880 g.

v=300m/s is velocity of colliding body.

V= ? V is velocity of combined system of m and M.

Therefore 20(g).300(m/s)=(880+20)( g)V(m/s.)OR

V=6000/900=6.67 m/s.