Question #8d6d9

Question

1788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-12T14:54:53

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
$=(p_f-p_I)/p_ixx100%=(3p_i-p_i)/p_i xx100=200%$