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Physics: Post your doubts here!

Suchal Riaz

Suchal Riaz

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Thought blocker said:
"there are two vectors, P and P' where P is initial momentum and P' is final momentum.
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"

Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here.. :(
Click to expand...
thanks my problem is now solved and the planet master has received enough lectures from me.
 
S

sagar65265

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..sacrifice4Revenge.. said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
q9


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf
q11
Click to expand...

Q9)

Either ways, the working for this question can be very easy if you work in a particular manner; first point is that Force = Δ(mv)/Δt, and if we take Δt to be 1 second, the equation is reduced to it's simplest form - Force = Δ(mv) in 1 second, i.e. force is equal to change in momentum in 1 second.

So, in 1 second, the velocity of some mass of air goes from 33 ms^-1 to 0 ms^-1. Therefore, the change in momentum = (mass of air) * (0-33) = -33(mass of air)
Okay. Now we know the change in momentum, but there still a few pieces of the puzzle left - how much mass is there in that "some mass of air"?

We have taken the time period over which this change in momentum occurs to be 1 second. Therefore, the total change in momentum of the air in 1 second is what we need.

What we can do is write m = ρV (by the definition of density) and calculate the volume of air that comes to a stop in 1 second. Multiplying this by ρ which they have given is 1.2 kgm^-3, we can find out how much air has lost it's momentum.

In that one second, air traveling at 33ms^-1 slows down and comes to a stop. This happens to all the air within 33 meters of the wall - if a small cross section of air is 33 meters from the wall, in one second it will travel 33 meters and come to a stop, right? Air that is 40 meters from the wall will not stop in that one second. Air that is 25 meters from the wall will stop before that one second is over. But totally, 33 meters length of air will come to a stop.

We have the length of the air column that stops in 1 second. But that doesn't give us the volume; we need the area as well.
Luckily, the question tells us that the area of the wall is 12m^2. Correspondingly, the volume of air that comes to a stop is 12m^2 * 33 m = 396 m^3 (which is the units for volume; we are on the right track, then!).

So now we can find out the mass. This mass = 1.2 kgm^-3 * 396 m^3 = 475.2 kg (which is the unit of mass; still on the right track!)
And since we have change in momentum in 1 second= -33(mass of air), change in momentum in 1 second = Force exerted on air BY wall = -15,681.6 Newtons.

By Newton's Third Law, the force exerted by the air on the wall acts in the opposite direction, with an equal magnitude. Therefore, the force is approximately 16,000 Newtons = D.

Q11)

In 1 second, 5.0 × 10⁴ α-particles collide into the lead sample. Let's take these 50,000 particles as our system and work from there.

Imagine the collisions happening; initially, the particles each have a certain amount of momentum, by virtue of their mass and velocity. They keep moving closer and closer to the lead sample over time, and eventually they smash into the lead sheet. Their momentum seems to vanish since they come to a stop, but something does happen here, something very important - there is a force on our system (the particles) from the lead sheet, and that force is large enough to bring all these particles to a halt in one second.

So how large is this force that acts on our system? By Newton's Second Law, this force is equal to the rate of change of momentum of our system (again, our system consists of the particles).
So they go from moving at a high speed to coming to a stop. Their speed changes, and that is why their momentum changes!

Let's find out this change in momentum. In 1 second, 50,000 α particles, each with a mass of 6.6 × 10⁻²⁷ kg and a speed of 1.5 × 10⁷ ms⁻¹ collide with the lead sheet over an area of 1.0 cm². After the collision, their velocity becomes zero. Therefore, we can write that the initial momentum is:

Initial total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(1.5 × 10⁷ ms⁻¹) = 4.95 * 10⁻¹⁵ kg ms⁻¹.
Final total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(0 ms⁻¹) = 0 kg ms⁻¹
Change in momentum (of system) = 0 - 4.95 * 10⁻¹⁵ = -4.95 * 10⁻¹⁵ kg ms⁻¹

This change in momentum takes 1 full second to occur, therefore the force on the system is

Force = (Change in Momentum)/(Time taken for change in momentum) = (Change in momentum)/1 = Change in momentum/second = -4.95 * 10⁻¹⁵ kg ms⁻².

By Newton's Third Law, this is the same force exerted by the system (all the particles) on the lead. Therefore, the pressure on the lead per second is equal to

Pressure = (Magnitude of Force)/(Area) = (4.95 * 10⁻¹⁵ kg ms⁻²)/(1.0/10000) = 4.95 * 10⁻¹¹ kg m⁻¹s⁻² = 5.00 * 10⁻¹¹ Pascals = C.

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

amal sharkawi

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Thought blocker said:
vertical velocity= u cos 45
horizontal velocity = u sin 45
so, initial K.E. : the vertical and horizontal components of kinetic energy = 1/2 m ( u cos 45 ) + 1/2 m ( u sin 45 ) ......... ( 1/2 m v^2 formula )

so, sin 45 = 0.71
....cos 45 = 0.71 ( check them in the calculator )

so they will give the same kinetic energy for both velocities ( horizontally and vertically )

so at the highest point, (theta) = 90 ( for the vertical velocity ) , horizontal velocity doesn't change
so final kinetic energy = 1/2 m ( u cos 90 ) + 1/2 m ( u sin 45 )..................... ( horizontal velocity is constant ) ( cos 90 is 0 )
final kinetic energy = 1/2 m ( u sin 45 ) + 0 only which is half of the initial kinetic energy.
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ZaqZainab

ZaqZainab

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chishtyguy

chishtyguy

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selena said:
Answer is B but I don't know how
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My mistake... Answer is B... It's the cell with which the voltmeter is connected not the resistor... So Emf coming out of the cell and across the 15 ohm resistor is 7.5V... Then by using the formula I=V/R =7.5/15 , we will get .5 A current...
 
Hadi Murtaza

Hadi Murtaza

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Last edited:
Hadi Murtaza

Hadi Murtaza

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Hadi Murtaza said:
q.9

Displacement is distance from starting point regardless of how much da body has travelled.

Distance from 0 to 3 s is (1/2) * 3 * 30 = 45 m

Distance from 3 to 5 is (1/2) * 2 * -20 = -20 m

So final distance from starting point = 45 - 20 = 25 m
Answer: B
Click to expand...

Der is another way of solving dis too but its slightly more complicated
 
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