Question

A bag of sand of mass m is suspended by a rope. A bullet of mass m is fired at it with speed v

Answer 1

Answer:

Explanation:

Please refer to attachment .

Approach

1. We can see that we will have to find final velocity by applying laws of conservation of momentum.

2. Since initially , the momentum of bullet was maximum & bag had zero momentum so we will subtract the kinetic energy of (bullet-bag) system from kinetic energy of bullet .

Note : I have substituted the value of v' (final velocity) while calculating lost kinetic energy.

Mmv^2/2(m+M).

Explanation:

We know that for change in energy we have the initial momentum = final momentum which will be (M+m)u=mv.

The ratio of the final kinetic energy to the initial kinetic energy is given by the 1/2(m+M)u^2/1/2mv^2/ which will be (M+m)/m(u/v)^2 which will be (M+m)/m*/(m/M+m)^2 or m/(m+M).

So, finally the kinetic energy loss will be initial kinetic energy - the final kinetic energy or initial kinetic energy(1- final kinetic energy/initial kinetic energy) which will be (1/2mv^2)*M/m+M or Mmv^2/2(m+M).

Answer 2

Answer:

Explanation:

Given that

Mass of sand bag = M

Mass of bullet = m

Initial velocity of bullet before striking the bag = v

Lets the final velocity of bullet and bag is u because they will move together.

We know that there is not any external force so the linear momentum will be conserve.

m v =(M+m)u

So  u=m v/(m+M)

Initial kinetic energy  of system

Final kinetic energy  of system

So the loss in kinetic energy