Question

If kinetic energy of a body is increased by 300% then percentage change in momentum will be

Answer 1

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bt I m giving u a a simple...answer so that u can
solve in less time √n+1 - 1 *100%
where n = like here 3...
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Answer 2

Given:  kinetic energy of a body is increased by 300%.

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 × 300%

           K.E.f = K.E.i + K.E.i × 300/ 100

           K.E.f = K.E.i + K.E.i × 3

            K.E.f = 4 K.E.i  _____(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

⇒ (√4K.E.i - √K.E.i / √K.E.i ) × 100

⇒ √K.E.i / √K.E.i ( √4 - 1/ 1) × 100

⇒ ( 2 - 1/1 ) × 100

⇒ 100%

Hence, the % change in momentum is 100%.