Two particles with masses 2m and 4m are moving toward each other along the x axi
ID: 1588728 • Letter: T
Question
Two particles with masses 2m and 4m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m is traveling to the left, while particle 4m is traveling to the right. They undergo an elastic glancing collision such that particle 2m is moving downward after the collision at right angles from its initial direction.
(a) Find the final speeds of the two particles.
____________ vi
(b) What is the angle at which the particle 4m is scattered?
__________ degrees
particle 2m ____________ vi particle 4m____________ vi
Explanation / Answer
a)Initial horizontal momentum=final horizontal momentum of system
4m.vi - 2m.vi = 2m.0 + 4m. Vxf.....here Vxf is x component of final velocity of 4m
2mVi=4m.Vxf
Vxf =0.5 Vi
Initial vertical momentum=final vertical momentum of system
0 = -2m.Uyf + 4m. Vyf .....here Vyf is y component of final velocity of 4m, and Uyf is final speed of 2m
Uyf =2Vyf
Balancing kinetic energy, which is conserved because it is elastic collision
0.5*2mvi2 + 0.5*4mvi2 = 0.5*2mUyf2 + 0.5*4mVf2
3Vi2 = Uyf2 + 2 (Vyf2 +Vxf 2 )
3Vi2 = Uyf2 + 2 (Uyf2/4 +Vi2/4)
2.5 Vi2= 1.5 Uyf2
Uyf = 1.29 Vi
Vxf=0.5 Vi
Vyf= 0.645 Vi
Vf= square root of ( Vxf2 + Vyf2) = 0.816 Vi
Hence speed of 4m is 0.816*Vi and speed of 2m is 1.29*Vi
b) tan theta = Vyf / Vxf = 0.645/0.5 =1.264
theta = 52.21 degree
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