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

answer

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

$m_{k}\times g\times H_{k}= m_{a}\times g\times H_{a}$......(1)

here g is gravitational acceleration

put values of $m_{k} , g, H_{k}, m_{a}$ and $H_{a}$ in equation 1

$.25kg\times H_{k}=.6kg\times .5m\\H_{k}=\frac{.3kg\times m}{.25kg} \\H_{k}=1.2m$

therefore, Height at which kiwi reach=1.2m