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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 :

Answer:

The average acceleration of the bearings is [tex]0.77\times10^{3}\ m/s^2[/tex]

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval [tex]t=14.86\times10^{-3}\ s[/tex]

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Put the value into the formula

[tex]9.8\times1.94=\dfrac{1}{2}\times v^2[/tex]

[tex]v=\sqrt{2\times9.8\times1.94}[/tex]

[tex]v=6.166\ m/s[/tex]

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

[tex]mgh=\dfrac{1}{2}mv^2'[/tex]

Put the value into the formula

[tex]9.8\times1.48=\dfrac{1}{2}\times v^2'[/tex]

[tex]v'=\sqrt{2\times9.8\times1.48}[/tex]

[tex]v'=5.38\ m/s[/tex]

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

Using formula of acceleration

[tex]a=\dfrac{v-v'}{t}[/tex]

Put the value into the formula

[tex]a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}[/tex]

[tex]a=776.98\ m/s^2[/tex]

[tex]a=0.77\times10^{3}\ m/s^2[/tex]

Hence,The average acceleration of the bearings is [tex]0.77\times10^{3}\ m/s^2[/tex]

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