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Suppose I am throwing horizzontally a ball with mass "m" against a wall (whose mass and stiffness can be assumed to be infinite).

I accelerate the ball with a constant force "F" along a distance "s" before letting go. So, calling ""$v_f$" the final velocity of the ball, I can write (work - energy theorem):

$$F\cdot s = 1/2 \cdot m \cdot v_f^2$$

Hence I can express the final velocity in the following way:

$$v_f = \sqrt{ \frac {2 \cdot F \cdot s} {m} }$$

During the collision with the wall, the ball will reach a stop point (just before bouncing back). At this point, all the original kinetic energy of the ball has been converted to elastic energy (if we assume the collision to be totally elastic). Calling "k" the ball stiffness and "$x_f$" the ball deformation at this point, we can write

$$1/2 \cdot m \cdot v_f^2 = 1/2 \cdot k_f \cdot x^2$$

From this expression we can calculate the peak force $P_{max} $ of the collision:

$$P_{max} = k \cdot x_f = v_f \cdot \sqrt{m \cdot k} $$

So far so good. I 'm pretty sure this expression is correct, since I read it somewhere else too.

But here comes my doubt. If I substitute $v_f$ with the expression derived earlier, I end up with

$$P_{max} = \sqrt{ \frac {2 \cdot F \cdot s} {m} } \cdot \sqrt{m \cdot k} = \sqrt {2 \cdot F \cdot s \cdot k}$$

This expression suggests that the peak impact force does not depend on the mass of the object I am throwing! Where did I go wrong? (it seems counterintuitive to me, since I would expect the force to grow with the mass, even if the velocity decreases)

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Your derivation is already neat, whole and intact. The final two expressions are simply alternative forms of the peak force. While the second one suggests it's independent of the mass, you might want to consider that $F=m\,a.$ And since you accelerated the ball from rest at a constant pace, we can use one of the equations of uniformly accelerated motion to relate the acceleration "a" and distance covered "s". Using $$\begin{align}v^2 &=v_0^2+2as\\s &={v^2\over 2a}\;given\; v_0=0ms^-1 \end{align}$$ Putting $s$ back into your second expression for $P_{max}$ and using $F=m\,a$, we have $$\begin{align}\sqrt{2.F.s.k}&=\sqrt{2.m.a.{v^2\over 2a}.k}\\\therefore P_{max}&=v\,\sqrt{m.k}\end{align}$$ as required.

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I don't see what is necessarily wrong with leaving it in the form with $v_f$ and concluding that it is proportional to $m^\frac{1}{2}$, but nonetheless we can at least come up with some physical intuition as to why the final answer in independent of mass.

If you think about what happens when you increase the mass, but don't increase the force, the final velocity $v_f$ is going to be lower, so the ball hits the wall at with less velocity, and the total momentum change is going to be smaller. In light of this, it doesn't at all seem unreasonable that changing the mass without changing the energy imparted to the ball would leave the peak reaction force unchanged.

This is a specific example of something that comes up a lot in physics - forces and energies and velocities and all sorts of other things can often be seen as having different proportionalities depending on what we choose to hold constant. In this case, when you hold energy constant, there is no proportionality between mass and peak force, but when other variables, such as final velocity is held constant, the is a proportionality relation.

I am also going to mention that I'm not entirely sure that there can actually be an elastic collision with a compressible ball, even allowing for the other infinite assumptions, based on some oscillation that will occur when the ball re-expands after leaving the wall.

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  • $\begingroup$ Thank you for the reply. I see your point, but I'm still not sure about the correctness of my last expression (even within the ideal model it is based on). If you wanted to damage a wall by throwing an object against it with full force, would you choose a 5 kg ball or a 0,5 kg ball? According to the expression derived, your choice should not matter a lot. But that's clearly counterintuitive $\endgroup$ Jun 10, 2020 at 0:44
  • $\begingroup$ Possible explanation: since the lighter ball should require higher velocity, this would imply that you accelerate your arm and hand to a higher velocity, too. This would require a higher effort. But that's the only possible reason I see to choose the heavier ball instead, provided my derivation is correct. What do you think? $\endgroup$ Jun 10, 2020 at 0:49

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