Miser's Gold: A Frozen Pond Rescue Plan

  • Thread starter Thread starter crosbykins
  • Start date Start date
  • Tags Tags
    Gold Plan Pond
AI Thread Summary
The discussion revolves around a scenario where a miser is stranded on a frozen pond with a bag of gold coins. To escape, he can throw the gold in a direction parallel to the ice, utilizing the conservation of momentum to propel himself in the opposite direction. This method relies on the principle that the net momentum of the system remains constant, allowing the miser to move towards the edge of the pond. Additionally, the conversation shifts to a diving scenario, exploring the vertical momentum of a diver and the forces exerted by water upon entry. The implications of excessive force upon entering the water and the dangers posed by underwater rocks are also examined.
crosbykins
Messages
50
Reaction score
0

Homework Statement


A miser sits alone in the middle of a frozen pond holding a bag of gold coins. He is unable to walk on the slippery ice. Explain how he could use the gold to rescue himself.


Homework Equations



Ft=change in p

The Attempt at a Solution


-i'm really not sure,but maybe he could push the gold forward and then bring it back towards himself so it collides with him. Then he will be pushed back and can move.
 
Physics news on Phys.org
Note that you have an isolated system. (Of course, you have gravitational force, but it's balanced with the reaction force of the ground.) There's no friction or any other forces.

That means that the momentum is conserved. So, if he just throws the gold in any direction parallel to the ground (ice), it will set him in motion. Net momentum is zero in the beginning, but after he throws the gold, gold will have its momentum and, since the net momentum must be zero again (law of conservation), miser must also move (in opposite direction of the gold).
 
thanks for your help, any chance you can help me with this qs (i need help on c) & d)

Divers at Acapulco, Mexico, regularly dive head first from a heigh of 36m into water 3.6m deep. [assuming g=9.8m/s^2) and that air resistance is negligible.]
a) What would be the vertical momentum of a 50kg diver when he reached the water?
b)If he lost his vertial velocity by the time he reached a depth of 3.0, what average force must the water exert on him?
c)Comment on the likely results if the inital force as he entered the water was ten times the averag force.
d)The rocks at the base of the cliff jut more than 6m out into the water. If a diver hit these would he be killed. Why?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .

Similar threads

Back
Top