Answer:
a. 1.12m/s
b. 0.34m/s
Explanation:
Let Vgi = the velocity of the girl relative to the ice
Let Vgp = the velocity of the girl relative to the plank
Let Vpi = the velocity of the plank relative to the ice
Vgi = Vgp + Vpi
From the question, Vgp = 1.46/s
So, Vgi = 1.46 + Vpi
Conservation of Momentum (Relative to the ice), we have
Initial Momentum = Final Momentum
Because the girl and the plank are at rest initially, the initial Momentum = 0
So, 0 = Mg * Vgi + Mp * Vpi
Where Mg = Mass of the girl = 45.8kg
Mo = Mass of the Plank = 151kg
Make Vpi the subject of the formula, we then have
-Mg*Vgi = Mp*Vpi ---------- Divide through by Mp
Vpi = -Mg * Vgi/Mp
Vpi =( -45.8 * Vgi)/151
Vpi = -45.8Vgi/151
Remember that, we have (Vgi = 1.46 + Vpi)
We then substitute the expression of Vpi in the above equation
That is;
Vgi = 1.46 + (-45.8Vgi/151) ------- Open the bracket
Vgi = 1.46 - 45.8Vgi/151 ----------- Multiply through by 151
151 * Vgi = 151 * 1.46 - 45.8Vgj
151Vgi = 220.46 - 45.8Vgi --------- Collect like terms
151Vgi + 45.8Vgi = 220.46
196.8Vgi = 220.46 --------------- Divide through by 196.8
Vgi = 220.46/196.6
Vgi = 1.1213631739572736
Vgi = 1.12 m/s (Approximated)
So, the velocity of the girl relative to the ice is 1.12m/s
b. Velocity of the plank, relative to the ice
We can solve this using (Vgi = 1.46 + Vpi)
All we need to do is substitute 1.12 for Vgi in the equation
So, we have
Vgi = 1.46 + Vpi becomes
1.12 = 1.46 + Vpi -------- Collect like terms
1.12 - 1.46 = Vpi
-0.34 = Vpi
So, the Velocity of the plank is 0.34m/s to the left
