0
$\begingroup$

How can an object that's not accelerating cause another stationary object to move in a collision? This website: http://www.physicsclassroom.com/Class/momentum/U4l2a.cfm, says that a "rightward moving" 7-ball experiences a leftward force when it hits an 8-ball, and the 8-ball experiences a rightward force, but where is this force coming from when the ball was moving with a constant velocity before the collision?

How does Newton's third law explain any collisions that involve a constant velocity?

EDIT: Does it slow down because of the normal force? And then, is there a normal force action-reaction pair? And, if so, how do you calculate the magnitude of the normal force?

Improve this question
$\endgroup$
13
  • $\begingroup$ Will the velocities be constant upon impact? $\endgroup$ – Kyle Kanos Aug 13 '13 at 18:17
  • $\begingroup$ Yes, they will be. $\endgroup$ – glaba Aug 13 '13 at 18:19
  • $\begingroup$ So the 7-ball passes through the 8-ball unimpeded? $\endgroup$ – Kyle Kanos Aug 13 '13 at 18:24
  • $\begingroup$ The force originates from the 2nd law not the 3rd. $\endgroup$ – John Alexiou Aug 13 '13 at 18:58
  • $\begingroup$ @Kyle That's what I concluded, but that makes no sense $\endgroup$ – glaba Aug 13 '13 at 19:38
1
$\begingroup$

Yes, the normal force applied over an tiny amount of time changes the speed (momentum) of the objects. The exact magnitude of the force depends on the time it takes for the impact to happen.

A force $F$ applied for a time period $\Delta t$ has impulse of $ J = F \Delta t $. The impulse is equal to the change of linear momentum in an opposing fashion (3rd law)

$$ \begin{aligned} m_1 \Delta v_1 & = J \\ m_2 \Delta v_2 & = -J \end{aligned} $$

The impact is characterized by a certain coefficient of restitution $\epsilon$ definiting the final relative velocities $v_1^F$ and $v_2^F$ in terms of the initial velocities $v_1^I$ and $v_2^I$ as

$$ \begin{aligned} (v_2^F-v_1^F) & = -\epsilon (v_2^I-v_1^I) \\ (v_2^I + \Delta v_2-v_1^I - \Delta v_1) & = -\epsilon (v_2^I-v_1^I) \\ ( \Delta v_2 - \Delta v_1) & = -(\epsilon+1) (v_2^I-v_1^I) \\ \frac{-J}{m_2} - \frac{J}{m_1} & = -(\epsilon+1) (v_2^I-v_1^I) \end{aligned}$$

$$ J = \frac{(1+\epsilon) v_{rel} } { \frac{1}{m_2} + \frac{1}{m_1} } $$

where $v_{rel} = v_2^I-v_1^I$ is the impact relative velocity. Once $J$ is known, $\Delta v_1$ and $\Delta v_2$ are calculated, as well as the force $F = \frac{J}{\Delta t}$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I never knew about the coefficient of restitution! Which brings up a good question: why don't many beginner physics resources have any information about it? But, regardless, thank you for the answer; you've solved a problem about momentum and Newton's Third Law that's been bugging me for a long time. $\endgroup$ – glaba Aug 13 '13 at 22:31
  • $\begingroup$ Well to my experience most physics books that talk about collisions and momentum do have the COR. $\endgroup$ – John Alexiou Aug 14 '13 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.