1
$\begingroup$

In an elastic collision between two masses, if one mass is much heavier than the other, then the heavier mass will continue to move with same velocity while the lighter mass doubles its velocity. How is the Law of Conservation of Energy conserved in this?

I concluded that the assumption of neglecting the change in velocity of the heavier mass is responsible for an increase in kinetic energy of the whole system, thus failing to satisfy the Law of Conservation of Energy. Is this a correct explanation?

$\endgroup$
  • $\begingroup$ How?You are answering your own question in second paragraph $\endgroup$ – user227513 Apr 13 at 18:48
  • $\begingroup$ I was just cross checking my answer. $\endgroup$ – Shine kk Apr 13 at 18:49
  • 1
    $\begingroup$ Note that questions of the form "Is this right?" tend to be poor fits for this site because the answer, yes or no, is too short to be a valid answer. Consider making the question more open ended so a proper answer can be written $\endgroup$ – Kyle Kanos Apr 15 at 11:37
3
$\begingroup$

You should recognize this

"the heavier mass will continue to move with same velocity"

as an approximation.

The actual velocity will be reduced, and you can figure out how much from the conservation of momentum. $$ \Delta V \approx - 2 V \frac{m}{M} \tag{for $M \gg m$}\;.$$ (Strictly speaking this is another approximation., but now we're talking about a small correction to a small correction and I'm going to ignore it.)

Now that is a small change in speed but by hypothesis it is connected with a large mass so it results in a non-trivial change of kinetic energy \begin{align} \Delta K_M &= \frac{1}{2} M \left[ (V + \Delta V)^2 - V^2\right] \\ &= \frac{1}{2} M \left[ (2 V \Delta V + (\Delta V)^2\right] \;. \end{align} Now, we factor one power of $\Delta V$ out and notice that it cancels the $M$ to give \begin{align} \Delta K_M &= - m V \left[ 2V + \Delta V \right] \\ &\approx - m V \left[ 2V \right] \\ &= - \frac{1}{2} m (2V)^2 \end{align} The large mass losses approximately the same energy as the small one gains.

Keeping all the terms is more algebraically difficult, but follows exactly the same pattern.

$\endgroup$

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.