Answer:
We use Eq 9−67 and 9−68 to find the velocities of the particles after their first collision (at x=0 and t=0 )
ν
1f
−
m
1
+m
2
m
1
+m
2
ν
1i
=
0.30kg+0.40kg
0.30kg−0.40kg
(2.0m/s)=−0.29m/s
ν
2f
=
m
1
+m
2
2m
1
ν
1i
=
0.30kg+0.40kg
2(0.30kg)
(2.0m/s)=1.7m/s .
At a rate of motion of 1.7m/s,2x
w
=140cm (the distance to the wall and back to x=0 ) will be traversed by particle 2 in 0.82s . At t=0.82s , particle 1 is located at
x=(−2/7)(0.82)=−23cm ,
and particle 2 is " gaining " at a rate of (10/7)m/s leftward ; this is their relative velocity at that time . Thus , this "gap" of 23cm between them will be closed after an additional time of (0.23m)/(10/7m/s)=0.16s has passed . At this time (t=0.82+0.16=0.98s) the two particles are at x=(−2/7)(0.98)=−28cm
Step-by-step explanation: