In the figure, particle 1 of mass m1 = 3.0 kg slides rightward along an x axis o

Question

In the figure, particle 1 of mass m1 = 3.0 kg slides rightward along an x axis on a frictionless floor with a speed of 2.0 m/s.When it reaches x = 0, it undergoes a one-dimensional elastic collision with stationary particle 2 of mass m2 = 4.9 kg. When particle 2 then reaches a wall at xw = 74 cm, it bounces from the wall with no loss of speed. At what position on the x-axis does particle 2 then collide with particle 1?

see the image http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/q65.jpg

Explanation / Answer

m1v1i = m1v1f + m2v2f
3*2 = 3v1f + 4.9v2f   
6 = 3v1f + 4.9v2f   .............(1)

1/2m1v1i2 = 1/2m1v1f2 + 1/2m2v2f2
3*2^2 = 3v1f2 + 4.9v2f2 .............(2)
from (1), we have
v1f = (4.9v2f - 6)/3............(3)
putting this value in (2)

12 = 3[(4.9v2f - 6)/3]2 + 4.9v2f2   
36 = [(4.9v2f - 6)/3]2 + 4.9v2f2   
36 = 24.01v2f2 - 58.8v2f + 36 +
28.91v2f2 - 58.8v2f - 36 = 0

This gives
v2 = 2.53 m/s
putting this in (3)
v1f = (4.9*2.53 - 6)/3
v1f = 2.132 m/s

Now just do what distance (x) does particle 1 travel in the time it takes particle 2 to travel ( x + 1.4)
x1(t) = v1ft + x1
x2(t) = v2ft + x2........(4)
Now
x1(t) = x2(t)
v1ft + x1 = v2ft + x2
putting the values
2.132*t + 0 = 2.53*t + 0.74
t = 0.54 sec
put this in eq(4)
x2(t) = 2.132*0.54 + 0.74
x2(t) = 1.89 m

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