Maximum force?

Question

Superball Hits Wall Figure 7-24 shows an approximate plot of force magnitude versus time during the collision of a 60 g Superball with a wall. The initial velocity of the ball is 32 m/s perpendicular to the wall. It rebounds directly back with approximately the same speed, also perpendicular to the wall. What is Fmax, the maximum magnitude of the force on the ball during the collision?

Figure 7-24:
http://i234.photobucket.com/albums/ee9/locowise/10_32.gif

Thanks in advance for any help you can offer!

Answer 1

Av. impulse of force = F(max.) ( 1 + 2 + 1)*(10)^(-3) = 4*(10)^(-3)F(max.)
= change in momentum of the ball
= (0.06) [ 32 - (-32) ] kg m/s
= 3.84 kg ms

=> 4*(10)^(-3) F(max.) = 3.84
=> F(max.) = 3840/4 = 960 N.

Answer 2

Force is the rate of change of momentum. In calculus terms:

F = dP/dt (where "P" is momentum)

And that means, conversely, that the change in momentum is the integral of force with respect to time:

ΔP = integral(F·dt)

Which means:

ΔP = (area under the F vs. t curve)

You can figure out geometrically what the area under the curve is. It's just the area of a trapezoid. The answer will be an expression that has Fmax (the height of the trapezoid) in it. I leave that as an exercise for the reader.

You can figure out ΔP from the numbers given in the problem. The initial P is +(60g)(32m/s); and the final P is −(60g)(32m/s). So this means:

ΔP = 2(60g)(32m/s)

So, set that equal to the area of the trapezoid:

2(60g)(32m/s) = (area of trapezoid)

You should now have an equation that you can solve for Fmax.

Answer 3

i will let you know ok, just send me a new brain.