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Physics: Post your doubts here!

a_wiserME!!

a_wiserME!!

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Rizwan Javed said:
In this question we're told that the velocity is positive upwards. Also, we know that at highest & lowest points the mass will come momentarily to rest. So when mass is at lowest point, the velocity will be zero, and afterwards, when it'll again start moving upwards, it's velocity will be positive. this elimintes the chance of B being the answer as after coming to rest, the velocity is negative showing that the ball is moving downwards now.

Hope you understood.
Click to expand...

yup! got it (y)
 
Rizwan Javed

Rizwan Javed

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a_wiserME!! said:
View attachment 58724

and why is this B not A?? when it changes direction isnt it supposed to be subtracted?
Click to expand...
Sorry for getting late.

Look, if you take the downward direction to be negative, and the upward direction to be positive, then the momentum before collision will be : -m√(2gh1)
and the collision after collision will be m√(2gh2)

so then the change in momentum can be found by:

final momentum - initial momentum

= m√(2gh2) - (-m√(2gh1))
= m√(2gh2) + m√(2gh1))

So Answer is B.

P.S. You can also take the upward direction to be -ive and downward direction to be +ive, but this will result in the change in momentum to be :

- m√(2gh2) - m√(2gh1)) <--- this will be also correct, but this not an option given here.
 
a_wiserME!!

a_wiserME!!

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HumptyR said:
Let m√2gh1 = V1 and -m√2gh2= V2
So change in momentum will be,

V1-V2
which is equal to

m√2gh1-(-m√2gh2)
so,

m√2gh1+m√2gh2
is the ans.
Hope u got that
Click to expand...
Rizwan Javed said:
Sorry for getting late.

Look, if you take the downward direction to be negative, and the upward direction to be positive, then the momentum before collision will be : -m√(2gh1)
and the collision after collision will be m√(2gh2)

so then the change in momentum can be found by:

final momentum - initial momentum

= m√(2gh2) - (-m√(2gh1))
= m√(2gh2) + m√(2gh1))

So Answer is B.

P.S. You can also take the upward direction to be -ive and downward direction to be +ive, but this will result in the change in momentum to be :

- m√(2gh2) - m√(2gh1)) <--- this will be also correct, but this not an option given here.
Click to expand...

oh yeah...i got it!! thanx a lot :)
 
HumptyR

HumptyR

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2 A climber is supported by a rope on a vertical wall, as shown in Fig. 2.1
.Untitled.jpg
The weight W of the climber is 520 N. The rope, of negligible weight, is attached to the climber and to a fi xed point P where it makes an angle of 18° to the vertical. The reaction force R acts at right angles to the wall. The climber is in equilibrium. (a) Complete Fig. 2.2 by drawing a labelled vector triangle to represent the forces acting on the climber.
1.jpg
 
Z

zeejay

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AOA everyone,
Actually I'm in a really confused state, my problem is related to M1 mechanics and physics....Alright, so we all know that when a simple object like a block of wood moves in a forward direction there is kinetic friction between the block and the floor in the opposite direction.... But things get tricky when it comes to an individual wheel, my mind is actually boggled about *HOW A WHEEL MOVES FORWARD?*. Several pages on Google state that the axle produces two tangential forces on the top and bottom of the wheel and this produces a torque about the axle. Then they state that the friction is opposite to the rotation of the wheel and so it acts in the forward direction causing the wheel to roll forward. But how can friction (forward) ever be greater than the backward force on the wheel produced by the axle. Some articles even mention static friction and rolling friction acting opposite to the direction of motion and all these contradicting statements on the web are confusing me further. Please help me with my query with a detailed and sensible explanation in terms of whatever I stated above. It would be really benign of you. Thanks in advance!
 
Catherine_1

Catherine_1

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Hey,
Could anyone please help me out with the following doubt?:)
download (1).png
In this graph of X-ray spectrum, how is it possible for the photon to have maximum energy and yet zero relative intensity?:/
Since Intensity=Power/Area==>Energy/Area*Time?

Thanks.:)
 
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zeejay

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Hey can anyone tell me that why do we use average force for strain energy i.e 1/2×F×e2
.....i mean isn't the extension produced by the full force rather than half the force (average force)....pls help me asap...thanks in advance ✊
 
A

awesomaholic101

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zeejay said:
Hey can anyone tell me that why do we use average force for strain energy i.e 1/2×F×e2
.....i mean isn't the extension produced by the full force rather than half the force (average force)....pls help me asap...thanks in advance ✊
Click to expand...
The work done is basically the area under the graph ... And its a triangle so the formula works that way.
Practically speaking, if we use force at any given point, we'd only be calculating the energy held at that particular point ... Like the instantaneous energy or sth ... So we need to take avg force to calc the overall strain energy held by the spring.

I'm not 100% sure abt this ... So if anyone can confirm, it wud be good.
 
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