A spherical ball of mass 10-6 kg hits a wall 1000 times per second normally with a velocity of 1000 m/s and rebounds with same velocity along the initial direction. The force experienced by the wall is :
Answer & explanation
Correct answer: option 3
Force experienced by the wall can be described as the change in net momentum (transferred to wall) divided by the total time taken.
\(\Delta P_{ball} = m \Delta v\)
\(\Rightarrow \Delta P_{ball} = 10^{-6}(-1000-1000)\)
\(\Rightarrow \Delta P_{ball} = 10^{-6} (-2000)\)
\(\Rightarrow \Delta P_{ball} = -2\times 10^{-3} \text{ kg m/s}\)
Momentum change transferred to wall :
\(\Rightarrow \Delta P_{ball} = -\Delta P_{wall} = 2 \times10^{-3} \text{ kg m/s}\)
Net Momentum transferred to wall :
\(\Rightarrow P_{net} = 2\times 10^{-3} \times (\text{ no. of times hitting})\)
\(\Rightarrow P_{net} = 2\times 10^{-3} \times 10^{3}\)
\(\Rightarrow P_{net} = 2 \text{ kg m/s}\)
So, force experienced by wall:
\(\Rightarrow F = \frac{\Delta P_{net}}{dt}\)
\(\Rightarrow F = \frac{2}{1}\)
\(\Rightarrow F = 2 \text{ N}\)