Calculating Time Interval for a Perfectly Elastic Collision in Curling Match

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The discussion revolves around calculating the time interval between two collisions in a curling match involving perfectly elastic rocks. Rock A, with a mass of 2 kg and an initial velocity of 1.5 m/s, collides with stationary Rock B, which has a mass of 1 kg, while Rock C, also with a mass of 2 kg, is positioned 4.5 m away from B. The conservation of momentum equation is applied to find the velocities after the first collision, but the user struggles to derive the time interval for the second collision between B and C. Participants emphasize the importance of using conservation of energy due to the perfectly elastic nature of the collisions. The discussion highlights the need for clarity in applying physics principles to solve the problem effectively.
ElementUser
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Homework Statement



Matthew and Amanda have challenged Trevor and Julie to a curling match, played with perfectly elastic rocks sliding on level frictionless ice. Three rocks are lined up all in a row (perfectly straight line). B is a junior rock and its mass is exactly half of A or C. B and C are stationary while A approaches. Calculate the time interval between the collision of A&B and the collision of B&C. The distance between B and C is 4.5m. Rock A has an initial velocity of 1.5 m/s.

Diagram: 2M -----> M __________ 2M
Distance between M and the right-most 2M is 4.5m.

Homework Equations



m1v1+m2v2=m1v1^prime + m2v2^prime
d=vt, where v is equal to v2^prime
J=Ft
p=mv
Probably some equation that I missed (main problem of the question that I can't figure out)

The Attempt at a Solution



Variables: m1=2 kg
m2=1 kg
v1=1.5 m/s
v2=0 m/s
v1^prime=?
v2^prime=?

2(1.5)+0=2v1^prime+v2^prime
3=2v1^prime+v2^prime

Then I tried doing v2^prime=d/t, which got me to:

v2^prime=4.5/t

I tried subbing that into the other equation to make it 3=2v1^prime+v2^prime, but that got me nowhere. So basically I'm stuck trying to find v2^prime, which is what I need to figure out the time interval between the 2 collisions. I tried to think of the impulse&momentum equations, but they didn't seem to help (can't find the force for the impulse equation).

Any help is appreciated! Thanks :).

-ElementUser (first post in these forums)
 
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Welcome to PF!

ElementUser said:
… a curling match, played with perfectly elastic rocks sliding on level frictionless ice …

Hi ElementUser! Welcome to PF! :smile:

(no need to write ^prime … just write ' :wink:)

In exam questions, "perfectly elastic" means you can use conservation of energy. :wink:
 
Thanks tiny-tim :)

Took me a long time because my physics teacher today added a portion for this question, that you also have to find the time interval for the second timer Rock B will hit Rock A (after B bounces off of C).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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