Question

Question #13b3d

Answer 1

The kinetic energy increases by 125 %.

Explanation:

Let initial momentum be `p` and kinetic energy be `K`.

`p = mv`

If `p` increases by 50% (0.50`p`)

the new momentum `p' = p + 0.50p = 1.50p`

The initial `K = 1/2 mv^2 = (mv)^2/(2m) = p^2/(2m)`

The new kinetic energy `K' = (p')^2/(2m)`

So `(K')/K = (p'^2)/ p^2`

`(K')/K = (1.50p)^2/p^2 = 1.50^2/1 = 2.25`

`K' = 2.25 K`

% change = `(K' – K)/K × 100 % = (2.25 K – K)/K × 100 % = 125 %`

% increase in kinetic energy = 125%