Physics Help Forum would forward force lessen this angle?

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 Aug 4th 2017, 07:41 AM #1 Junior Member   Join Date: Apr 2017 Posts: 10 would forward force lessen this angle? Suppose you have a person swinging / pushing a racket at a ball at an angle that is between forward and down, so e.g. down to the ground in the distance. That's scenario A. Now scenario B suppose instead, the same person is running and then hitting it with the racket tilt and swing at the same angle.. Will the forward motion of them running cause the ball to go at a less steep angle? This video explains visually what I mean..
 Aug 4th 2017, 08:02 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,287 In the case of the person running forward the ball will leave the racket at a less steep angle. The reason is due to the way that the racket interacts with the ball. Any relative motion of the racket that is not strictly in line with the angle of the racket (i.e. off-axis relative to the face of the racket) imparts a component of velocity in that off-axis direction. This assume that there is friction that occurs between ball and racket (such as with a tennis racket and tennis ball, or ping pong paddle and ping pong ball) so that some of that off-axis motion can be transferred into the ball.
 Aug 4th 2017, 04:50 PM #3 Junior Member   Join Date: Apr 2017 Posts: 10 How would one calculate the new angle? Let's say it's a tennis ball, and the 45 degree move is 1 Kilo Newton of force, and the forward motion is 5 Kilo Newtons of force.. The new steepness would be somewhat less than 45 degrees.. But how much less?
 Aug 5th 2017, 06:33 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,287 if you actually had a wau to measure the forces that you're referring to, the angle would be determined by the angle of the sim of the force vectors. But in real lfe you probably don't know those forces - what you may know are components of velocity of the racket. Unless you know how friction operates between racket and ball it's impossible to predct the launch angle simply from velocity components.
Aug 17th 2017, 03:05 PM   #5
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 Originally Posted by ChipB In the case of the person running forward the ball will leave the racket at a less steep angle. The reason is due to the way that the racket interacts with the ball. Any relative motion of the racket that is not strictly in line with the angle of the racket (i.e. off-axis relative to the face of the racket) imparts a component of velocity in that off-axis direction. This assume that there is friction that occurs between ball and racket (such as with a tennis racket and tennis ball, or ping pong paddle and ping pong ball) so that some of that off-axis motion can be transferred into the ball.

Scenario A-

So a snooker ball is struck against an edge aka cushion of a snooker table.. the snooker table is rotated 45 degrees as in this pic, and so i'm sure the ball will rebound straight downwards.

And I think the harder the ball is struck, the faster it will go, but I'm sure it would always go in that direction.. the angle won't change depending on the speed.

Now consider this, Scenario B

This time instead of the ball hitting the cushion, the cushion hits the ball. The ball is sitting where the cushion was, but the cushion can be moved back and spring forwards along those blue lines back into position. As in this picture. (note- more text below the picture)

Now, in scenario B, will the ball still go straight down?

If so, then why should the racket case in the original post be any different? (like why should it be that when the racket has a forward momentum it makes the ball go at less of an angle)..

And if not, then why does it make a difference to the angle of the ball, if the ball hits the cushion vs the cushion hitting the ball?

Last edited by ralphza; Aug 19th 2017 at 01:59 AM.

 Aug 18th 2017, 12:06 PM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,287 You raise a valid point. For the case of the snooker ball its direction is dictated by the point of contact between the cushion and ball. In the two scenarios you describe that point of contact is at 45 degrees relative to horizontal, hence the force applied to the ball is at 45 degrees downward, and its leftward velocity is changed to strictly downward velocity. This happens in both scenarios A and B. But I want to note that this is strictly true only if there is no deformation of either ball or cushion at the moment of impact. Stated another way - this assumes that the duration of the impact is precisely 0 seconds. A snooker ball is very hard, and the velocity of impact is quite low compared to a tennis ball being hit by a racket, and in essence any deformation of the ball when it hits the cushion can be ignored. As a result it follows the "angle of incidence = angle of reflection" model. I feel this is different from your original scenario. When a tennis ball interacts with a tennis racket both the ball and the strings of the racket deform (ever see a slow-motion movie of a tennis ball being struck by a racket?), and hence there is not a single point of contact between racket and ball and some of the horizontal momentum of the racket is transferred into the ball. The result is that the ball takes on a component of velocity in the direction of the moving racket.
 Aug 19th 2017, 01:55 AM #7 Junior Member   Join Date: Apr 2017 Posts: 10 Thanks, that is fascinating

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