Average force exerted on ball by wall at angle

In summary, the problem involves an 8kg steel ball striking a wall with a speed of 11.7 m/s at an angle of 35.3 degrees. The ball bounces off with the same speed and angle after being in contact with the wall for 0.297 seconds. The question asks for the magnitude of the average force exerted on the ball by the wall. Using the equation Ft=mv, the force is calculated to be 315.15 kg*m/s^2 or N in the linear case. However, considering the angle, the horizontal and vertical components of velocity are calculated to be 9.549 m/s and 6.761 m/s respectively. Using these values in the equation, the force
  • #1
sensesfail
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0

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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  • #2
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.
 
  • #3
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?
 

1. What is the formula for calculating the average force exerted on a ball by a wall at an angle?

The formula for calculating the average force exerted on a ball by a wall at an angle is F = m * (Δv/Δt), where F is the average force, m is the mass of the ball, Δv is the change in velocity, and Δt is the change in time.

2. How does the angle at which the ball hits the wall affect the average force exerted?

The angle at which the ball hits the wall affects the average force exerted by changing the direction and magnitude of the change in velocity. The greater the angle, the larger the change in velocity and therefore the greater the average force exerted on the ball by the wall.

3. Can the average force exerted on a ball by a wall at an angle be negative?

Yes, the average force exerted on a ball by a wall at an angle can be negative if the ball is slowed down by the wall. This would occur if the ball bounces off the wall at an angle that causes it to lose speed.

4. What factors can affect the average force exerted on a ball by a wall at an angle?

The factors that can affect the average force exerted on a ball by a wall at an angle include the mass of the ball, the angle at which the ball hits the wall, the elasticity of the ball and the wall, and any external forces acting on the ball.

5. How is the average force exerted on a ball by a wall at an angle related to the ball's kinetic energy?

The average force exerted on a ball by a wall at an angle is directly proportional to the ball's change in kinetic energy. This means that the greater the average force, the greater the change in kinetic energy and vice versa. This relationship can be described by the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

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