Question

a sand bag of mass M is suspended by a rope . A bullet of mass m fired at it with speed v and the bullet gets embedded in it . the loss of kinetic energy in the collision is ? ​

Answer 1

Answer:

$Loss\ in\ KE=\dfrac{1}{2}mv^2-\dfrac{1}{2}(m+M)u^2$

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

$KE_i=\dfrac{1}{2}mv^2$

Final kinetic energy  of system

$KE_f=\dfrac{1}{2}(m+M)u^2$

So the loss in kinetic energy

$Loss\ in\ KE=\dfrac{1}{2}mv^2-\dfrac{1}{2}(m+M)u^2$

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