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Question: A ball weighing 10 gm hits a hard surface vertically with speed and rebounds at the same speed. The ...

A ball weighing 10 gm hits a hard surface vertically with speed and rebounds at the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on the ball is
A) 100 N
B) 10 N
C) 1 N
D) 0.1 N

Explanation

Solution

Here the force exerted on the ball can be easily calculated by the rate of change of momentum. As the ball is rebound from the surface the force can be calculated from it.

Complete step by step answer:
Force can be defined as the action done on an object to change its momentum.
Newton's second law of motion gives the definition of force as the multiplication of mass and acceleration. Acceleration on the other hand is the velocity change divided by time, this can be written as,
Force = mass×velocitytime\text{Force = mass} \times \dfrac{\text{velocity}}{\text{time}}
Force=mass×velocitytime\Rightarrow \text{Force} =\dfrac{\text{mass} \times \text{velocity}}{\text{time}}
Force=momentumtime\Rightarrow \text{Force} =\dfrac{\text{momentum}}{\text{time}}
Multiplying time to both sides, we get,
Force×time=momentum\text{Force} \times \text{time} = \text{momentum}
Thus, we can write,
F=ΔpΔtF = \dfrac{{\Delta p}}{{\Delta t}} ……..(1)
Where, pp is the momentum, and tt is the time period for which contact was made.
Now, we know that there was a change in momentum as the ball changed direction.
Thus, change in momentum of the ball can be written as,
Δp=m(vfvi)\Delta p = m({v_f} - {v_i})
Where, m= mass of the body, vf{v_f}= final velocity, and vi{v_i}= initial velocity.
As the ball is changing direction let us consider +y-direction as positive and -y-direction as negative.
Thus, we have
Δp=m(5(5)) Δp=0.01(10) Δp=0.1  \Rightarrow \Delta p = m(5 - ( - 5)) \\\ \Rightarrow \Delta p = 0.01(10) \\\ \Rightarrow \Delta p = 0.1 \\\
So, we have the momentum as 0.10.1.
Now, the time period of contact is given as t=0.01st= 0.01 s

Thus, Equation 1 gives us,
F=ΔpΔt F=0.10.01 F=10  \Rightarrow F = \dfrac{{\Delta p}}{{\Delta t}} \\\ \Rightarrow F = \dfrac{{0.1}}{{0.01}} \\\ \Rightarrow F = 10 \\\

Thus, the required effort exerted on the ball to bounce it back at 5ms15m{s^{ - 1}} from the wall after remaining in contact for 0.01s0.01s is 10N10 N.

Therefore, option (B) is the right answer to this question.

Note:
To solve a problem like this we always have to remember Newton’s third law which tells us that every action has an equal and opposite reaction. Here, if we calculate the reaction for then the solution to the problem is done.