Question

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

Answer 1

Answer: This may help you

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