Finding the Velocity of the Final Chunk in an Exploding Hockey Puck Scenario

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The discussion centers on calculating the velocity of the final chunk of a hockey puck that explodes into three pieces. The initial momentum is zero, and the pieces have different masses and velocities, with one piece moving at speed v and another at speed 2v at right angles. To find the velocity of the third piece, conservation of momentum principles must be applied in both the x and y directions. The user is encouraged to set up momentum equations based on the known velocities and masses of the first two chunks. Overall, the key to solving the problem lies in correctly applying the conservation of momentum to find the unknown velocity of the final chunk.
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Homework Statement



A hockey puck of mass 4m has been rigged to explode as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk of mass m slides across the ice at speed v. Another chunk of mass 2m slides across the ice at speed 2v, in a direction at right angles to the direction of the first piece. From this information, find the velocity of the final chunk.

Homework Equations



This is from the chapter on momentum, collisions and impulse, but I'm not sure which equations to use.

The Attempt at a Solution



I'm really not sure where to start.

I drew a diagram, and intuitively assumed that the third piece goes off at a 135 degree angle to the other two pieces, but I'm not sure about that or how to prove it if it is right. It obviously has mass m (4m - m - 2m), though.

I also assumed that the force of the explosion is equal on all chunks, but realized that's wrong since the higher mass chunk has the highest velocity.

I was thinking that change in momentum is the impulse, and since momentum is 0 initially for all chunks, the impulse should be mv for chunk 1 and 4mv for chunk two...but then I'm at a loss for how to relate that to chunk 3.

I'd really appreciate it if someone could point me in the right direction.
 
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The total initial momentum is equal to the total final momentum. Take the x-axis along the dirn of m and the y-axis along the dirn of the 2m mass. If the speed of the 3rd piece is v, then what are the momenta along x and y?
 
You could write conservation of momentum equations in the directions of the first two chunks ...
 
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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