Calculate Speed at Top of Roller Coaster 2nd Hill (5.5m)

AI Thread Summary
The discussion focuses on calculating the speed of a roller coaster at the top of a second hill, which is 5.5 meters high, after ascending an 18-meter lift hill at 2.50 m/s. The calculations involve potential and kinetic energy, leading to a speed of 15.9 m/s at the top of the second hill. Participants agree that the mass of the coaster does not affect the outcome due to the constant acceleration from gravity. The final answer is confirmed as correct by other users in the thread. The problem illustrates the principles of energy conservation in roller coaster dynamics.
cowgiljl
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The mass of the coaster is 800 kg
The lift hill is 18 m highabd it goes up the lift hill at 2.50 m/s
The second hill is 5.5m high.

Wants to know the speed at the top of the second hill.
do i really care about the weight of the car?

This is what i have so far

PEtop1 +KEtop1 = KE bott
2*9.80*18 +2.5^2 = v^2 bott
V bott = 18.9 m/s

second formula was
Ke bott - Pe top1 = KE top 2
18.9^2 - 2*9.80*5.5 = V^2 top 2

Vtop 2 = 15.9 m/s and that is my final answer is that look right

thanks
 
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Yep...answer looks right to me.
 
thanks

thanks hindi
 
I believe the mass doesn't matter in this problem because the coaster is under constant acceleration for the whole run (gravity).
 
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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