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Question: A 55gm superball travelling at a speed of \(29m{{s}^{-1}}\) bounces off a brick wall and rebounds at...

A 55gm superball travelling at a speed of 29ms−129m{{s}^{-1}} bounces off a brick wall and rebounds at 18.0ms−118.0m{{s}^{-1}}. A high-speed camera records this event. If the ball is in contact with the wall for 3.5ms , what is the magnitude of average acceleration of the ball during this time interval? (Note: 1s=103ms1s={{10}^{3}}ms)

Explanation

Solution

In the problem, we have been given the initial speed and the final speed of superball. Also, we have been given the time of contact of the superball with the brick wall. So, we can use the equation of “straight line motion” with constant acceleration to get our answer. We shall proceed in this manner to get our answer.

Complete answer:
Let us first assign some terms that we are going to use in our solution.
Let the initial speed of the ball before hitting the brick wall be ‘u’. Then, the value of ‘u’ has been given to us as:
⇒u=29ms−1\Rightarrow u=29m{{s}^{-1}}
Let the final speed of the ball after rebounding off the brick wall be ‘v’. Then, the value of ‘v’ has been given to us as:
⇒v=18ms−1\Rightarrow v=18m{{s}^{-1}}
Let the time of contact of the ball with the brick wall be ‘t’. Then, the value of ‘t’ has been given to us as:
⇒t=3.5ms ∴t=3.5×10−3s \begin{aligned} & \Rightarrow t=3.5ms \\\ & \therefore t=3.5\times {{10}^{-3}}s \\\ \end{aligned}
Now, let the average acceleration of the ball be ‘a’ over the time it is in contact with the ball. Then, we can write:
⇒v=u+at\Rightarrow v=u+at
Putting the values of all the known terms from above equations and calculating for ‘a’, we get:
⇒18=29+a(3.5×10−3) ⇒−11=a(3.5×10−3) \begin{aligned} & \Rightarrow 18=29+a\left( 3.5\times {{10}^{-3}} \right) \\\ & \Rightarrow -11=a\left( 3.5\times {{10}^{-3}} \right) \\\ \end{aligned}
⇒a=−11(3.5×10−3)\Rightarrow a=-\dfrac{11}{\left( 3.5\times {{10}^{-3}} \right)}
⇒a=−110003.5ms−2 ∴a=−3142.86ms−2 \begin{aligned} & \Rightarrow a=-\dfrac{11000}{3.5}m{{s}^{-2}} \\\ & \therefore a=-3142.86m{{s}^{-2}} \\\ \end{aligned}
Here, the negative sign in acceleration means that the ball is decelerating.
**Hence, the magnitude of average acceleration of the ball during this time interval comes out to be −3142.86ms−2-3142.86m{{s}^{-2}}.

Note: **
There has been no mention of, whether the acceleration is constant or time varying. But, the term that has been asked is “average acceleration”. The average acceleration over a time period is the sum of all irregular accelerations upon the total number of different accelerations. In simple terms it is the constant acceleration over that time period.