A pitcher throws a 150g baseball so that it crosses home plate with a speed of 40 m/s. The ball hit straight back towards the pitcher at 45 m/s. If the contact time between the ball and the bat is 2x10^-3 seconds, what is the magnitude of the force on the ball?

Momentum = (mass) x (velocity)

On the way to the plate, the momentum of the ball was

(0.15 kg) x (40 m/s this way <==) = 6 kg-m/s this way <==

After the hit, the momentum of the ball was

(0.15 kg) x (45 m/s that way ==>) = 6.75 kg-m/s that way ==>

The CHANGE in momentum was 12.75 kg-m/s that way ==>

That's the momentum that the bat gave it.

Change in momentum = Impulse

Impulse = (force) x (length of time the force lasts)

12.75 kg-m/s = (force) x (0.002 second)

Force = (12.75 kg-m/s) / (0.002 second)

Force = 6,375 Newtons, that way ==>

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west