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Let us consider two bodies (A and B) lying on a surface at a certain distance from each other. Let A be accelerating to the right, towards B and hence, the force experienced by A be F. When A collides with B, A would exert force F on B and thereby also experience a force F in the opposite direction by the third law of motion. This should bring A to rest immediately. However that isn't the case in reality. An accelerating body on collision doesn't come to rest immediately. Why?

Edit: The force that was accelerating A would remain there throughout the process.

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closed as unclear what you're asking by Bob D, Jon Custer, John Rennie, ZeroTheHero, Kyle Kanos Sep 17 at 11:24

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    $\begingroup$ When A collides with B will the same force that was accelerating A still be there? $\endgroup$ – Bob D Sep 15 at 17:09
  • $\begingroup$ Because there are no infinite forces. But for all practical purposes in this example it is fast enough to be negligible. $\endgroup$ – Wolphram jonny Sep 15 at 22:27
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    $\begingroup$ In order to answer your question properly, you need to state whether or not the force $F$ is removed from body A just prior to the collision. That is assumption in derbast answer. On the other hand, ABatt is assuming the force remains. You need to clear this up otherwise I will vote to close based on it is unclear what you are asking. $\endgroup$ – Bob D Sep 15 at 23:58
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Since the total momentum in a closed system is conserved, it follows that:

$p_A + p_B$ = $p'_A + p'_B$

where $p_{A,B}$ is the momentum of body A/B before the collision and $p'_{A,B}$ after the collision.

Since $p = m \cdot v$:

$m_A \cdot v_A + m_B \cdot v_B = m_A \cdot v'_A + m_B \cdot v'_B$

In your question you only mentioned, that body A is being accelerated, so we will set $v_B = 0$.

If you now want to achieve that body A will be standing still instantly after the collision, $v'_A$ has to be equal to $0$. Putting this into the above equation yields:

$m_A \cdot v_A = m_B \cdot v'_B$

As long as this equation is satisfied, body A will be at rest after the collision.

P.S. $v_A = \int_{t=0}^{t_{impact}} \frac{F(t)}{m_A} dt.$

Remeber, that this is a very simplified case, because in a real-world situation other effects -- e.g. drag, friction, etc. -- would have to be considered. Furthermore, I assumed that the force acting on body A will stop acting on A as soon as the collision happens.

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As you've stated, Newton's third law of motion would tell us that the forces experienced by A at the point of collision are equal and opposite (The original force being applied to A is still being applied, while B is also exerting the same force in the opposite direction). This would mean that there is no net force being applied to A because they cancel each other out.

However, this does not mean that the object will come to rest because there is no net force. Thinking independently of the collision itself, Newton's first law would tell us that A will remain in motion because there is no net force to slow it down, and it will stop accelerating because there is no net force to speed it up.

Thinking with the collision, what actually happens at that point will depend on the masses of A and B, and applying the conservation of momentum to them, but if the same force that was being applied to A is still being applied after the collision, it will continue to accelerate after the collision as it was before it.

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  • $\begingroup$ So can I conclude likewise: If no force act on body A after the collision, it would remain in the velocity it had 'during' collision. ? $\endgroup$ – Zam Sep 16 at 3:01

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