Momentum of a Tennis Ball at an Angle

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SUMMARY

The discussion centers on the momentum change of a tennis ball with a mass of 0.060 kg and an initial speed of 26 m/s striking a wall at a 45° angle and rebounding at the same speed. The initial momentum is calculated as 1.1028 kg·m/s, while the final momentum is -1.1028 kg·m/s, leading to a total change in momentum (Δp) of -2.2056 kg·m/s. The participant seeks clarification on the direction of the impulse imparted to the ball, having eliminated downward and leftward options.

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Homework Statement



A tennis ball of mass = 0.060 kg and speed = 26 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° .

Homework Equations



Δp = pfinal - pinitial
horizontal velocity = (26 m/s)cos 45°

There's a picture if you all care.

The Attempt at a Solution



pintial = mv
pintial = (0.060 kg)cos 45°
pinitial = 1.1028 kgm/s

pfinal = - pintial

I made this relationship because it is now going in the opposite direction.

pfinal = -1.1028 kgm/s

Δp = -1.1028 kgm/s - 1.1028 kgm/s = -2.2056 kgm/s

The answer is positive and I cannot understand why.

What is the direction of the impulse given to the ball?

A. Upward
B. Downward
C. Left
D. Right

I've eliminated B and C but I'm still unsure. The sine and cosine of forty-five degrees are the same.
 

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Momentum is a vector. It's magnitude is m*v, where v is the speed.

The change in momentum would be found by finding the momentum vectors for before and after and finding the difference.
 

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