Calculate the initial recoil spped of the boat

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The discussion revolves around calculating the initial recoil speed of a boat when a man throws a rock from it, as well as determining the loss in mechanical energy due to friction and the effective coefficient of friction. The conservation of momentum is emphasized, noting that the momentum of the rock and the boat-man system must balance. The recoil distance of 4.2 meters raises questions about the energy loss and friction involved in stopping the boat. Participants also discuss a separate collision scenario involving two pucks, reinforcing the principles of momentum conservation and energy loss. Overall, the focus is on applying physics concepts to solve problems related to motion and energy.
HurricaneH
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1. A 75KG man stands in a 100kg rowboat at rest in still water. He faces the back of the boat and throws a 5kg rock out of the back at a speed of 20 m/s. The boat recoils forward 4.2 m..
Calculate the initial recoil spped of the boat, the loss in mecahnical energy due to friction, the effective coefficient of friction.

Ok, what throws me off is that 4.2 meters that it recoils...


2. A 0.3kg puck, at rest, is struck by a 0.2 kg puck moving along the xaxis with a speed of 2 m/'s. After the collision, the 0.2 kg puck has a speed of 1 m/s at an angle of 53 degrees to the xaxis. Find the velocity of the 0.3kg puck after the collision, find the fraction of KE lost in the collision.

Momentum would be equal before and after right?
 
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1. What's the mechanical energy in this case? What's the relation between mechanical energy and external work (friction in this case)? What's the relation between force of friction and mass?

2. Yup, momentum is always conserved.
 
HurricaneH said:
1. A 75KG man stands in a 100kg rowboat at rest in still water. He faces the back of the boat and throws a 5kg rock out of the back at a speed of 20 m/s. The boat recoils forward 4.2 m..
Calculate the initial recoil spped of the boat, the loss in mecahnical energy due to friction, the effective coefficient of friction.

Ok, what throws me off is that 4.2 meters that it recoils...


2. A 0.3kg puck, at rest, is struck by a 0.2 kg puck moving along the xaxis with a speed of 2 m/'s. After the collision, the 0.2 kg puck has a speed of 1 m/s at an angle of 53 degrees to the xaxis. Find the velocity of the 0.3kg puck after the collision, find the fraction of KE lost in the collision.

Momentum would be equal before and after right?

1. Remember that there is conservation of momentum to consider. Then you can use that information to calculate the loss of energy too.

2. Same principles :-)
 
HurricaneH said:
1. A 75KG man stands in a 100kg rowboat at rest in still water. He faces the back of the boat and throws a 5kg rock out of the back at a speed of 20 m/s. The boat recoils forward 4.2 m..
Calculate the initial recoil spped of the boat, the loss in mecahnical energy due to friction, the effective coefficient of friction.

Ok, what throws me off is that 4.2 meters that it recoils...

Conservation of momentum - a 5kg rock goes one direction, the boat and man conserve momentum by going the opposite direction at a given speed (initial recoil speed). The boat and man decelerate back to zero velocity (how much of a loss in mechanical energy is that). What coefficient of friction decelerates your boat from initial recoil speed to zero in 4.2 meters?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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