question is from physics

Question

Q A ball of the mass m is moving with velocity v. It strikes the floor and rebounce with the same velocity. If n such balls strike the floor per sec. Then exerted by the balls on the floor is....

A. mv.               B.mnv

C.2mnv.            D.mn/v

Answer 1

Hi Hemang!

 

Here, we can use the impulse-momentum theorem to relate change in momentum to average applied force (by the floor on the balls).  In that case:

 

FΔt = Δp

 

or

 

F = Δp/Δt

 

where

 

F = (average) applied force

Δt = time of collision

Δp = change in momentum

 

For this case, Δp = mΔv = m(v_f-v_o), with v_f and v_o as, respectively, the final and initial velocities (direction included).  If the balls rebound with the same speed (not the same velocity, technically, because they are going in the opposite direction), and we define down as negative, then v_f = v and v_o = -v.  Then we have:

 

Δp = m(v - (-v)) = 2mv

 

for each ball.  Since F = Δp/Δt, then, if n balls hit in 1 second, we have

 

Δp_tot/Δt = (2mv·n)/1 s = F

 

Thus, the average force exerted by balls on the floor (equal to the magnitude of the average force exerted by the floor on the balls, is:

 

F = 2mnv

 

I suspect that is what the question is asking for.  But if it is not, please let me know.  I hope this helps!  Also, let me know if you have any questions about this.