Question

A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0156 s, what is the magnitude of the average force on the floor from the ball?

Answer 1

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Step-by-step explanation:

Given that

`V_1=21 m/s,V_2=9.8m/s`

We know that impulse(J) impulse is a vector quantity

`J=P_2-P_1`

We know that P=mV

So
`J=m(V_2-V_1)`

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

`F=(dP)/(dt)`

or we can say that

`F=m\left((v_2-V_1)/(dt)\right)`

So
`F=(40.04)/(0.0156)`

F=2566.66 N