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Impulse and Momentum: Small steel ball moving with speed v0

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A small steel ball moving with speed v0 in the positive x-direction makes a perfectly elastic noncentral collision with an identical ball originally at rest. After impact, the first ball moves with speed v1 in the first quadrant at an angle theta1 with the x-axis, and the second moves with speed v2 in the fourth quadrant at an angle theta2 with the x-axis.

a) what would be the equations expressing conservation of linear momentum in the x-direction and in the y-direction? What is the outcome if I square and add these?

b.) How can I prove that theta 1 + theta 2 = pi/2

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Solution Summary

The impulse and momentum of a small steel ball moving with speed v0 is determined. With good explanations and calculations, the problems are solved.

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a.) because, by law of conservation of momentum in x-direction:
m*v0 + m*0 =m*v1*cos(Q1) + m*v2*cos(Q2)
here, read Q1 as theta1 and Q2 as theta2.
=> v1*cos(Q1) + ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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