Two particles with masses m and 8m are moving toward each other along the x axis

Question

Two particles with masses m and 8m are moving toward each other along the x axis with the same initial speeds vi. Particle m is traveling to the left, while particle 8m is traveling to the right. They undergo an elastic, glancing collision such that particle m is moving in the negative y direction after the collision at a right angle from its initial direction. (a) Find the final speeds of the two particles in terms of vi. particle m: x vi particle 8m: x vi (b) What is the angle ? at which the particle 8m is scattered?

Explanation / Answer

The momentum of m before collision = m vi The momentum of 4m before collision = - 7m vi (Negative sign because it is moving toward right). Total momentum before collision = m vi – 7m vi = - 6mvi. --------------------------------------… Let w1 be the velocity of m after collision in the downward direction Let w2 be the velocity of 4m after collision at an angle ? to the initial direction. In the up and down direction, taking downward as negative, Total momentum of both after collision is -mw1 + 7m w2 sin ?. This must be equal to the initial momentum of both in this direction which was = 0. Hence 7m w2 sin ? = m.w1 w2 sin ? = w1 / 7 .---------------1 --------------------------------------… In the left and right direction Total momentum of both is 0 + 7m w2 cos ?. Initial momentum in this direction was 6mvi. Hence 6vi. = 7 w2 cos ?. w2 cos ?. = 6 vi/7 .----------------2 From 1 and 2 w2^2 = [w1 /7]^2 + [6 vi/7]^2 -----------3 --------------------------------------… Considering the kinetic energies of the particles Initial k.e = 8m vi^2 . Final k.e = m w1^2 + 7m w2^2 Hence 8m vi^2 = m w1^2 + 7m w2^2 8vi^2 = w1^2 + 7 w2^2 Eliminating w2 using 3 8vi^2 = w1^2 + 7{[w1 / 7]^2 + [6 vi/7]^2} 8vi^2 = w1^2 + w1^2 / 7 + 36vi^2 / 7 2.857 vi^2 = 1.14 w1^2 w1 = ± 1.583 vi From 3 w2^2 = [w1 /7]^2 + [6 vi/7]^2 w2^2 =[ 1.583 vi / 7] ^2 + [6 vi/7] ^2 w2^2 = 0.05114vi^2 ^2 + 0.73469vi ^2 w2 = ± 0.88647 vi. w2 sin ? = w1 / 7 = 1.583vi /7 =0.226 vi w2 cos ?. = 6vi/7 . Tan ?.= 0.2638 ? = 14.78°

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