Question

Kinetic energy of body is increased by 44 what is the percentage increase in the momentum

Answer 1

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:

Answer 2

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%.