Question

if the momentum is increased by 20%, what is the % change in its kinetic energy?​

Answer 1

Answer:

p=mv

ke=p^2/2m

p'=p+ 0.2p

=1.2p

ke'=p'^2/2m

=1.44p/2m

∆ke%=(ke'-ke) *100

i. e. 44%

Answer 2

Answer:

Explanation:

%change in K. E is given by

1-{kE(initial)/kE(final))*100

We need to cal. Initial and final kE

Relation is p(square)/2m=kE

P(initial)= 1 (100%)ie 100/100

P(final)=1+0.2=1.2 (100+20)% ie 100+20)/100

KE props P(square)

K(initial) props (1)square

K(final) props 1.2(square)=1.44

From 1st formula

1-{1.44/(1)}=0.44*100

44%