Discover the Recoil Speed of a Cannon at the Circus - Momentum Formula Help

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Homework Help Overview

The discussion revolves around calculating the recoil speed of a cannon when a clown is fired from it, as well as analyzing the motion of two carts connected by a cord. The subject area includes concepts of momentum and impulse in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the meaning of 'recoil speed' and its relation to momentum. Questions arise regarding which momentum formula applies, including conservation of momentum and impulse. There is also discussion about the initial conditions of the system and how to approach the problem of the two carts after the cord is cut.

Discussion Status

Some participants have provided insights into the concepts of momentum and impulse, while others are seeking clarification on the appropriate formulas and initial conditions. Multiple interpretations of the problems are being explored, particularly regarding the two carts and their motion after the cord is cut.

Contextual Notes

Participants note that the clown and cannon start at rest, and there is an emphasis on equal and opposite forces and momenta. The discussion includes assumptions about the identical nature of the carts and their initial velocities.

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1. At the circus, a 100 kilogram clown is fired at 15 meters per second from a 500 kilogram cannon. What is the recoil speed of the cannon? My question is what does 'recoil speed' mean? that's it.
2. There are two identical carts, attached by a cord moving to the right at speed V. If the cord is cut, what would be the speed of cart A? Just imagine that there are two carts, and between them there is a cord attached to each other, and i think this cord moves to the right at the speed of V. How can i find cart A, what formula can i use? momentum?
thanks.
 
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1. The recoil speed is how fast the cannon will move in the opposite direction after the blast.

2.Yes, momentum will work for this problem.
 
but which formula of momentum does it relate to the question? change in momentum, as in impulse? or conservation of momentum? Can you sort of give me some ideas, i really don't understand it.
 
Heres my 2 cents about the 2nd question.

If the two cars are connected with a cord (such that the first car is pulled by the second), then consider their mass to be say "m", and since the two cards are identical, they have the same mass.

Find the momentum of the whole system, i.e. MV. Where M is the sum of the masses of individual cars i.e. 2m. And the total velocity of the system is given (V). If the cord is cut, the cars are no longer attached. But by conservation of momentum, the sum of their individual momentums should be equal to the total momentum of the system. I hope this helps.

Regards,
Sleek.
 
so i use this formula: m1v1i+m2v2i=m1v1f+m2v2f? what's the initial velocity?
 
MIA6 said:
so i use this formula: m1v1i+m2v2i=m1v1f+m2v2f? what's the initial velocity?
The clown and cannon start at rest. It is an impulse and the time is the same for the clown and cannon.

Equal and opposite force. Equal and opposite momentum.
 
Astronuc said:
The clown and cannon start at rest. It is an impulse and the time is the same for the clown and cannon.

Equal and opposite force. Equal and opposite momentum.

I meant the second question about the cord.
 
MIA6 said:
I meant the second question about the cord.

You start with both objects as part of a bigger object at some speed,V. The total mass will be the mass of the two carts, or 2m, since the carts are identical. So, you start with on object, whose momentum is , p=2mV, and end with 2, whose momenta are: p_A=mv_A and p_B=mv_B.

The big hint here is that the carts are identical and moving at the same speed. If you think about this, it should make the problem much easier.
 
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