Linear momentum of particle is increased by
(a) 100% (b) 1%
without changing its mass. Find percentage increase in its kinetic energy in both cases.

Answer

B, C

Explanation

(a) Reaction between kinetic energy K and momentum p is given by:
`K=(p^2)/(2m)` …(i)
Now, momentum is increased by 100%
So, new momentum, `p^'=2p`
`:. K^'=((p^')^2)/(2m)=((2p)^2)/(2m)=4[[p^2)/(2m)]=4K` [From Eq. (i)]
Now, percentage change in Kinetic energy,
`=(("Final value"-"Initial value")/("Initial value"))xx100`
`=((K^'-K)/(K))xx100=((4K-K)/(K))xx100`
`=+300%`
Plus sign indicates that, with increase in linear momentum, kinetic energy will also increase.
(b) `K=(p^2)/(2m)` or `Kpropp^2`
Here, power of `K` is `1` and power of `p` is `2`. For small changes, we can write it like this
`(1)(% "change in" K)=(2)(% "change in p")`
or `% "chang e i n" K=(2)(1%)=+2%`