Question

A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size. The superball bounces off the table and rises back to the same height. The clay ball sticks to the table. For the superball in the previous question, if it was in contact with the table for 34.3ms, calculate the average force exerted on the ball by the table. Hint: First calculate the momentum before and after hitting the table. Don't forget the gravitational force.

Answer 1

Answer:4.08 N

Step-by-step explanation:

Given data

superball dropped from a height of 10 cm

Mass of ball
`\left ( m\right )`=0.05kg

time of contact
`\left ( t\right )`=34.3
`* 10^(-3)` s

Now we know impulse =
`Force* time\ of\ contact`=Change in momentum

`F_(average)* t`=
`m\left ( v-(-v)\right )`

and velocity at the bottom is given by

v=
`√(2gh)`

`F_(average)* 34.3* 10^(-3)`=
`0.05\left ( 1.4-(-1.4)\right )`

`F_(average)`=4.081N