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Physics: Post your doubts here!

chishtyguy

chishtyguy

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sagar65265

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Asad Moosvi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf

Question 10 anyone?
Click to expand...

The instantaneous velocity (at any instant, instead of an average) of any moving object is equal to the rate of change of displacement.

In this case, the displacement is only along one axis, the up-down (vertical) axis. Therefore, the only measurements needed to qualify the displacement are the magnitude of the displacement and the sign (if the object moved in 3 dimensions, we'd need a vector arrow. Here, the object moves in only 1 dimension so we do not need that).

So, the instantaneous velocity of the object is simply equal to the gradient of the displacement-time graph.
Take a look at the graph they have given for displacement versus time - initially, at the very beginning, the gradient is zero, i.e. the gradient is a flat line. Therefore, the displacement is barely changing with time at t = 0. This means that the velocity of the object at t = 0 is also zero.

This narrows the answer down to A, C or D. Let's look closer at the curve.

After the initial moment, the displacement is negative but the gradient slowly becomes more and more positive - that means the object is moving upwards, which is a positive velocity (since upwards is arbitrarily given the positive direction). So, since the gradient is positive and the change in displacement is upwards, the speed just after t = 0 is also positive (and upwards). Let's move on.

At the second scale-bar on the time axis, you can see on the displacement graph that the gradient becomes zero; this means that the velocity also becomes zero, which means that at the second bar on the time axis the velocity must be zero. This still doesn't get us any further, so we'll go a last bit along the time axis.

Right after the flat-line at the second bar on the time axis, we see that the gradient on the displacement-time graph becomes negative, i.e. the displacement becomes more negative than before. So, the velocity right after the second bar must also be negative. The only option that has a negative velocity after the second bar (from A, C and D) is A, our final answer.

Hope this helped!
Good Luck for all your exams!
 
chishtyguy

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Asad Moosvi

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Hadi Murtaza said:
If der r 2 or more wires protruding from a single point, then da current is split. When current is split, da value of current is not da same everywhere in da circuit. Da only circuit arrangement, where dis is not da case is arrangement D. In D, a single wire is connecting every meter, so current is same throughout circuit
Click to expand...

I don't understand it.
 
maryam fatima

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Shaoli Hassan

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Snowysangel said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf8:B
8: B
9: D
11:B
12:A
Could someone please explain why!
Click to expand...

Q-8: We know, s=ut+0.5gt2
So, L=0.5gT2
since initial speed is 0
Now, for 0.5T,
s=0.5(0.5T)2
=0.5(0.25)T2
=0.25L

Q-9: For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.

Q-11: Since the collision is elastic, relative velocity of approach is equal to relative velocity of separation. So, the speed of each mass after collision will be v. Kinetic energy after collision is mv2.

Q-12: For the barrel,
F=ma
120g-T=120a where T is the tension in the string
T=120g-120a

For the stake,
F=ma
T-80g=80a
T=80g+80a

Equate them to obtain,
120g-120a=80g+80a
a=0.5g

v2=u2+2as
v2=0+2(0.5g)(9)
v=6
 
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Snowysangel

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Shaoli Hassan said:
Q-8: We know, s=ut+0.5gt2
So, L=0.5gT2
since initial speed is 0
Now, for 0.5T,
s=0.5(0.5T)2
=0.5(0.25)T2
=0.25L

Q-9: For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.

Q-11: Since the collision is elastic, relative velocity of approach is equal to relative velocity of separation. So, the speed of each mass after collision will be v. Kinetic energy after collision is mv2.

Q-12: For the barrel,
F=ma
120g-T=120a where T is the tension in the string
T=120g-120a

For the stake,
F=ma
T-80g=80a
T=80g+80a

Equate them to obtain,
120g-120a=80g+80a
a=0.5g

v2=u2+2as
v2=0+2(0.5g)(9)
v=6
Click to expand...
How do know s is 9m? Won't the distance moved by the heavier object he greater? Oh and also, what happens to the acceleration of an object in both the presence and the absence if air resistance?
 
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Shaoli Hassan

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Snowysangel said:
How do know s is 9m? Won't the distance moved by the heavier object he greater? Oh and also, what happens to the acceleration of an object in both the presence and the absence if air resistance?
Click to expand...
No, the distance travelled won't be different since both of them are connected by the same string. The question says what will be the speed when the man's head is level with the bottom of the barrel. This means, the distance travelled by both the objects must be half of the length of the string=(18/2).
If an object if falling vertically (not connected to a string), in the absence of air resistance, the acceleration remains unchanged. But, in the presence of air resistance the acceleration decreases. If connected by a string as in question 12, just remember that the acceleration is the same for the whole system.
 
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