Question

A bullet of mass 0.1kg is fired and gets embedded into a block of wood of mass 1kg initially at rest. The velocity of before collision is 110m/s.
a)What is the velocity of the system after collision?
b)Calculate the kinetic energies before and after collision.
c)Is it an elastic collision or inelastic collision?
d) How much energy is lost in collision?​

Answer 1

Explanation:

Given : m

1

​

=0.1kg m

2

​

=2kg u

1

​

=500 ms

−1

u

2

​

=0

Let the velocities of the bullet and the block just after the collision be v

1

​

and v

2

​

respectively.

The height upto which the block is raised h=2 m

∴ From work-energy theorem for the block,

2

1

​

m

2

​

v

2

2

​

=m

2

​

gh

⟹v

2

​

=

2gh

​

v=

2×10×2

​

=2

10

​

ms

−1

Applying conservation of momentum just before and just after the collision :

m

1

​

u

1

​

+0=m

1

​

v

1

​

+m

2

​

v

2

​

0.1(500)=0.1v

1

​

+(2)×2

10

​

⟹v

1

​

=373.51 ms

−1