Answer:
a) [tex]v_{1f}=-6.67\: cm/s[/tex]
[tex]v_{2f}=13.33\: cm/s[/tex]
b) [tex]n=88.84\: \%[/tex]
Explanation:
a) Applying the conservation of momentum, we have:
[tex]p_{i}=p_{f}[/tex]
p(i) is the initial momentum. In our case is due to the 5 g object.
p(f) is the final momentum. Here, both objects contribute.
[tex]m_{1i}v_{1i}=m_{1f}v_{1f}+m_{2f}v_{2f}[/tex]
Where:
To find both final velocities we will need another equation, let's use the conservation of kinetic energy.
[tex]m_{1i}v_{1i}^{2}=m_{1f}v_{1f}^{2}+m_{2f}v_{2f}^{2}[/tex]
So we have a system of equations:
[tex]5*0.2=5v_{1f}+10v_{2f}[/tex] (1)
[tex]5*0.2^{2}=5v_{1f}^{2}+10v_{2f}^{2}[/tex] (2)
Solving this system we get:
[tex]v_{1f}=-6.67\: cm/s[/tex]
[tex]v_{2f}=13.33\: cm/s[/tex]
b) The fraction of the initial kinetic energy transferred is:
[tex]n=\frac{m_{2}v_{2f}^{2}}{m_{1}v_{1i}^{2}}[/tex]
[tex]n=\frac{10*13.33^{2}}{5*20^{2}}[/tex]
[tex]n=88.84\: \%[/tex]
I hope it helps you!
