How Do You Calculate the Compression of a Ball Hitting a Wall?

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In summary, the conversation discusses finding the average force exerted on a rubber ball hitting a wall, the change in kinetic energy of the ball during the collision, and estimating the amount of compression the ball experiences. The solutions provided involve using momentum, kinetic energy, and Hooke's law. However, it is suggested that the problem may require a simple F=ma solution instead.
  • #1
RK455
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Homework Statement
i) Find the average force exerted on a rubber ball mass m hitting a wall at speed v. The ball spends a time t in contact with the wall before bouncing off at speed v.
ii) The change in KE of the ball from when it hits the wall to when it is instantaneously at rest
iii) Assuming a steady force is acting on the ball, estimate the amount by which the ball is compressed during the collision

The attempt at a solution
i) This is easy enough
Average force = change in momentum / time taken = 2mv/t
ii) Again, easy: 0.5mv²
iii) If we assume the ball obey's Hooke's law (very rough, I know)
F=kx
=>2mv/t=kx or equivalently 2mvx/t=kx²
If all the KE of the ball is converted to EPE internally: mv²=kx²
Sub in kx²=2mvx/t giving 2mvx/t=mv²
Cancel and rearrange for x, the compression, giving: x=vt/2

The answers give vt/4, is there something I'm doing wrong? I'm guessing they made similar assumptions to me to get a similar answer but I can't see what I've done wrong.

Cheers.
 
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  • #2
RK455 said:
Homework Statement
i) Find the average force exerted on a rubber ball mass m hitting a wall at speed v. The ball spends a time t in contact with the wall before bouncing off at speed v.
ii) The change in KE of the ball from when it hits the wall to when it is instantaneously at rest
iii) Assuming a steady force is acting on the ball, estimate the amount by which the ball is compressed during the collision

The attempt at a solution
i) This is easy enough
Average force = change in momentum / time taken = 2mv/t
ii) Again, easy: 0.5mv²
iii) If we assume the ball obey's Hooke's law (very rough, I know)
F=kx
=>2mv/t=kx or equivalently 2mvx/t=kx²
If all the KE of the ball is converted to EPE internally: mv²=kx²
Sub in kx²=2mvx/t giving 2mvx/t=mv²
Cancel and rearrange for x, the compression, giving: x=vt/2

The answers give vt/4, is there something I'm doing wrong? I'm guessing they made similar assumptions to me to get a similar answer but I can't see what I've done wrong.

Cheers.

Note that the questioner said assume a steady force. This is certainly not the case in a typical Hookes Law situation, so perhaps you are supposed to do a simple F=ma etc solution.
Part (i) you calculated the average force - so the steady force needed for all this to happen.
Part (ii) you calculated the energy change.

I think it is now just work [Fx] = change in energy.
 

Related to How Do You Calculate the Compression of a Ball Hitting a Wall?

1. How does the angle of the ball affect its trajectory when hitting a wall?

The angle of the ball can greatly affect its trajectory when hitting a wall. If the ball hits the wall at a shallow angle, it will bounce off at a sharper angle. If the ball hits the wall at a steeper angle, it will bounce off at a more shallow angle.

2. What is the relationship between the speed of the ball and its bounce off the wall?

The speed of the ball can greatly impact its bounce off the wall. The faster the ball is moving, the more force it will exert on the wall, causing it to bounce off with more speed and momentum. On the other hand, a slower moving ball will have less force and will bounce off with less speed and momentum.

3. How does the material of the ball and the wall affect the bounce?

The material of the ball and the wall can greatly impact the bounce. A softer ball and a softer wall will result in a less bouncy bounce, whereas a harder ball and a harder wall will result in a more bouncy bounce. This is because the harder materials can absorb and transfer more energy during the impact.

4. Why does the ball sometimes bounce off at an angle after hitting the wall?

This is due to the conservation of momentum and the law of reflection. When the ball hits the wall, it transfers its momentum to the wall, causing it to bounce off in the opposite direction. However, the angle of reflection will be equal to the angle of incidence, resulting in the ball bouncing off at an angle.

5. What factors can affect the distance the ball travels after hitting the wall?

The distance the ball travels after hitting the wall can be affected by several factors such as the initial speed of the ball, the angle at which it hits the wall, the material of the ball and the wall, and any external forces acting on the ball, such as air resistance. Additionally, the surface of the wall (smooth or rough) and any imperfections or irregularities on the surface can also impact the distance the ball travels.

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