Answer:
a) [tex] V_{CM}=2.89 m/s [/tex]
b) [tex] v_{1}'=1.11 m/s [/tex] [tex] v_{2}'=-1.89 m/s [/tex]
c) [tex] v=2.89 m/s [/tex]
Explanation:
The equatiom of velocity of the center mass it is given by:
[tex] V_{CM}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}[/tex]
m₁=12 kg, v₁=4 m/s.
m₂=7 kg, v₂=1 m/s.
[tex] V_{CM}=\frac{55}{19}=2.89 m/s[/tex]
b) The velocity of the first and second particle in the center of mass reference frame is given by:
[tex] v_{1}'=v_{1}-V_{CM} [/tex] (1)
[tex] v_{2}'=v_{2}-V_{CM} [/tex] (2)
So we will have:
[tex] v_{1}'=4-2.89=1.11 m/s [/tex]
[tex] v_{2}'=1-2.89=-1.89 m/s [/tex]
c) Using the conservation of momentum in a perfectly inelastic collision:
[tex] m_{1}v_{1}+m_{2}v_{2}= (m_{1}+m_{2})v [/tex] (3)
Solving (3) for v:
[tex] v=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}} [/tex] (4)
Let's recall a perfectly inelastic collision means that the colliding particles stick together afte the collision.
If we see, the equation 4 is the same as the velocity of the center of mass. So v = 2. 89 m/s.
I hope it helps!
Have a nice day!
