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This question is an exact duplicate of:

If a 2 times bigger mass collides with an object in which a 2 times smaller mass collides but with 2 times the speed which one will produce more force on the object?

The object is stationary before and its mass is comparable to the 2 mases.

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marked as duplicate by Gert, John Rennie newtonian-mechanics Mar 1 at 16:49

This question was marked as an exact duplicate of an existing question.

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This is kind of a loaded question because there are a lot of specifics you're leaving out. However, let's try to constrain the problem:

Assumptions:

  1. The two moving masses collide with a stationary object. The object does not move before, during, or after the collision.
  2. The time that it takes for the moving masses to come to a complete stop is the same. Let's call it $\Delta t$.

Moving mass 1 has velocity $v_1$ and mass $m_1$, and mass 2 has velocity $v_2$ and mass $m_2$. If they stop within $\Delta t$, then $a_1=\frac{v_1}{\Delta{t}}$, and $a_2=\frac{v_2}{\Delta{t}}$ (since they both came to rest).

Force is mass * acceleration. Therefore, $F_1=m_1a_1$ and $F_2=m_2a_2$.

Or:

$F_1=m_1\frac{v_1}{\Delta{t}}$, $F_2=m_2\frac{v_2}{\Delta{t}}$

But we know that $m_1=2 m_2$ and $v1=\frac{v_2}{2}$.

Plugging those in to the equation for $F_2$, we get:

$F_2=\left(\frac{m_1}{2}\right)\frac{(2 v_1)}{\Delta{t}}=m_1\frac{v_1}{\Delta{t}}=F_1$

So they produce the same force, given the assumptions described.

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  • $\begingroup$ The point is that the object moves too and it's mas is comparable to other 2 . Otherwise it's trivial. $\endgroup$ – Schaurberger Mar 1 at 18:25
  • $\begingroup$ The real point is that $F=\Delta p/ \Delta t = m\Delta v / \Delta t$. You need to worry about the product of mass multiplied by the change in velocity, divided by the time interval involved in the collision. This calculation for the colliding object is independent of the final motion of the object that it collided with, as Newton's third law guarantees that each object feels the same force of collision. $\endgroup$ – David White Mar 1 at 19:00

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