Question

A ball falls on surface from \(10 \mathrm{~m}\) height and rebounds to $2.5 \mathrm{~m} .\( If duration of contact with floor is \)0.01 \mathrm{sec}$. then average acceleration during contact is \(\mathrm{ms}^{-2}\) (A) 2100 (B) 1400 \(\begin{array}{ll}\text { (C) } 700 & \text { (D) } 400\end{array}\)

Short answer

The average acceleration during contact is approximately \(1900 ms^{-2}\), which is closest to option (A) 2100 ms².

Step-by-step solution

Calculate the velocity before impact

First, we need to find the velocity of the ball just before it makes contact with the floor. We can use the equation: \(v^2 = u^2 + 2as\) where \(v\) is the final velocity just before hitting the floor, \(u\) is the initial velocity (which is 0 since the ball is initially dropped), \(a\) is the acceleration due to gravity (approximately 9.81 m/s²), and \(s\) is the distance fallen (10 m). \(v^2 = 0 + 2 * 9.81 * 10\) \(v^2 = 196.2\) \(v = \sqrt{196.2} \approx 14 ms^{-1}\) So, the velocity just before impact is approximately 14 m/s.

Calculate the velocity after impact

Next, we need to find the velocity of the ball right after the contact. We can use the ratio of rebound height to initial height and the conservation of energy principle: \(\frac{1}{2} m v_{after}^2 = \frac{1}{2} m g H_{rebound}\) where \(v_{after}\) is the velocity after impact, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(H_{rebound}\) is the rebound height (2.5 m). Since the masses and the factor 1/2 in the equation cancel out, we get the following equation: \(v_{after}^2 = g * H_{rebound}\) \(v_{after}^2 = 9.81 * 2.5\) \(v_{after}^2 = 24.525\) \(v_{after} = \sqrt{24.525} \approx 4.95 ms^{-1}\) So, the velocity right after impact is approximately 4.95 m/s.

Calculate the average acceleration during contact

The average acceleration during contact (\(a_{avg}\)) can be calculated using the formula: \(a_{avg} = \frac{\Delta v}{\Delta t}\) where \(\Delta v\) is the change in velocity (in this case, the difference between the velocity after impact and the velocity before impact) and \(\Delta t\) is the duration of contact (0.01 s). Since the ball changes direction during impact, we get: \(\Delta v = -(14 + 4.95) ms^{-1} = -18.95 ms^{-1}\) Now, we can find the average acceleration: \(a_{avg} = \frac{-18.95}{0.01}\) \(a_{avg} = -1895 ms^{-2}\) Since we are asked to find the magnitude of the average acceleration (ignoring the negative sign), we get: \(a_{avg} = 1895 ms^{-2}\) The answer is approximately 1900 ms², which is not exactly one of the listed options. However, it's closest to option (A) 2100 ms².