Calculating Momentum Change of a Snooker Ball Collision

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A snooker ball with a mass of 0.350 kg collides with a table at right angles, initially traveling at 2.8 m/s and rebounding at 2.5 m/s. The change in momentum is calculated using the formula p = mv, but the initial approach was flawed due to a misunderstanding of the collision direction. The ball moves forward and back along the same line, not at angles. After clarification, the correct momentum change can be determined. The discussion highlights the importance of accurately interpreting physics problems.
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Homework Statement


A snooker ball of mass 0.350kg hits the side of a snooker table at right angles, and bounces off also at right angles. If its speed before collision is 2.8ms^-1 and its speed after is 2.5ms^-1, calculate the change in its momentum.
(The answer to the question is not 0.105kg ms^-1)

Homework Equations


p = mv (well, that's what I've tried answering it with)

The Attempt at a Solution


I'm assuming my method is flawed, but I thought it might be worth using pythagoras to work out the actual (?) velocity.
I got 3.75 but if you do 3.75*0.35 i reach 1.3 which according to the answers in the back of the book are incorrect.

Could someone enlighten me please? I'm really stumped.

Thanks
 
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Crosshash said:
I thought it might be worth using pythagoras to work out the actual (?) velocity.
I got 3.75 but if you do 3.75*0.35 i reach 1.3 which according to the answers in the back of the book are incorrect.

Hi Crosshash! :smile:

You're misreading the question … the ball goes forward and back along the same line! :smile:
 
tiny-tim said:
Hi Crosshash! :smile:

You're misreading the question … the ball goes forward and back along the same line! :smile:

argh! you're kidding me? I've been stuck on this question for almost 30 minutes and it's that simple!?

I thought it deflects off the side at a right angle :P

I get the right answer now, thankyou!
 
Now if you could just figure out what a snooker is :smile:
 
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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