Canoeist Weight Distribution: How Far Can He Reach Without Getting Wet?

  • Thread starter Thread starter gtaylorMD
  • Start date Start date
AI Thread Summary
A 56kg canoeist in a 23kg canoe attempts to reach a dock from the front of a 2.3m long canoe. The discussion centers on whether he can do this without getting wet, considering the principles of conservation of momentum. As he moves forward, the canoe will shift backward, affecting his ability to reach the dock. The center of mass remains stationary, indicating that his movement will result in the canoe moving away from the dock. Therefore, without precise calculations, it is unlikely he can reach the dock without getting wet.
gtaylorMD
Messages
1
Reaction score
0
A 56kg canoeist sits at rest at the back end of a 23kg canoe. The canoe is at rest in still water with its front end about 0.50m from the dock. He realizes that he left his paddle on the dock, so he carefully works his way to the front of the 2.3m long canoe. Is he likely to reach the dock from the front of the canoe (without getting wet)? Explain with some estimates

I am completely confused on this one, I believe he can make it without getting wet but I cannot seem to accurately prove that. Thanks for any help
 
Physics news on Phys.org
Sounds like conservation of momentum problem. Which means the center of mass won't move since the initial momentum was zero.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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...
Back
Top