Answer:
a). 1.218 m/s
b). R=2.8[tex]^{-3}[/tex]
Explanation:
[tex]m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg[/tex]
[tex]v_{bullet}=341\frac{m}{s}[/tex]
Momentum of the motion the first part of the motion have a momentum that is:
[tex]P_{1}=m_{bullet}*v_{bullet}[/tex]
[tex]P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529[/tex]
The final momentum is the motion before the action so:
a).
[tex]P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}[/tex]
[tex]P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}[/tex]
[tex]P_{1}=P_{2}[/tex]
[tex]2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}[/tex]
b).
kinetic energy
[tex]K=\frac{1}{2}*m*(v)^{2}[/tex]
Kinetic energy after
[tex]Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J[/tex]
Kinetic energy before
[tex]Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J[/tex]
Ratio =[tex]\frac{Ka}{Kb}[/tex]
[tex]R=\frac{1.14}{406.4}\\R=2.8x10^{-3}[/tex]
