How Do Angles Affect Puck Movement in a Non-Head-On Collision?

AI Thread Summary
The discussion revolves around a collision between two pucks on an air hockey table, where puck A, weighing 0.031 kg and moving at +5.5 m/s, collides with stationary puck B, which has a mass of 0.066 kg. The collision is described as non-head-on, leading to both pucks moving apart at specific angles. Participants express confusion over the absence of a drawing and a clear question regarding the angles' impact on puck movement. The conversation highlights the need for visual aids to better understand the dynamics of the collision. Understanding these angles is crucial for analyzing the resulting velocities and directions of the pucks.
shawonna23
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There is a collision between two pucks on an air hockey table.











Puck A has a mass of 0.031 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.066 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.
 
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I don't see a drawing.
 
I don't see a question either.
 
yea, I don't see either either :)
 
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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