Question

Question #8d6d9

Answer 1

The kinetic energy `E` of a body of mass `m` and velocity `v` is given by the following relation.

`E = 1/2mv^2`

`=>E = 1/2mv^2=1/2(mv)^2/m=p^2/(2m)`

`"where " p=mv=" linear momentum"`

Mass remaining constant Kinetic energy `Epropp^2`

For two cases (final and initial) the relation becomes

`E_f/E_i=p_f^2/p_i^2`

Given `E_f=E_i+800% of E_i=9E_i`

So `p_f/p_I=sqrt(E_f/E_I)=sqrt9=3`

Hence percent increase of momentum
`=(p_f-p_I)/p_ixx100%=(3p_i-p_i)/p_i xx100=200%`