0
$\begingroup$

If a ball of mass 2kg that is moving, hits a stationary ball of mass 1kg, in an inelastic collision. How would we find which ball moves faster after the collision?

We have

mass of 2kg ball * velocity of 2kg ball (intial) = mass of 2kg ball * velocity of 2kg ball (after) + mass of 1kg ball * velocity of 1kg ball (after).

$m_{2kg}*v_{initial} = m_{2kg}*u_{2kg} + m_{1kg}*u_{1kg}$

How would we work out the velocity for each ball after collision?

Are they just distributed in proportion to the masses, i.e. would the velocity of the 1kg ball be twice the velocity of the 2kg ball?

$\endgroup$
  • $\begingroup$ You can't work out the velocity of each ball. Simply not enough information. But you can determine which ball moves faster. You do need some out of the box consideration though. Imagine each scenario and see if that scenario is viable/ feasible. $\endgroup$ – npojo Feb 8 '18 at 5:18
  • $\begingroup$ I believe the 1kg ball would move faster, if the same amount of momentum went into each ball (as the 1kg ball is lighter, so must have higher velocity, for a given amount of momentum). However, I don't really know if the same amount of momentum does enter each ball or not. Thank you. $\endgroup$ – K-Feldspar Feb 8 '18 at 5:27
  • $\begingroup$ Impulse $F*\delta t$ is the same so change in momentum is the same. But this is only change. It does not imply that final velocity of the smaller mass is higher than final velocity of the larger mass. There is one equation with two variables. Infinite solutions. Think for example if $u_{1kg}$ can be negative while $u_{2kg}$ is still positive in a real life situation. Then take it from there. $\endgroup$ – npojo Feb 8 '18 at 6:01
1
$\begingroup$

Let's assume that $$V_{2~\rm kg(before)}=e\left(V_{1~\rm kg(after)}-V_{2~\rm kg(after)}\right)$$ We called $e$ Recovery coefficient

Then you can get that $$ V_{1~\rm kg(after)}={2+2e\over 2e-1}V_{2~\rm kg(after)}$$

Of course there is $0\le e\le 1$ and while $e=0.5$ velocity of $2~\rm kg$ ball is zero, so $1~\rm kg$ ball runs faster.

$\endgroup$

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.