A1.3 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?

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

b)F=2566.66 N

Explanation:

Given that

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

We know that P=mV

So

Now by putting the values

J=1.3(9.8-(-21))

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

Force is given as

or we can say that

So

F=2566.66 N

iron would move it is magnetic there is a mag field around the current in the wire