How Do Speed and Direction Change for Two Colliding Pool Balls?

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    2d Ball Collision
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When two pool balls collide, their speeds and directions change based on conservation of momentum and energy principles. The initial speeds and directions, represented as angles in radians, will need to be recalculated post-collision. The positions of the balls must be the same at the point of collision. The discussion emphasizes the need to apply formulas for elastic collisions, which account for both x and y momentum conservation. Understanding these concepts is crucial for accurately determining the new speeds and directions of the balls after they collide.
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I have two balls with d1 and d2 as their direction, and I have the speed s1 and s2, d1 and d2 are in radians showing the angle made with positive direction of x-axis and in anticlockwise direction i.e 0<=d1 and d2<2*pi
their positions are (x1,y1) and (x2,y2) when they collided

after collosion
what would be their speed, and direction?
 
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shouldn't their positions be the same when they collide?

basically you want to split the problem up into two problems: momentum is conserved in both the x and y directions, independently. also it looks like you're assuming completely elastic collisions.
 
Sorry--you will need to use conservation of energy, too.
this plus both equations for conservations of momentum should be enough to solve the problem.
 
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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