Momentum and the Conservation Laws

AI Thread Summary
The discussion centers on calculating the average force exerted on a steel ball by a wall after it strikes at an angle. The initial calculations incorrectly consider both the x and y components of velocity, leading to an erroneous result. It is clarified that only the x-component of velocity changes direction, while the y-component remains constant. The correct approach involves focusing solely on the change in the x-component of velocity to determine the average force. This highlights the importance of vector analysis in momentum calculations.
closer
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A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

p8-09.gif


F\DeltaT = \Deltap
p = mv

\Deltap = pf - pi = mvf - mvi = m(vf - vi)
\Deltap = m(vf - vi)
\Deltap = 4(-9 - 9)
\Deltap = -72
Faverage\Deltat = \Deltap
Faverage(.2) = -72
Faverage = -360

Final answer is incorrect, any ideas?
 
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closer said:
A 4.00 kg steel ball strikes a wall with a speed of 9 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P8.9) If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?

F\DeltaT = \Deltap
p = mv

\Deltap = pf - pi = mvf - mvi = m(vf - vi)
\Deltap = m(vf - vi)
\Deltap = 4(-9 - 9)
\Deltap = -72
Faverage\Deltat = \Deltap
Faverage(.2) = -72
Faverage = -360

Final answer is incorrect, any ideas?


The only thing that reversed is the x-component of velocity.
The y-component remained the same and since you are taking differences in Vectors - Velocity is a vector after all - then the force should bbe determined off of the change in x-component of velocity.
 
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