What is the relationship between ramp height and final velocity of a marble?

AI Thread Summary
The discussion centers on the relationship between ramp height and the final velocity of a marble. A marble released from a 10cm ramp achieves a final velocity of 1.4 m/s, prompting the question of its velocity from a 20cm ramp. The relevant equation used is the conservation of energy, equating gravitational potential energy to kinetic energy. The initial calculations suggested a final velocity of 2 m/s for the 20cm ramp, but there was uncertainty about its accuracy. Ultimately, the poster resolved the confusion and found the correct relationship between height and velocity.
16Candles
Messages
6
Reaction score
0

Homework Statement


A marble is released from a 10cm ramp. Its final velocity is 1.4 m/s. If it is released from a 20cm ramp, twice the height, what will its final velocity be?

Homework Equations


E = mgy + mv^2/2
mgy = mv^2/2 (Like initial gravitational potential energy will equal final kinetic energy? Does that make sense?)

The Attempt at a Solution


I tried to use mgy = mv^2/2, where I get 2gy = v^2. For g, our physics teacher has us use 10 m/s^2, and if the height is 20cm, then (2*.2m*10m/s^2) = v^2, 400 = v^2, v = 2m/s, but I don't think that is right.

I asked my teacher for help, and he said to think about the relationship between the height and the velocity (which I think has to do with square roots, but other than that, I am lost).
 
Last edited:
Physics news on Phys.org
Never mind, I figured it out! :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top