A superball of mass 0.05 kg and a clay ball of the same mass aredropped from a h

Question

A superball of mass 0.05 kg and a clay ball of the same mass aredropped from a height 10 cm above a table top. The superballbounces and rises to the same height; the clay ball sticks to thetable. If the superball was in contact with the table for 30milliseconds, calculate the average force exerted on the ballby the table (Hint: first calculate themomentum before and after hitting the table. Don't forget thegravitational force.) Then, calculate the impulse exerted onthe clay ball and compare it to that with the superball. Which is larger or are they both the same? Explain.

Explanation / Answer

Given : .             Mass of the super ball (m1) = 0.05 kg .             Initial velocity is : u = 2gh1 = (2 * 9.8 *0.1m) = 1.4 m/s .              Finalvelocity is : v   = 2gh2 =  (2 * 9.8*0.1m) = 1.4 m/s .             As the ball bounces back ; v = -1.4   m /s .              Timeof contact (t) = 30 ms   = 0.03s . Impulse on superball is : .              J1    =    m1 ( v - u ) =   0.05 ( -1.4 -1.4 ) = - 0.14 kg- m/s   -----------(1) . Hence average force exerted on super ball is : .               F     = m1 ( v - u ) / t    = 0.05( -1.4 -1.4 ) / 0.03 .                                                    =  - 4.667 N . Hence magnitude is :   F = 4.667 N . (b) Mass of clay ball (m2) = 0.05 kg .       Initial velocity(u1) = 2gh1 = (2 *9.8 *0.1m) = 1.4 m/s .       Final velocity (v1)  = 0 m /s .       Impulse exerted on clay ball(J) = m2 ( v1 - u1)   = 0.05 * ( 0 -1.4 )   = - 0.07 Kg - m /s -----------(2) . From (1) and (2)  JSuperball    >  Jclayball   . Hope this helps u!

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