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Physics MCQs thread.

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Alright, everyone so here's the explanation for hassam's oil tanker question:

Using density = mass/volume you'll get mass of oil in the tanker

Volume of oil cross-sectional area into height.

Weight = mass times g

So weight of oil = Volume of oil times its density times g

This weight of oil will be equal to weight of water displaced.

Let height, h, for which tanker will move up.

We'll get mass of water displaced by using density= mass/volume

Volume of water displaced = height, h, times area of tank's cross-section.

So weight of water displaced = its density x h x its cross sectional area times g

Equate these two equations and you'll get h as 18.5. So the answer is D.
 
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Zishi remember u said speed of separation v2-v1= u2-u1 speed of approach ?????? so do we have to use negative and positive signs for the values of u and v , like in this case for example
 
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Ok, so the obvious answer is B ( since there is a normal contact force and friction) But if the friction is acting backwards, then isn't the actual diagram supposed to be as shown?
 
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@Hateexams93

Since momentum must be conserved:
mu = mv1 + mv2
u = v1 + v2 --> (i)

Since kinetic energy must be conserved:
mu^2/2 = (mv1^2 / 2 ) + ( mv2^2 / 2)

u^2 = v1^2 + v2^2

Substitute in the value of u from (i)
(v1 + v2)^2 = v1^2 + v2^2
v1^2 + 2v1v2 + v2^2 = v1^2 + v2^2
2v1v2 = 0

Therefore, either v1 or v2 is 0. Using equation (i) if v1 is zero, then v2 = u, and vice versa. However, v1 cannot be the same as u as this would mean that the first ball had to move through the second ball! So, the only physical solution is that the first ball stops, and the second ball continues moving with the first ball's original velocity.
 
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Shooingstar is there is a shorter way ???? or u have to do all of this ? i'm so confused ..and my teacher sucks ..he is never serious ..and can't explain things properly
 
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Hateexams93 said:
Shooingstar is there is a shorter way ???? or u have to do all of this ? i'm so confused ..and my teacher sucks ..he is never serious ..and can't explain things properly

Yes you do have to take signs because velocity is a vector quantity. You can always solve questions like this by simply using speed of approach = speed of separation

There's no need to learn all that provided in some previous posts. Here's a simple example.

Two balls are moving towards each other with the speeds of u and -v(minus sign indicates opposite direction). Now when we'd be finding speed of approach, we'd have to add them because both are moving towards each other and so it's overall a positive effect. So speed of approach = u-v

Now when they collide, they move in opposite directions, velocity of one will be negative and of the other one will be positive. And as both are moving oppositely, it will have positive effect on their separation, and hence, again, their respective velocities would add up to give u-v...

The negative effect is like when both things are moving in same directions - it means that one thing is moving with the higher speed than other and the other one is trying to chase the other(trying to get closer, but won't because of other's higher velocity). This time, their velocities would be subtracted.
 
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Saturation said:
I really don't have any idea about this qs! please explain in detail! thanks!

U first consider the whole system..
F=ma where m=m+3m=4m
F=4ma
make a the subject,
a=F/4m

U can then consider y separately and solve(substitute a in the below eq)
F=ma
=3m x (F/4m)
= 3F/4

The ans is D
 
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Zishi said:
Two balls are moving towards each other with the speeds of u and -v(minus sign indicates opposite direction). Now when we'd be finding speed of approach, we'd have to add them because both are moving towards each other and so it's overall a positive effect. So speed of approach = u+v
so if u have to add them, won't it be U + (-V) =u-v ???????
 
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Hateexams93 said:
Zishi said:
Two balls are moving towards each other with the speeds of u and -v(minus sign indicates opposite direction). Now when we'd be finding speed of approach, we'd have to add them because both are moving towards each other and so it's overall a positive effect. So speed of approach = u+v
so if u have to add them, won't it be U + (-V) =u-v ???????

Oh, sorry - that was a typo
 
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Zishi said:
Now when they collide, they move in opposite directions, velocity of one will be negative and of the other one will be positive. And as both are moving oppositely, it will have positive effect on their separation, and hence, again, their respective velocities would add up to give u-v...

.
SO when they move in opposite directions , we will have -u and +v right ? and if its gonna add up ..the resuling will V-U ? but its not the same as the one before collision :eek:
 
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Hateexams93 said:
Zishi said:
Now when they collide, they move in opposite directions, velocity of one will be negative and of the other one will be positive. And as both are moving oppositely, it will have positive effect on their separation, and hence, again, their respective velocities would add up to give u-v...

.
SO when they move in opposite directions , we will have -u and +v right ? and if its gonna add up ..the resuling will V-U ? but its not the same as the one before collision :eek:

No, they won't exchange their speeds unless they have same masses, and final velocity may not be equal to initial. They'll be the velocities which would add up to give the expression of U-V.
 
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Hateexams93 said:
saturation the answer to the first Q is given in the previous posts ..look for it

I know the answer, but I want to know, why the resultant of the forces acting on the car by the ground, isn't in the direction I showed!
 
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Saturation said:
Hateexams93 said:
saturation the answer to the first Q is given in the previous posts ..look for it

I know the answer, but I want to know, why the resultant of the forces acting on the car by the ground, isn't in the direction I showed!
according to Newton's third law of motion, for every action there is equal and opposite reaction. the force exerted by the wheel acts backwards, as a result there is a force exerted by the road on the wheels in the FORWARD DIRECTION. another force is the normal force which is acting UPWARDS so resultant of the two forces should be (B)
 
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Zishi said:
No, they won't exchange their speeds unless they have same masses, and final velocity may not be equal to initial. They'll be the veracities which would add up to give the expression of U-V.
can u plz just solve the question which i posted on the previous page .. using the speed of approach and separation
 
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