Question

the momentum of a particle is increased by 300% find percentage increase in kinetic energy

Answer 1

Given:-

→Momentum of a particle is

increased by 300%

To find:-

→Percentage increase in kinetic energy of

the particle.

Solution:-

We know that, relationship between kinetic energy and momentum is :-

=>K.E. = p²/2m. ----(1)

Now,when momentum of the particle is increased by 300%, then p' :-

=>p' = p + 300/100×p

=>p' = p + 3p

=>p' = 4p

Hence,the new kinetic energy i.e. K.E.', will be:-

=>K.E.' = (p')²/2m. ----(2)

On diving eq.2 by eq.1, we get:-

=>K.E.'/K.E = [(p')²/2m/p²/2m]

=>K.E'/K.E = [(4p)²/p²]

=>K.E.'/K.E. = 16p²/p²

=>K.E.'/K.E. = 16

=>K.E.' = 16K.E.

Now,percentage increase in kinetic energy will be:-

=>[K.E.' - K.E./K.E ]× 100

=>[16K.E. - K.E./K.E.] ×100

=>[15K.E./K.E.]×100

=>15×100

=>1500%

Thus,kinetic energy of the particle

increases by 1500%.