Calculating Plank's Velocity on Frictionless Ice Surface

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To find the plank's velocity on a frictionless ice surface when a person walks on it, the principle of conservation of momentum is applied. The person, weighing 47.7 kg, walks at a velocity of 2.06 m/s relative to the plank, which weighs 77 kg. By considering the system's initial and final momentum, where the initial state is at rest, the plank's velocity can be determined as the person moves. The plank will move in the opposite direction to conserve momentum, allowing for the calculation of both the plank's and the person's velocities relative to the ice. This approach effectively illustrates the relationship between the velocities of the person and the plank.
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A 47.7 kg person is standing on a 77 kg plank. The plank is originally at rest on a frozen, frictionless lake. The person begins to walk along the plank at a constant velocity of 2.06 m/s to the right relative to the plank.

What is the plank's velocity relative to the ice surface in m/s?

I realize this is a momentum problem, I just don't know how to solve for the velocity of the plank with respect to the ice or of the person with respect to the ice or what a "right relative to the plank" is...

Any help is appreciated as always :smile:
 
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Draw a diagram. If the Planck's velocity is v2 (relative to ice) into the opposite direction, what is the person's velocity (relative to ice)?
 
One way to do this problem is to consider it in the moving board frame. That is, if the board is moving with a speed v to the left (with respect to the ice, since the person is moving to the right), then in the frame with speed v to the left you know the initial and final momenta (ie, before the person starts walking, the person and board move right (in the frame) at a speed v, and afterwards just the person is moving at the given speed) and you know momentum is conserved, so you can solve for v.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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