Dynamics problem no.3 (Mimix)

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Alright this one’s a tricky one A fluffy throws a tennis ball with velocity of 10 m/s against a wall, as shown. If e = 0.7, what is the magnitude of the rebound velocity?
1st person to solve this gets to choose what art style I’'ll make my next dynamics problem. Good luck!!

Hint: the answer isn’t 7.85

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:kissing_heart::smiling_face_with_three_hearts::heart_eyes::star_struck::partying_face:
Adorable!

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7 mayhaps :face_with_monocle:

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-10 m/s since it’s moving in the opposite direction.

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I get 9.35 m/s.

To find the magnitude of the rebound velocity, we need to consider the components of the initial velocity and the coefficient of restitution (E).

The initial velocity (v) is 10 m/s at an angle of 30 degrees above the horizontal. We can break this down into its horizontal (v_x) and vertical (v_y) components:

v_x = v * cos(30) = 10 * 0.866 = 8.66 m/s
v_y = v * sin(30) = 10 * 0.5 = 5 m/s

When the ball hits the wall, the horizontal component of the velocity remains unchanged, but the vertical component is reversed and reduced by the coefficient of restitution (E).

The rebound velocity (v’) has the same horizontal component (v_x) and a new vertical component (v_y’):

v_y’ = -E * v_y = -0.7 * 5 = -3.5 m/s

The magnitude of the rebound velocity (v’) is the square root of the sum of the squares of its components:

v’ = sqrt(v_x^2 + v_y’^2) = sqrt(8.66^2 + (-3.5)^2) = sqrt(75.0956 + 12.25) = sqrt(87.3456) = 9.35 m/s

So, the magnitude of the rebound velocity is approximately 9.35 m/s.

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Well actually the answer was 9.34 so sorry!!

jk jk although I am a bit confused as to why the horizontal component wouldn’t change considering its moving at an opposite direction, if you could clarify that part pls?

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When the ball hits the wall:

  • The wall pushes the ball back in the direction it came from (vertical direction). This slows down and reverses the vertical part of the ball’s motion.
  • The wall doesn’t push the ball sideways (horizontal direction). So, the horizontal part of the ball’s motion remains the same.

Think of it like a bounce on a trampoline. When you jump up and land on the trampoline, it pushes you back up, but it doesn’t make you move sideways. The sideways motion (horizontal) remains the same, but the up-and-down motion (vertical) changes.

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Ok that makes a bit more sense, I just assumed the horizontal component would become negative bc it would prob bounce off the wall, the vertical part makes a lot more sense now

You’re the winner again!! Whose art style should I try next??

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Well, yes, it would be moving in the opposite direction. The equations give a positive velocity though, even if going away from the wall.

You might ask your teacher on this one.

How about trying @Fluffus?

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Heck if I know because mathemathics & physics ain’t for me… but I can bet the real question is how much hurt the fluffy will be & how loud it will cry when the ball inevitably smacks it in the face when it bounces back. :troll:

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