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Question: A ball weighing \(10g\) hits a hard surface vertically with a speed of \(5m/s\)and rebounds with the...

A ball weighing 10g10g hits a hard surface vertically with a speed of 5m/s5m/sand rebounds with the same speed. The ball remains in contact with the surface for(0.01)s\left( {0.01} \right)s. The average force exerted by the surface on the ball is:-
(A) 100N100N
(B) 10N10N
(C) 1N1N
(D) 0.1N0.1N

Explanation

Solution

Hint
We know that the average force can be calculated from the rate of change of momentum of the ball, firstly we need to calculate the change in momentum. So, you should know the relation of mass and change in velocity.

Complete step by step answer
Weight of ball =10g = 10g
Time at which ball and surface get in contact =0.01s = 0.01s
Initial velocity of ball Vi=−5m/s{V_i} = - 5m/s(As per question, initial and final velocities are same)
We are assuming that when the ball hits a hard surface its velocity is negative and when returns back it will be positive.
Final velocity of ball =Vf=5m/s = {V_f} = 5m/s
Step by step solution:
Let, Average force =Favg = {F_{avg}}
Change in momentum =ΔP = \Delta P
Change in time =Δt = \Delta t
We will find the average velocity from the rate of change of momentum.
⇒Favg=ΔPΔt......(1)\Rightarrow {F_{avg}} = \dfrac{{\Delta P}}{{\Delta t}}......(1)
Now, we will calculate change in momentum, as we know that change in momentum can be found from the product of mass and change in velocity.
⇒ΔP=m(Vf−Vi)\Rightarrow \Delta P = m({V_f} - {V_i})
As per question, initial and final velocity are the same which is 5m/s5m/sbut their signs are opposite, as initially the ball is going to hit a hard surface and then return back with the same velocity, so their signs will get opposite.
Put value ofmm, Vi{V_i}and Vf{V_f}
⇒ΔP=0.01(5−(−5))=0.1kgm/s\Rightarrow \Delta P = 0.01(5 - ( - 5)) = 0.1kgm/s
Here, we the change in momentum has calculated
Put value of ΔP\Delta P and Δt\Delta tin equation (1)(1)
⇒Favg=0.10.01=10N\Rightarrow {F_{avg}} = \dfrac{{0.1}}{{0.01}} = 10N
Hence, we get the average force exerted on the ball.
Hence option (B) is correct.

Note
The point to be noted is, as per question a ball hits a hard surface and bounce back with the same velocity, means we should use these same velocities which are initial and final velocity but the sign of initial velocity will be negative.