Question

A bullet of mass 20g fired from a gun is moving with a velocity of 300m/s. [3]
It gets embedded in a wooden block kept suspended at a distance. The mass
of the wooden block is 880gm. What is the velocity with which the block will
move when the bullet pierces in it?

Answer 1

Answer:

6.67 m/s.

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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.

Answer 2

Answer:

6.82 m/s

Explanation:

p= linear momentum

m1= 20g

m2 = 880g

v1= 300m/s

v2=?

=>    by law of comservation of linear momentum,

         p1=p2

  m1v1=m2v2

20x300=880xv2

v2=6000/880

v2=75/11 m/s

v2= 6.82 m/s