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Physics: Post your doubts here!

Hadi Murtaza

Hadi Murtaza

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MYLORD

MYLORD

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sagar65265 said:
Q11)
The object hits the wall with a speed v, and rebounds with the same speed, but a velocity in the opposite direction.
Kinetic Energy does not depend on the direction of an object's velocity, just the magnitude of that objects velocity, i.e. the speed. Therefore, since the speed remains the same even after the collision, the Kinetic Energy of the object is conserved.

By the same logic applied above, we can say that the speed is conserved, so that option is eliminated.

Lastly, the mass of the object doesn't change, and therefore the mass also remains conserved.
By elimination, we can confidently say that the momentum of the object is the only value that is not conserved - momentum, unlike the other values, is a vector quantity; since the direction of the object's speed changes, the momentum also changes, and is therefore not conserved.

Q12)

Two concepts are very important here -
i) momentum is conserved in any collision (as far as this level is concerned, external forces play very little part in collisions, so yes, momentum can be assumed to be conserved in all collisions as long as the momenta of all the colliding bodies are taken into account. This is why in the question above, the momentum of the object is not conserved, but the momentum of the object+wall is conserved - the wall is just so heavy it barely moves, that's all),

ii) and the relative velocity/ the velocity of approach of the bodies involved in an elastic collision are the same before and after the collision. This does not apply to an inelastic collision, so it is good that this collision is an elastic one.

For momentum to be conserved:

Initial momentum = mv
Final momentum = mv(X) + mv(Y) where v(X) is the final velocity of ball X and v(Y) is the final velocity of ball Y.

Equating them and cancelling out the mass m, we get

v = v(X) + v(Y)

if we take the left side to be the negative direction and the right side to be the positive direction, we can see that C and D are wrong, and so we can eliminate them.
Now for the relative velocity - in the beginning, ball X is going towards ball Y at an apparent rate of v ms^-1 (and Y is stationary, so it doesn't matter either ways). This is the initial relative velocity.
After the collision, suppose B were the answer and both X and Y were travelling at the same speed, v/2. The distance between them will remain the same; that distance won't increase or decrease, so the relative velocity of approach will be zero (because from the point of view of either ball, the other one is not coming any closer to them, and so it is equivalent to a situation where both X and Y are stationary).

Therefore, A must be the answer, since the rate of change of distance between the two balls is v before the collision, and v after the collision.

Q14)

Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.

So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)

Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.

This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,

(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.

I'll try out the rest afterwards, let me know in the meantime if there are any things you didn't understand.

Hope this helped!
Good Luck for all your exams!
Click to expand...
tHNX SOOOOO MUCH !!!
 
Thought blocker

Thought blocker

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MYLORD said:
Not Now !!
But i am still doing so i ll disturb u again !!:):D
Click to expand...
Well your one doubt was left I guess, that is w12_13.... Q11
so here you go
P initial = 2mu - mu =mu
And for momentum to be conserved, P ini = P final :)
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option b)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf :)
So answer is option B :)
 
MYLORD

MYLORD

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Thought blocker said:
Well your one doubt was left I guess, that is w12_13.... Q11
so here you go
P initial = 2mu - mu =mu
And for momentum to be conserved, P ini = P final :)
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option b)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf :)
So answer is option B :)
Click to expand...
thnx !!
 
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MYLORD said:
ok here's one more doubt !:rolleyes:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf Q 13, 15 ,17,
Click to expand...
we will solve it through two simultaneous equations
W-T=ma (for the 2.0 kg mass) As this mass is pulled down due to its weight. The resultant force acting on it will be W-T as W is greater. There is no component of weight acting in the direction of motion for the 8 kg mass, therefore it will move when the other mass moves down.
T-F=ma (for the 8 kg mass)
F is the frictional force. T is cancelled out. Solve for a which is 1.36. Closest answer is A.
 
MYLORD

MYLORD

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Thought blocker said:
we will solve it through two simultaneous equations
W-T=ma (for the 2.0 kg mass) As this mass is pulled down due to its weight. The resultant force acting on it will be W-T as W is greater. There is no component of weight acting in the direction of motion for the 8 kg mass, therefore it will move when the other mass moves down.
T-F=ma (for the 8 kg mass)
F is the frictional force. T is cancelled out. Solve for a which is 1.36. Closest answer is A.
Click to expand...
THNX
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf q22
 
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