Question

If an 0.6kg apple is dropped from 0.5m height onto a seesaw and flips up a 0.25kg kiwi. To what height does that kiwi reach. Assume all the energy of the apple transfers perfectly to the kiwi. show work

keep the tip. ; )

Answer 1

Height at which kiwi reach=1.2m

Solution

In this question we have given

mass of apple,
`m_(a) =.6kg`

mass of kiwi,
`m_(a) =.25kg`

height from which apple is dropped,
`H_(a) =.5m`

Let the height at which kiwi reached is
`H_(k)`

here,

We will apply law of conservation of energy

it means,

potential energy of kiwi= potential energy of apple

`m_(k)* g* H_(k)= m_(a)* g* H_(a)`......(1)

here g is gravitational acceleration

put values of
`m_(k) , g, H_(k), m_(a)` and
`H_(a)` in equation 1

`.25kg* H_(k)=.6kg* .5m\\H_(k)=(.3kg* m)/(.25kg) \\H_(k)=1.2m`

therefore, Height at which kiwi reach=1.2m