Answer:
1) She would be moving at a speed of 1.4 m/s
2) She would be moving at a speed of 1.4 m/s of she throws both the bricks at once.
Explanation:
Since the system is isolated we shall conserve the linear momentum of the system to solve the system for required quantities:
We have by conservation of momentum,
[tex]\overrightarrow{p_{i}}=\overrightarrow{p_{f}}[/tex]
Since initially the system has no momentum thus
After throwing the first brick we have
[tex]\overrightarrow{p_{i}}=\overrightarrow{p_{f}}\\\\(40+10+5)v_{2}+5\times (-7m/s)=0\\\\\therefore v_{2}=\frac{35}{55}=0.63m/s[/tex]
Again conserving the momentum when she throws the second brick we have
[tex]\overrightarrow{p_{i}}=\overrightarrow{p_{f}}[/tex]
[tex](40+10+5)\times 0.63m/s=(40+10)v_{f}+5\times (-7m/s)\\\\\therefore v_{f}=\frac{55\times 0.63+35}{50}=1.4m/s[/tex]
b) If she had thrown both the bricks at the same time analyzing the system in the same manner we have
[tex]\overrightarrow{p_{i}}=\overrightarrow{p_{f}}[/tex]
[tex]0=(40+10)v_{f'}+10\times -7m/s\\\\\therefore v_{f'}=\frac{70}{50}=1.4m/s[/tex]
