In the figure, particle 1 of mass m1 = 3.1 kg slides rightward along an x axis on a frictionless floor with a speed of 5.0 m/s. When it reaches x = 0, it undergoes a one-dimensional elastic collision with stationary particle 2 of mass m2 = 4.6 kg. When particle 2 then reaches a wall at xw = 79 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?

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

Explanation:

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: