- #1
Ishfa
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- Homework Statement
- Classical Mechanics
- Relevant Equations
- s= v0t + 1/2 gt^2
Minimum mass of a block
- Thread starter Ishfa
- Start date
- #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
- 15,776
- 8,958
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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- 10,435
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:
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
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}...
Here we know that if block B is going to move up or just be at the verge of moving up
##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ##
Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
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