Kinetic energy of body is increased by 44 what is the percentage increase in the momentum
Answers
Answer:
The relation is
Ke=p^2÷2m
Now,
Ke'=ke+44%ofke
=Ke+0.44ke
Ke' =1.44ke
P'^2÷2m=1.44p^2÷2m
P'2=1.44p^2
P'=1.2p
Now %increase in momentum,
={(P'-p)÷p}×100
={(1.2p-p)÷p}×100
={0.2p÷p}×100
=0.2×100
=20%
Hence momentum will increase by 20%xplanation:
Given: kinetic energy of a body is increased by 44%.
To find: percentage change in momentum.
Solution: Let the initial kinetic energy be K.E.i
Then the final kinetic energy will be given by as per the question:
K.E.f = K.E.i + K.E.i × 44%
K.E.f = K.E.i + K.E.i × 44/ 100
K.E.f = 144 K.E.i/100 _____(1)
Now, through the concept of error analysis, we know that the % change in momentum is given by ;
% change in p =( Pf - Pi / Pi) × 100
Putting the given values in the formula, we get:
⇒ ( √2mK.E.f - √2mK.E.i / √2mK.E.i ) × 100
⇒√2m/ √2m ( √K.E.f - √K.E.i / √K.E.i ) × 100
⇒ (√144K.E.i/100 - √K.E.i / √K.E.i ) × 100
⇒ √K.E.i / √K.E.i ( √144/100 - 1/ 1) × 100
⇒ ( 12 - 10/10/1 ) × 100
⇒ 2/ 10 × 100
⇒ 20%
Hence, the % change in momentum is 20%.