Calculating Impulse in a Tennis Ball Collision with a Wall

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A tennis ball of mass m=0.060kg and speed v=25m/s strikes a wall at a 45 degree angle and rebounds with the same speed at a 45 degree angle. What is the impulse (magnitude and direction)?

F=deltaP=mv=deltaP/deltaT

Not sure what I am doing wrong. mv=P so 0.060kg x 25m/s = 1.5Nm/s
The book says the answer is 2.1kg x m/s
How do I solve this. What equations am I forgetting?
 
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Couple of things to consider.

Impulse is a vector quantity, so direction is important. That's where the 45 degrees comes in.

What does "delta" stand for in deltaP?
 
Break the initial and final velocities into x and y components to find Δv. Use v2-v1 on to find that change in velocity on both x and y, then use trig to find hypotenuse and that will be your Δv. Good luck
 
deltaP equals change in P, or change in momentum. I just now noticed the quick symbols on the right. The ball is going up to the right and hits the wall at 45 degrees to the wall. The ball bounces off the wall and is going left 45 degrees from the wall.
 
Exactly, so assign directional convention to your problem and you will get it!
 
Try using vectors here :

what is the initial velocity of the ball ?

what is the final velocity of the ball ?

what is the change in velocity ?

Multiply the change in velocity by the mass of the particle . You got the magnitute.

The direction of the impulse is the same as the direction of the force.
A force was applied by the wall on the particle PERPENDICULAR to its surface.
Thus, the direction of impulse is ?
 
You have to resolve the velocity into its component. For me, I resolved it in the horizontal component. Then, you have to consider the direction. After that, just use deltaP=P(final)-P(initial), Take note of the signs. You can use P=mv and then you can find impulse. The direction of the impulse will be in the same direction as the force applied. Hope I helped!