Question

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.50 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?

Answer 1

K = 0.076 J

Step-by-step explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

`h=ut+(1)/(2)at^2\\\\t=\sqrt{(2h)/(g)}`

Put all the values,

`t=\sqrt{(2* 0.860 )/(9.8)} \\\\=0.418\ s`

The velocity of the ball is :

`v=(1.5)/(0.418)\\\\= $$3.58\ m/s`

The kinetic energy of the ball is :

`K=(1)/(2)mv^2\\\\K=(1)/(2)* 0.0120* 3.58^2\\\\=0.076\ J`

So, the required kinetic energy is 0.076 J.