A proton is traveling to the left at 2.8 107 m/s. It has a head-on perfectly ela

Question

A proton is traveling to the left at 2.8 107 m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

Explanation / Answer

Let the final velocity of proton be v(in opposite direction) and Carbon atom be V Velocity Of approach = 2.8*10^7m/s Velocity of separation = v+V For elastic collision we have velocity of approach = velocity of separation =>2.8*10^7 = v+V eq 1 My momentum conservation we have m*2.8*10^7 = -mv + 12mV => 2.8*10^7 = 12V-v eq2 Adding Eq 1 and Eq2 we have 5.6*10^7 = 13V => V = 4.3*10^6 in the direction same as the initial direction of proton => v = 5.17*10^7 in direction opposite to the initial direction of proton

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