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When an in-motion object collides with a stationary object, does that necessarily mean that the in-motion object will always become stationary after the collision because of the equation:

$$ v_1=(m_1-m_2)u_2 + \frac{2 m_2 u_2}{m_1+m_2}$$

Where $v_1$ = velocity of the in-motion object after the collision, $m_1$ = mass of the in-motion object, $m_2$ =mass of the stationary object, $u_2$ = velocity of the stationary object. And because $u_2 = 0$, doesn't that mean that $v_1$ will always be $0$?

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3 Answers 3

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If the masses of the two colliding bodies are equal and the moving body collides with the stationary body, they will exchange their velocities according to the equation you stated. It is a consequence of the elastic nature of the collision. However, this equation does not apply if the collision is not perfectly elastic, i.e. the kinetic energy isn't conserved.

I would like to add that the equation you mentioned has a small mistake. The correct equation is:

$$v_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2} + \frac{2 m_2 u_2}{m_1 + m_2}$$

You miswrote the first fraction. Not that it changes anything in this situation as ($m_1 - m_2$) evaluates to zero.

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  • $\begingroup$ I think OP mentioned the correct equation but wasn't obvious because not using mathjax. (And your suggested edit had made another big mistake, which I corrected) $\endgroup$
    – ACB
    Sep 11, 2021 at 12:12
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    $\begingroup$ I believe your equation is not dimensionally correct @ACB $\endgroup$
    – Mechanic
    Sep 11, 2021 at 12:15
  • $\begingroup$ Well, I am not allowed to change what OP has written, despite applying mathjax. You can view the edit history and compare the original equation and mine.(and your suggested edit) $\endgroup$
    – ACB
    Sep 11, 2021 at 12:17
  • $\begingroup$ I agree that the equation is not correct, but it is what OP has written. (The reason we are not allowed to correct formula is that it may be the source of confusion of the asker) $\endgroup$
    – ACB
    Sep 11, 2021 at 12:21
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When an in-motion object collides with a stationary object, does that necessarily mean that the in-motion object will always become stationary after the collision ?

Not necessarily. It depends on the relative masses of the objects and the coefficient of restitution of the collision.

Suppose the in-motion object has mass $m_1$ and initial velocity $u_1$ and becomes stationary after the collision. Suppose the initially stationary object has mass $m_2$ and velocity $v_2$ after the collision. Then by conservation of momentum we have

$$m_1u_1 = m_2 v_2$$

If the coefficient of restitution is $e$ then we also have

$$v_2 = eu_1 \\ \Rightarrow m_1u_1 = em_2u_1$$

But $u_1$ is non-zero (no collision can take place if both objects are stationary) so we can divide both sides by $u_1$ to get a necessary and sufficient condition:

$$m_1 = em_2$$

In a perfectly elastic collision we have $e=1$, so the condition becomes $m_1=m_2$.

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Head to head elastic collision

$$v_{1f}={\frac { \left( m_{{1}}-m_{{2}} \right) { v_{1i}}}{m_{{2}}+m_{{1}}}}+2 \,{\frac {m_{{2}}{\,v_{2i}}}{m_{{2}}+m_{{1}}}}$$

$$v_{2f}=2\,{\frac {m_{{1}}{\,v_{1i}}}{m_{{2}}+m_{{1}}}}+{\frac { \left( m_{{2}} -m_{{1}} \right) {v_{2i}}}{m_{{2}}+m_{{1}}}}$$

  • $v~$ velocity
  • index i intial
  • index f final

you can check that with those equations the conservation of the energy $$\frac 12\,m_1\,v_{1f}^2+\frac 12\,m_2\,v_{2f}^2= \frac 12\,m_1\,v_{1i}^2+\frac 12\,m_2\,v_{2i}^2$$ and the conservation of the linear momentum $$m_1\,v_{1f}+m_2\,v_{2f}=m_1\,v_{1i}+m_2\,v_{2i}$$ is fulfilled.

Those:

if $~m_1=m_2~$ and $~v_{2i}=0~$ then $~v_{1f}=0~$ and $v_{2f}=v_{1i}$

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