A proton of mass m collides obliquely withanother proton. The first proton is mo
ID: 1676165 • Letter: A
Question
A proton of mass m collides obliquely withanother proton. The first proton is moving with a speed of 6.0 x106 m/s before it hits the second, stationary proton.Assuming the collision is perfectly elastic (no kinetic energy islost), and using the fact that the first proton is moving30o from its initial path after the collision, figureout the speed and direction of each proton after the collision.
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Let the first moving proton be object A, with a speed ofVA; let the second proton be object B
VA2=VA'2+VB'2
VA=VA'cos(30)+VB'cos()
0=VA'sin(30)+VB'sin()
Explanation / Answer
In elastic collision linear momentum as wellas the enery remains conserved. as youdid vA2 = vA'2 + vB'2 --------(1) and m *vA + m *0 = m * vA' * cos300 + m *vB' * cos ( formomentum over x - axis) vA = vA'* cos300 + vB' *cos ---------(2) m *0 + m *0 = m * vA' * sin300 - m *vB' * sin 0 = vA' * sin300 - vB' * sin --------(3) Square and add (2) and (3) vA2 = vA'2*(cos2 300 + sin2300) + vB'2 * ( cos2 + sin2)+ 2 * vA' * vB' * cos300 *cos - 2 * vA' *vB' * sin 300 * sin vA2 = vA'2 + vB'2 + 2* vA' * vB' * ( cos 300 * cos - sin 300 * sin) Substituting from equation(1) vA2 = vA2 + 2* vA' * vB' * cos ( +300) 0 = 2 *vA' * vB' * cos ( + 300) => cos (+300) = 0 + 300 = 900 = 600 hence the second proton will move at600 below x -axis. Now equation (3)becomes 0 = vA'* sin300 - vB' *sin 600 vA' = vB'sin 600 / sin 300 = 1.732* vB' Substitute this value in (1) vA2 = 1.7322*vB'2 + vB'2 vB' = {(6.0* 106)2 / (3.00 + 1)} = 3.0* 106 m/s vA' = 1.732* 3.0 * 106 = 5.196* 106 m/s = 5.196* 106 m/sRelated Questions
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