Calculating Impulse for 0.5kg Ball Bouncing off Wall: Theta = 50

[ted_chou12]

Hi I am not sure about the idea of momentum p=mv:

3: A 0.5kg rubber ball bounces off a wall (below left). The speed of the ball is 8m/s before
and after the collision. Use theta = 50.
a: Determine the impulse given to the ball. In other words, find (delta)p.

The diagram looks like
bounce up to here

|    /
|t /
|/
|\
|t \
|    \

start here
the two t's represents theta.
Thanks,
Ted

 

[Encephalon]

The initial momentum of the ball is $$p = mv = (.5)(8)=4$$ and this is at a fifty degree angle with the vertical. Therefore, the y-component is $$psin(\theta)=4sin(50)$$ and the x-component is $$pcos(\theta)=4cos(50)$$ When the ball bounces off the wall, the y-component of momentum is staying the same (not accounting for gravity in this case). Meaning, that only the x-component of momentum has changed. Since the final angle is the same as the initial angle, the change in momentum is enough for the x-component of momentum to have the same magnitude in the opposite direction, meaning it was two times the original x-component, so your $\Delta P$ should actually be $$2pcos(\theta)=8cos(50)$$

I'm pretty sure you're not supposed to account for gravity in this problem and assume that the y momentum is zero, but I could be thinking through this wrongly.

 

[ted_chou12]

thank you very much! this answer is perfectly matching the answer key 6.13kgm/s(x).