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Physics: Post your doubts here!

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sagar65265

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not.maria said:
View attachment 43738
kindly someone help me with Q9 mj13 qp12
how to we solve this?
Click to expand...

On checking the marking scheme the answer is D - where did you find the answer to be A?

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.

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

mehria

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Thought blocker said:
k
Click to expand...
Thought blocker said:
I did like this :¬
View attachment 43780
Click to expand...

dnt say "k"
n ya they r talkng abt the power dissipated in the device nt the circuit...so first of all we will find the V through the electric device... i.e. P=V^2/R
4=V^2/20 => V=8.9V
now the current across the circuit will be I=V/R => I= 8.9/20 = 0.445 A
the Voltage across the variable resistor will be 16-8.9=7.1V
as we have both current n voltage so now we cn easily find the resistance
R=V/I = 7.1/0.445 = 15.9 ohms... which is approximately 16
 
Thought blocker

Thought blocker

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mehria said:
dnt say "k"
n ya they r talkng abt the power dissipated in the device nt the circuit...so first of all we will find the V through the electric device... i.e. P=V^2/R
4=V^2/20 => V=8.9V
now the current across the circuit will be I=V/R => I= 8.9/20 = 0.445 A
the Voltage across the variable resistor will be 16-8.9=7.1V
as we have both current n voltage so now we cn easily find the resistance
R=V/I = 7.1/0.445 = 15.9 ohms... which is approximately 16
Click to expand...
Ty :)
 
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sagar65265

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MYLORD said:
If any1 could help me with these doubts :Do_O


View attachment 43767 View attachment 43768 View attachment 43769 View attachment 43770 View attachment 43771 View attachment 43772
Click to expand...

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!
 
Hadi Murtaza

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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 !!!
 
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