Average force exerted on object by wall

[shadowice]

[solved]Average force exerted on object by wall

Handball
A 301.0 g handball moving at a speed of 5.3 m/s strikes a wall at an angle of 29.0° to the normal of the wall and then bounces of with the same speed at the same angle. It is in contact with the wall for 0.002 s. What is the average force exerted by the ball on the wall?
m= 3.01kg
v int = 5.3 m/s
v final = -5.3 m/s
theta = 29 deg
time = 0.002s
quick pic i drew of how i set it up

Homework Equations

F avg = (Pfinal - P initial)/time
P = m*v

The Attempt at a Solution

P int = 3.01kg*5.3m/s = 15.953
P final = 3.01kg*-5.3m/s = -15.953
F avg = (-15.953-15.953)/0.002 = -15953 N Lon capa declined this answer is exceptionally large for a tenis ball so i tried to include the angle it gave us.

Pint = -2mvcos29=-27.906
Pfinal = 2mvsin29=15.468

(15.468-27.906)/.002 = -6218.64 which is wrong as well

next i tried (-5.3cos(29)-5.3cos(29)*3.01)/.002 = -9294 N

 

[CompuChip]

So trying every combination with sine and cosine is not going to work.
What happens to the horizontal and vertical components separately, and how do you calculate those components?

 

[shadowice]

ok so id take
Px = 3.01(5.3cos(29)-5.3cos(151)= 27.9
Py = 3.01(5.3sin(29)-5.3sin(151)= 0

then dP^2 = Px^2 + Py^2
dP = Px

F avg = dp/dt = 27.9/.002s thnx for tip i got it finally

 

[CompuChip]

Yes. Note that

sin(151) = sin(29)​

cos(151) = -cos(29)​

So basically you are saying that the vertical component does not change, because there are no forces acting in that direction. The horizontal component of the velocity has the same magnitude but opposite direction after the collision, so it goes from v to -v (a difference of 2v) - hence the horizontal component of the momentum simply doubles.