- #1
Ishfa
- 3
- 0
- Homework Statement
- Classical Mechanics
- Relevant Equations
- s= v0t + 1/2 gt^2
What Determines the Minimum Mass of a Block in a Collision Scenario?
- Thread starter Ishfa
- Start date
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The discussion centers on determining the minimum mass of a block in a collision scenario involving two cans. Participants emphasize the importance of clearly explaining the process and providing sketches to illustrate height versus time. Key points include deriving expressions for the velocities of both cans over time and understanding their relationship at the moment of collision. The conversation also highlights the need to focus on eliminating the correct variables to solve for time rather than velocity. The minimum mass is linked to ensuring a specific time of flight after the collision when the masses combine.
- #2
BvU
Science Advisor
Homework Helper
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Hi,
See rules etc: explain (in words) what you are doing; make a sketch (Height versus time) of what happens.
We are a bit reluctant to reverse engineer your writings, so you have to help us help you
##\ ##
See rules etc: explain (in words) what you are doing; make a sketch (Height versus time) of what happens.
We are a bit reluctant to reverse engineer your writings, so you have to help us help you
##\ ##
- #3
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- 8,996
The collision time happens once and cannot be double-valued.
You can make your life simpler and avoid quadratics if you consider the velocities.
1. Write expressions for the velocity of each can as a function of time.
2. How is the velocity of the first can v1 related to the velocity of the second can v2 at the time of the collision?
- #4
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In your fifth equation, the one ending …16-1)=0, check where that -1 comes from. It should be something else.
Also, having obtained a quadratic in t that involves the unknown ##v_0##, bear in mind that what you want to find is t, not ##v_0##. So which should you be trying to eliminate?
Btw, what has this to do with "minimum mass of a block"?
Also, having obtained a quadratic in t that involves the unknown ##v_0##, bear in mind that what you want to find is t, not ##v_0##. So which should you be trying to eliminate?
Btw, what has this to do with "minimum mass of a block"?
Last edited:
- #5
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The minimum mass may have something with another part of the problem, e.g (b) Find the minimum value of the second mass so that the time of flight after the collision is ##t_2## seconds if the masses stick together (or some such thing.)haruspex said:
My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...