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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?

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  • $\begingroup$ 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
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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.

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