Elastic Collision of Blocks on a Half-Pipe: How to Determine the Final Heights?

AI Thread Summary
Two blocks on a frictionless half-pipe collide elastically after being released from different heights. Block B is more massive and starts from a lower height than Block A. The discussion centers on determining the final heights of the blocks post-collision, with various potential outcomes presented. Participants emphasize the importance of using conservation of momentum and energy to solve the problem, though some express confusion over how to apply these concepts without specific numerical values. A clear understanding of these principles is essential for determining the final heights of the blocks.
fightboy
Messages
25
Reaction score
0
Two blocks are released from rest on either side of a frictionless
half-pipe. Block B is more massive than
block A. The height HB from which block B is released is less
than HA, the height from which block A is released. The blocks
collide elastically on the flat section. After the collision, which
is correct?
A. Block A rises to a height greater than HA and block B
rises to a height less than HB.
B. Block A rises to a height less than HA and block B
rises to a height greater than HB.
C. Block A rises to height HA and block B rises to
height HB.
D. Block A rises to height HB and block B rises to
height HA.
E. The heights to which the blocks rise depends on where
along the flat section they collide.
I honestly didn't understand where to start with this problem, and got confused on the solution walkthrough. It basically wrote out two equations based on the conservation of momentum and conservation of energy, which said could be used to calculate the final speeds and then the final heights. Is there a more intuitive, less complicated way of figuring out this problem or does it require multiple equations? If someone could kindly give me an explanation for this problem it would be much appreciated!
 
Physics news on Phys.org
Well we can't really give you an explanation, that would be doing your homework for you. Conservation of energy and momentum are good places to start, do you know why?
 
BiGyElLoWhAt said:
Well we can't really give you an explanation, that would be doing your homework for you. Conservation of energy and momentum are good places to start, do you know why?
Well since it's an elastic collision I know Kf=Ki and due to the conservation of momentum Pf=Pi but i don't know how to put these two equations together to determine which height each of the blocks rises too. I can't really understand how the book uses the equations since the explanation is very vague. I guess I'm more confused since they don't give number values in this problem.
 
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