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This is an exceptionally good. It has questions that can solved only if you understand the topic (or funda) in depth. I'd say that if you are aspiring for JEE, then its a must for you
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Pannaguma is right. We consider the body as well as the earth as the system.
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Look at the figure below ( download it if not clear). The blocks A, B & C of mass 'm' each have accelerations a1 , a2 & a3 . F1 & F2 are external forces of magnitudes 2mg and mg respectively. Then,
a) a1 = a2 = a3
b) a1 > a2, a2 = a3
c) a1 > a3 > a2
On first look, this appears to be a question for beginners but here's an advise, don't be too, confident.
Well those who got (b) as the answer, you need to check again and so do those who got answer as (a).
So, the answer is (c).
How?????/
On first look it appears that all these systems are subjected to the same forces and hence the accelerations should be the same. Well, I'd suggest that you make your equations and check again.
Surprised. Well we must no forget that "acceleration depends on FORCE AND MASS"
Hope you find it intersting
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A good question to test your basics
if a body has an initial velocity 'u' and its acceleration changes with time as
"at" where 'a' is a constant. then,
a) v = u + at2
b) v = at
c) v = u + at2 /2
If you think that the answer should be (a) and that this is an idiotic ques. then you maybe right.
But you are not.
The answer is (c).
Why?????
Because the equations of motion are valid only in cases of constant acceleration
But here 'a' changes with time.
Give it another try now (after the hint)
Well,for those who still didn't get it
use integration.
integrate v =  at dt
You will get,
v = c + at2/2
As initial velocity 'u', then c = u
PS : RATE ME IF YOU FIND THIS HELPFUL. NUDGE ME IF YOU NEED HELP.
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I think you are going at a pretty good pace but you can surely increase your pace as you also need to solve other books. I'd suggest that you cover up the topics from other books also, like DC PANDEY or Irodov, etc.
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Take a particle of length dx at a distance 'x' from the end of the table.
Therefore the friction acting is  mg(dx)/L and the distance moved against friction is x.
Hence integrating ,
[0 ][2L/3] (-mg(dx)/L. x) = -2mgL/9
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The vertical component of the velocity is derived from Work Energy Theorem.
vy =  2gh
While vx should not affect the final velocity vector as it becomes vertical in the end.
hence it should be equal to vy =  2gh.
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40.
As the block goes up with a constant velocity , no acc. therefore following eqns.
Along the plane of incline,
mg sin  = ma
 a = g sin 
By second eqn. of motion.
t = (2l/gsin)1/2
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35.
Eqns.
0.5g - T1 = 0.5a ............................................(1)
T2 - 0.05g = 0.05a ..............................................(2)
T1 - T2 - 0.1g sin(30) = 0.1 a ................................(3)
add (1), (2) & (3)
0.45g - 0.05g = 0.65a
a = 0.4g/0.65 = 8g/13
Note : All the bodies are constraint ( or bound) to move with the same acceleration. hence all have the same "a" . Also, directions can be seen from the F.B.D or from the figure itself. The heavier block moves down, lighter moves up and the one on the incline moves up. You can assign any direction to a initially ( but all should be in the same direction in this case) and solve. If the direction is wrong, you'll get a negative sign
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Another important to keep in mind is that, static friction can vary from 0 to  s N .
But Kinetic friction will always be constant and equal to k N.
Hope you find this interesting as well as important
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Check this out.
We can't use parallel axes theorem directly as C is not the centre of mass
The theorem states that,
I = Icm + Md2
Hence first we need to find the COM
According to me, The soln. is (MR2 /  )*( - 2)
Plz. tell me if I'm correct
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Is it (t 2 + t0 ) after the acceleration changes
therefore,
2t0 + t 2 after the beginning.
The method i used is :-
Distance covered in first t0 seconds is at02 /2 . Time reqd. to stop after acc. changes will also be the same. Now calculate time reqd. to come back to original posn. covering dist. = at02 .
Please Rate me if I'm correct. Point out mistake (if any)
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Yeah, in a rigid body, the relative displacement is zero and hence it the relative velocity
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We can also get the same answer by comparing the given eqn. to the general eqn. of trajectory.
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Here are the solutions.
Download pics. for better view
Nudge me if you aren't satisfied or if you need help.
Rate me if you are satisfied
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Hope you find this useful.
Nudge me if you need any explanation or if you are not satisfied.
PS ; Rate me if you find this helpful.
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Rate me if you find this useful.
Nudge me if you have any probs. in deciphering my handwriting
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it appears that others have answered the 18th question.
Well i guess I'm answering the correct one.
Rate me if you find it helpful.
Nudge me if you need help in deciphering my handwriting
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Firstly, the argument that co-efficient of friction is the same is wrong. For kinetic friction is lesser than in case of static friction.
Now when a body is at rest, you need to apply more force to make it start moving, hence we conclude that the static frictional force is higher than kinetic friction as we need to apply comparatively lesser force to keep a body in motion.
Though I'm not sure of the reason, I think it is because :
1. When at rest, the electromagnetic forces of attraction between the two bodies in contact is pretty large and to overcome it you need to apply a lot of force.
2 But when the body is in motion, the electromagnetic forces must be lesser as the distance between the molecules change ( between two objects) and hence less force is needed to oppose the electromagnetic forces.
Hope this is correct and you find this useful. Rate me if convinced
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Download picture for better view
Rate me if you find this useful
Nudge me if you don't understand anything or you aren't satisfied.
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Vriti Edustore
