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Average force exerted on ball by wall at angle

  • Thread starter sensesfail
  • Start date
  • #1

Homework Statement



a 8kg steel ball strikes a wall with a speed of 11.7 m/s at an angle of 35.3 degrees with the normal to the wall. it bounces off with the same speed and angle as shown in the picture.

http://img405.imageshack.us/img405/7733/asdfkf7.jpg [Broken]

If the ball is in contact with the wall for 0.297 seconds, what is the magnitude of the average force exerted on the ball by the wall? Answer in N

Homework Equations





The Attempt at a Solution



Ft = mv
F(0.297) = 8 * 11.7
F = 315.15 kg* m/s^2 or N but thats if its a linear problem

Angle wise

Horizontal speed of 11.7 m/s
Cos 35.3 = x / 11.7
x = 9.549 m/s

Extra info
Vertical speed of 11.7 m/s
sin 35.3 = y / 11.7
y = 6.761 m/s

Ft = mv
F(0.297) = 8 * 9.549 m/s
F = 257.2069 N

is that right?
 
Last edited by a moderator:

Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
18,704
1,718
Ft = mv
F(0.297) = 8 * 11.7
F = 315.15 kg* m/s^2 or N but thats if its a linear problem

. . .

Ft = mv
F(0.297) = 8 * 9.549 m/s
F = 257.2069 N
Be careful here. Note that in the y direction, the y-component of velocity does not change. On the other hand, the x-component of velocity does change, in direction, but not in magnitude. The magnitudes in the x- and y- directions are correct.
 
  • #3
Be careful here. Note that in the y direction, the y-component of velocity does not change. On the other hand, the x-component of velocity does change, in direction, but not in magnitude. The magnitudes in the x- and y- directions are correct.
Does it matter if I include the y direction in the equation? Would I need to make two equations one going toward the wall and another going away? But then if I equal to them wouldn't it be zero?
 

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