What Was the Angle Between Two Objects in an Inelastic Collision?

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In a completely inelastic collision involving two objects of equal mass, both initially moving at speed v, they move together post-collision at speed v/3. To determine the angle between their initial directions, momentum conservation principles are applied. The relationship derived from the analysis shows that the cosine of half the angle (θ) is equal to 1/3. This leads to the conclusion that the angle between their initial directions can be calculated using this cosine relationship. Understanding this concept is crucial for solving similar physics problems involving inelastic collisions.
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This problem's got me stumped.

"After a completely inelastic collision between two objects of equal mass, each having initial speed v, the two move off together with speed v/3. What was the angle between their initial directions?"

Any help would be appreciated!
 
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Use momentum conservation. You should find that

\cos \frac {\theta}{2} = \frac {1}{3}
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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