Average force exerted on ball by wall at angle

AI Thread Summary
An 8kg steel ball strikes a wall at 11.7 m/s and an angle of 35.3 degrees, bouncing off with the same speed and angle. The average force exerted by the wall can be calculated using the change in momentum over the contact time of 0.297 seconds. The horizontal component of the ball's velocity changes direction, while the vertical component remains constant. The calculations yield a force of approximately 257.21 N for the horizontal component, while the vertical component does not contribute to the force change. Understanding the direction of velocity components is crucial for accurately determining the average force.
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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

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 that's 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?
 
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Ft = mv
F(0.297) = 8 * 11.7
F = 315.15 kg* m/s^2 or N but that's 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.
 
Astronuc said:
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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