1
$\begingroup$

An object of mass m moves with velocity $v$ towards a stationary object of same mass. Impact is an elastic collision.

$v_1$ is the velocity after impact of the mass originally moving

$v_2$ is the velocity after impact of the mass originally stationary

Elastic collision means K.E. is conserved, so: $$\frac{1}{2}mv^2=\frac{1}{2}m(v_1)^2+\frac{1}{2}m(v_2)^2$$ Thus, $$v^2=(v_1)^2+(v_2)^2$$

However, using relative velocities:

$$-v=(v_1)-(v_2)$$

Squaring both sides gives a different value for $v^2$. Instead of $v^2=v_1^2+v_2^2$, $v^2=v_1^2-2v_1v_2+v_2^2$. How come?

$\endgroup$
  • $\begingroup$ What is the velocity of the mass originally moving after the collision? $\endgroup$ – user28355 Mar 7 '14 at 1:18
  • $\begingroup$ The relative velocities equation comes from both energy conservation and momentum conservation. So why should squaring the last get the first velocity equation? $\endgroup$ – jerk_dadt Mar 7 '14 at 1:26
  • $\begingroup$ Please explain how u came up the relative velocitiess $\endgroup$ – mcodesmart Mar 7 '14 at 6:20
1
$\begingroup$

There is no reason squaring equation (2) should give equation (1), because they are independent equations. You can use this fact to solve for $v_1$ and $v_2$; if this weren't so then using both conservation of energy and momentum would be rather useless.

$\endgroup$

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.