Impulse and work done by a tennis racquet on a 0.06 kg ball

[Satoy]

My question is Atennis player receives a shot with the ball (0.06 kg)
traveling horizontally at 50 m/s and returns the shot with the ball traveling the opposite direction at 40 m/s (a) What is the impulse delivered to the ball by the racquet?
(b) What work does th racquet do on the ball?

Ok I have done a force diagram but I can't figure out the formula for (a) but on (b) The work done on the ball would be the change in velosity squared times half of the weight of the ball. Because of the Kentic engery therom. Right?

 

[stunner5000pt]

Impulse is $$\Delta p = F \Delta t$$
what is the change in momentum here?

the work is the change is kinetic nad potential energies
since there is no heigh difference here, just care about difference in kinetic energies

so w = Final - initial kinetic energies and you know kinetic energy is given by 1/2 mv^2, right?

 

[Satoy]

The change in momentum would be 90 m/s right 50-(-40). And I am not quite sure I understand your equation because I don't have a delta t. Do I?

 

[stunner5000pt]

The change in momentum would be 90 m/s right 50-(-40). And I am not quite sure I understand your equation because I don't have a delta t. Do I?

dont worry about the t the impulse is simplk the change in the momentum and omentum is p = MV

 

[Satoy]

the units would be kg*m/s for (a) right? And for (b) J

 

[stunner5000pt]

the units would be kg*m/s for (a) right? And for (b) J

yes those are the correct units

 

[dextercioby]

For the last part,apply the theorem of variation of KE:$$\Delta KE=W$$,where W stands for work done by forces acting on the ball...

Daniel.

 

[Doc Al]

The change in momentum would be 90 m/s right 50-(-40).

That's the change in velocity, not momentum.

 

[Satoy]

Ok I've got it now...thanks for all the help.