Answer to Question #132628 in Mechanics | Relativity for max

Question #132628
A ball hits the surface with initial momentum p repeatedly find the momentum transferred to the surface.
1
Expert's answer
2020-09-16T21:05:05-0400

The initial momentum of the ball is p:

The momenta of the ball after each successive collision be "p_1\n\n\u200b, p_2\n\n\u200b, p_3" and so on till it comes to rest  

For 1st collision :

"\u27f9p _\n1\n\u200b\t\n =ep"

Thus net momentum imparted to the floor by 1st collision, 

"P _\n1^\n\u2032\n\u200b\t\n =p _\n1\n\u200b\t\n \u2212(\u2212p)"

"\u2234 P _\n1^\n\u2032\n\u200b\t\n =ep+p=p(1+e) .............(1)"


For 2nd collision : 

"\u27f9p_ \n2\n\u200b\t\n =ep _\n1\n\u200b\t\n =e \n^2\n p"

Thus net momentum imparted to the floor by 2nd collision,

"P _\n2\n^\u2032\n\u200b\t\n =p _\n2\n\u200b\t\n \u2212(\u2212p_ \n1\n\u200b\t\n )"

"\u2234 P _\n2\n^\u2032\n\u200b\t\n =e \n^2\n p+ep=pe(1+e) .............(2)"


Similarly momentum imparted to the floor by each other successive collisions are -

"pe ^\n2\n (1+e), pe \n^3\n (1+e)," and so on


Total momentum imparted "P=p(1+e)+pe(1+e)+pe \n^2\n (1+e)+pe \n^3\n (1+e)+.........\u221e\\ term"

"P=p(1+e)[1+e+e \n^2\n +e \n^3\n +............\u221e]"

"P=p\\times\\frac{1+e}{1-e}"


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