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Physics: Post your doubts here!
- Thread starter XPFMember
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thx a lot physicist for the link, it really will be useful!!!
not everyone out there would be wiling to share such resources, thx a lot bruh!
have similar stuff for chemistry? i know this is the physics section bt just inquiring whether u have stuff like that for chem?
thx once again
not everyone out there would be wiling to share such resources, thx a lot bruh!
have similar stuff for chemistry? i know this is the physics section bt just inquiring whether u have stuff like that for chem?
thx once again
- Messages
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http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_2.pdf
Q1 b (ii) 2
Q3 c(ii) onward
Q6 (C)
thanks a lot in advance
Q1 b (ii) 2
Q3 c(ii) onward
Q6 (C)
thanks a lot in advance
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ok for part b, u have to find the speed of the bullet before hitting the block, so u have to apply rule of conservation of momentum which, if i paraphrase, states that the total momentum before collision must equal total momentum after collision in a closed system, hence:
momentum before collision: mass of bullet* velocity of bullet + mass of block* velocity of block.
so we change masses to the right SI unit in this case kg and remember, the block won't have momentum initially as it is at rest so 0 velocity before the collision.
so momentum before collision will be momentum of only the bullet as its fired so mv= 0.002kg* v
momentum after collision would be the total mass of both bodies times their velocity, bcd, remember they stuck together as u are told in the question so u treat it as one mass in that case, so total mass will be (0.6+0.002)= 0.602kg and velocity is 1.3 as told in the previous part
so if we equate the mv before collision and after, we get:
0.002v=0.602*1.3, and hence, we get v as 391mtrs per sec.
c.) here u have to find kinetic energy of the bullet before impact, and k. energy is given by 1/2mv^2 and we already found the velocity as 391 so just a matter of substituting so 1/2*0.002*391*391=153 joules
the second part is asking to state the type of collision, u should know that there is two types of collision, the inelastic collision and elastic collision. inelastic is where kinetic energy is not conserved that is, it is different before and after the collision while elastic is where the kinetic energy remains exactly the same before and after collision so in this case, if u find the kinetic energy after collision it will be 1/2*0.602*1.3^2= 0.5 joules so clearly u can see that the kinetic energy before was 153 and after was 0.5 so its not the same so it is inelastic.
momentum before collision: mass of bullet* velocity of bullet + mass of block* velocity of block.
so we change masses to the right SI unit in this case kg and remember, the block won't have momentum initially as it is at rest so 0 velocity before the collision.
so momentum before collision will be momentum of only the bullet as its fired so mv= 0.002kg* v
momentum after collision would be the total mass of both bodies times their velocity, bcd, remember they stuck together as u are told in the question so u treat it as one mass in that case, so total mass will be (0.6+0.002)= 0.602kg and velocity is 1.3 as told in the previous part
so if we equate the mv before collision and after, we get:
0.002v=0.602*1.3, and hence, we get v as 391mtrs per sec.
c.) here u have to find kinetic energy of the bullet before impact, and k. energy is given by 1/2mv^2 and we already found the velocity as 391 so just a matter of substituting so 1/2*0.002*391*391=153 joules
the second part is asking to state the type of collision, u should know that there is two types of collision, the inelastic collision and elastic collision. inelastic is where kinetic energy is not conserved that is, it is different before and after the collision while elastic is where the kinetic energy remains exactly the same before and after collision so in this case, if u find the kinetic energy after collision it will be 1/2*0.602*1.3^2= 0.5 joules so clearly u can see that the kinetic energy before was 153 and after was 0.5 so its not the same so it is inelastic.
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_2.pdf
ok so for question one part b 2, the first one is after 0.7 sec right, so here u just have to count the number of seven successive dots starting from the first one, coz u are told each dot is a flash of the object after 0.1 sec so for 0.7 sec it will be 0.1*7 which is 0.7 sec, so u count upto the 7th dot and read the value given and u will find it to be 132 cm.
now the one for after 1.1 seconds is abit challenging, check eh:
if u notice the trend, the thing starts moving with constant speed, how do i know that? well if u see the spacing of the ball, it becomes the same spacing after
around the fourth photograph, so this same spacing means that the distance covered in every 0.1 sec is the same and since distance is the same and the time is also the same i.e every 0.1 sec, so u can definitely tell tat the speed is constant since speed is distance/time. am trying to go step by step so u get it....
so now, what is this distance? well if u see, the distance is 25 meters from each flash starting the fourth photograph, for instance, take the fourth photograph, what distance can u read? u get 58cm, then take the fifth photo, the distance is 82 s if u minus 58 from 82 u get the distance as 25cm, then take the sixth photo u see, the distance is 108 and the fifth one is 82 so if u minus again, u still get 25 cm, so this shows that now the distance after each photograph will have a difference of 25cm from the preceding phograph so if we understand that concept, then we can find the distance for 1.1 sec as follows:
we know that till 0.7 sec the distance is 132 cm as calculated earlier and now we know that for 0.8 sec the distance will be 132cm + 25= 157, then for 0.9, it will be 157+25= 182cm then for 10.0sec it will be agin 182 plus 25 u get, 207cm and finally for 1.1, it will be 207cm plus 25 and u get 232xm and thats ur answer....
c ok for part c we have to find the number of photographs when it is a lead sphere, here we need to first use the equations of linear motion and i am gong to use s=ut+1/2at^2 and we know it is falling under gravity so the value for acceleration will be 9.8ms^-2, i hope u r familiar with this idea of gravity, if not just know that when a body is falling vertically =, then it falls under the influence of gravity and the value of acceleration is hence 9.8 or some books give 10 so using the equation now.
we know the distance is 160cm as told in the question, so in meters thats 1.6m, the value of u, the initial velocity is 0 as it started from rest.
so now we have: s=ut+0.5at^2 , hence, 1.6= 0t + 0.5*9.8*t^2 so if we now find t, we will get it to be 0.57seconds,
but we are told to find the number of photographs, but we know that each photo is after 0.1 seconds so
0.1sec= 1 photograph
0.57sec= x photogrpahs, if we cross multiply, we get 5.7 photographs.
bt we can't have a fraction of a photograph if we think of it practically so we take the next number which is 6 so photographs.
ok so for question one part b 2, the first one is after 0.7 sec right, so here u just have to count the number of seven successive dots starting from the first one, coz u are told each dot is a flash of the object after 0.1 sec so for 0.7 sec it will be 0.1*7 which is 0.7 sec, so u count upto the 7th dot and read the value given and u will find it to be 132 cm.
now the one for after 1.1 seconds is abit challenging, check eh:
if u notice the trend, the thing starts moving with constant speed, how do i know that? well if u see the spacing of the ball, it becomes the same spacing after
around the fourth photograph, so this same spacing means that the distance covered in every 0.1 sec is the same and since distance is the same and the time is also the same i.e every 0.1 sec, so u can definitely tell tat the speed is constant since speed is distance/time. am trying to go step by step so u get it....
so now, what is this distance? well if u see, the distance is 25 meters from each flash starting the fourth photograph, for instance, take the fourth photograph, what distance can u read? u get 58cm, then take the fifth photo, the distance is 82 s if u minus 58 from 82 u get the distance as 25cm, then take the sixth photo u see, the distance is 108 and the fifth one is 82 so if u minus again, u still get 25 cm, so this shows that now the distance after each photograph will have a difference of 25cm from the preceding phograph so if we understand that concept, then we can find the distance for 1.1 sec as follows:
we know that till 0.7 sec the distance is 132 cm as calculated earlier and now we know that for 0.8 sec the distance will be 132cm + 25= 157, then for 0.9, it will be 157+25= 182cm then for 10.0sec it will be agin 182 plus 25 u get, 207cm and finally for 1.1, it will be 207cm plus 25 and u get 232xm and thats ur answer....
c ok for part c we have to find the number of photographs when it is a lead sphere, here we need to first use the equations of linear motion and i am gong to use s=ut+1/2at^2 and we know it is falling under gravity so the value for acceleration will be 9.8ms^-2, i hope u r familiar with this idea of gravity, if not just know that when a body is falling vertically =, then it falls under the influence of gravity and the value of acceleration is hence 9.8 or some books give 10 so using the equation now.
we know the distance is 160cm as told in the question, so in meters thats 1.6m, the value of u, the initial velocity is 0 as it started from rest.
so now we have: s=ut+0.5at^2 , hence, 1.6= 0t + 0.5*9.8*t^2 so if we now find t, we will get it to be 0.57seconds,
but we are told to find the number of photographs, but we know that each photo is after 0.1 seconds so
0.1sec= 1 photograph
0.57sec= x photogrpahs, if we cross multiply, we get 5.7 photographs.
bt we can't have a fraction of a photograph if we think of it practically so we take the next number which is 6 so photographs.
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ok to be honest i have forgotten about moments or maybe am too tired to do it right now so maybe tomorrow or f someone else can help, that is number 3.
also number 6 for some reason i used to not have an explanation as to why the answer was so but always when asked such questions, u just have to describe the atom like its empty space and it is dense etc.
i used to not know the real reason tbh bt i used to just write that coz whenever such questions come, that is usually the answer.
u will find the notes for the atom on ur text book, bt reason why thats the answer, i really font kn =w maybe if someone can assist.
thx.
also number 6 for some reason i used to not have an explanation as to why the answer was so but always when asked such questions, u just have to describe the atom like its empty space and it is dense etc.
i used to not know the real reason tbh bt i used to just write that coz whenever such questions come, that is usually the answer.
u will find the notes for the atom on ur text book, bt reason why thats the answer, i really font kn =w maybe if someone can assist.
thx.
Hello, and thanks in advance : D
I am taking AS physics and have some questions.
1. What are the conditions for a formation of a stationary wave: e.g same speed and frequency but what else?
2. What is the difference between a potential divider circuit and a potentiometer?
iiwhat is the difference between a rheostat and a variable resistor
3. If we have a variable resistor and a fixed resistor connected to a e.m.f, How do we find which resistance of the variable resistor will give MAXIUM power usage to it?
I am taking AS physics and have some questions.
1. What are the conditions for a formation of a stationary wave: e.g same speed and frequency but what else?
2. What is the difference between a potential divider circuit and a potentiometer?
iiwhat is the difference between a rheostat and a variable resistor
3. If we have a variable resistor and a fixed resistor connected to a e.m.f, How do we find which resistance of the variable resistor will give MAXIUM power usage to it?
Help with june 2013 no. 6 b (i) and c (ii)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_42.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_42.pdf
Thank you
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Q5 a ii and iii) With proper explanation.
Q6 c i and ii)2. with explanation again pleaseeee
ty
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
Q6 c i and ii)2. with explanation again pleaseeee
ty
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_22.pdf
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Woww, this site is awesome. Thanks so much
Can you send me a link to ALL of the papers done?
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V2-U2 = 2as
Initial is zero as it is dropped from rest. you need the final velocity
Acc of free ( or you may call it g) is always -9.81 whether No Air resistance or Air resistance
Distance is 4.5
so input it
V2= 2 (9.81) (4.5)
Dont forget to take root.. I even missed the V2 would go on the other side and Square root
V=9.4 ms-1
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(b) The line of action of the weight W of the rod passes through the cord at point P.
Explain why, for the rod to be in equilibrium, the force F produced at the hinge must also
pass through point P?
How do we solve this?If all forces pass through a specific point,they have zero moment?How to explain this?
Awesome12 F.Z.M. 7 Dark Destination
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How to do this question?
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Could you elaborate?If all the forces pass through a specific point the net moments or forces are zero?Can you explain this much?
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So if all forces pass through one point,the system will be in equilibrium?regardless of force magnitude?
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Ok,so at AS level we just have to look at the direction and if they intersect its in equilibrium?
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Can u explain how are the wires in parallel?
thank you : D. for the formation of stationary waves, do the waves need some kind of phase difference to form, like if it is formed from reflection, what would you call the phase thhat they meet at if this makes sense