Question

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m into a thick steel plate. The bearings bounce back up to a height of 1.48 m. If the bearings are in contact with the steel plate for a time interval of 14.86 x 10-3 s, what is the magnitude of the average acceleration of the bearings while they are in contact with the plate

Answer 1

The average acceleration of the bearings is
`0.77*10^(3)\ m/s^2`

Step-by-step explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval
`t=14.86*10^(-3)\ s`

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

`mgh=(1)/(2)mv^2`

Put the value into the formula

`9.8*1.94=(1)/(2)* v^2`

`v=√(2*9.8*1.94)`

`v=6.166\ m/s`

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

`mgh=(1)/(2)mv^2'`

Put the value into the formula

`9.8*1.48=(1)/(2)* v^2'`

`v'=√(2*9.8*1.48)`

`v'=5.38\ m/s`

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

`a=(v-v')/(t)`

Put the value into the formula

`a=(5.38-(-6.166))/(14.86*10^(-3))`

`a=776.98\ m/s^2`

`a=0.77*10^(3)\ m/s^2`

Hence,The average acceleration of the bearings is
`0.77*10^(3)\ m/s^2`