Kinetic energy of a particle is increased by
(a) 50% (b) 1%
Find percentage change in linear momentum.

Answer

B

Explanation

(a) `p=sqrt(2Km)` …(i)
Kinetic energy is increased by `50%`. So, the new value of kinetic energy is
`K^'=1.5K`
`:. P^'=sqrt(2K^'m)`
`=sqrt(2(1.5K)m)=sqrt(1.5)(sqrt(2Km))`
`=1.22sqrt(2Km)=1.22p` [From Eq. (i)]
So, the percentage change in momentum is
`=((p^'-p)/(p))xx100=((1.22p-p)/(p))xx100=+22%`
(b) `p=sqrt(2Km)` or `ppropsqrtK` or `ppropK^(1/2)` (as `m=const ant)
Here, power of p is 1 and the power of K is `1//2`.
For small percentage changes we can write as
`(1)(% chang e i n P)=(1/2)(% chang e i n K)`
or `% chang e i n p=1/2(1%)=+0.5%`