Physics: Post your doubts here!

[alevel]

please i need help on how DC circuit works + emf in physics AS

 

[OakMoon!]

Check out the notes in revision section. And pick up a book-the best help available out there.

 

[rijabukhari]

i need answers for AS mcq june 07 paper.question 7,8,10,11,12,13.

 

[abcde]

i need answers for AS mcq june 07 paper.question 7,8,10,11,12,13.

Did you check the mark scheme? Or do want the explanations of the answers?

 

[Jasmine01]

i need explanations for the answers of AS Qno. 13, 14 of paper 1 May/june 2005, Qno.15 of paper 1 of May/June 2006 and Qno. 7,10 and 13 of paper 1 of May/june 2007.

 

[fko]

in the photoelectric effect Plx. explain me the difference between Photons and Photo electrons , I'm confused that when the electromagnetic radiation strikes the Cathode plate so is the radiation carrying photons or photo electrons and when -ve charges are emitted from this plate WHAT ARE THEY CALLED!!!!

 

[ffaadyy]

in the photoelectric effect Plx. explain me the difference between Photons and Photo electrons , I'm confused that when the electromagnetic radiation strikes the Cathode plate so is the radiation carrying photons or photo electrons and when -ve charges are emitted from this plate WHAT ARE THEY CALLED!!!!

Electromagnetic Radiation is composed of quanta (packets) of energy called Photons. So the electromagnetic radiation which strikes the cathode plate is carrying photons. And the -ve charges that are emitted from the plate are Photoelectrons. This can be figured out from the definition of work function which says that 'Work function is the minimum energy required to emit 'electrons' from a surface'. Electrons and Photoelectrons are the same thing over here. Once the electrons gain the energy from photons and are emitted from the surface, then they are called Photoelectrons.

 

[unique840]

i need explanations for the answers of AS Qno. 13, 14 of paper 1 May/june 2005, Qno.15 of paper 1 of May/June 2006 and Qno. 7,10 and 13 of paper 1 of May/june 2007.

june '05 q13) moments = force*perpendicular distance
5N and 10N force have anticlockwise moments. (5*2) + (10*2) = 30 N
20N force is clockwise. (20*3) = 60N
resultant = 60 - 30
= 30N

 

[fko]

thnxx alot.ffaadyy

 

[unique840]

i need explanations for the answers of AS Qno. 13, 14 of paper 1 May/june 2005, Qno.15 of paper 1 of May/June 2006 and Qno. 7,10 and 13 of paper 1 of May/june 2007.

june '07 q10) force = momentum*time
means area under graph
f = [p1 + (-p2)] / (t2 - t1)
f = (p1 - p2) / (t2 - t1)

 

[Jasmine01]

Thanks a lot for helping me understand those questions Now I'll wait for the other explanations as well...

 

[rijabukhari]

i have checked the answers already,i need the explanations.

 

[Mustehssun Iqbal]

i need answers for AS mcq june 07 paper.question 7,8,10,11,12,13.

Q.8) u=0, s=30, a=10
v^2=u^2+2as
v^2=(0)^2+2(10)(30)
(v)^2=600
v=24.4949 (taking square roots on both sides of the equations to obtain the value of v in m/s)
u=24.4949,s=10, a=10
s=ut+(0.5)at^2
10=(24.4949)(t)+(0.5)(10)t^2
10=(24.4949)(t))5t^2
5t^2+24.4949t-10=0
(Solve the quadratic equation)
t=0.3710s Ans A
The stone is dropped from rest so u=0.Distance travelled is (40-10)=>Distance travelled=30m.
Acceleration of free fall is almost 10m/s^2.
Final velocity of the motion(indicated as v) is calculated.
Final velocity v of the previous motion becomes the initial velocity (indicated as u) of the next motion.
Distance travelled = 10m and acceleration of free fall is approximately 10 m/s^2.
Q.11) Change in momentum=m1u1 - m2u2
Change in momentum=(20000)(20)-(900)(30)
Change in momentum=373000 kg m/s Ans B
Q.13)Torque=(8sin60) * (0.60) => Torque=4.1569 Nm Ans B
The diagram in Q.13 is about a couple of forces.The resulting Torque(moment of a force) is the product of
one of the forces in the couple of forces and the perpendicular distance between the point of action of the other force.
The force exerted to the beam perpendicular to the the point of action of the other force in the couple of
forces is 8sin60.
Q.12)Momentum of the system of trolleys=m1v1+m2v2
Momentum of the system of trolleys=2(4) + (4)(1)
Momentum of the system of trolleys=12 kg m/s
m1v1+m2v2=m'v'
12=(2+4)v'
12=6v'
v'=12/6
v'=2
Kinetic Energy=0.5m'v'^2
Kinetic energy=0.5(2+4)(2)^2
Kinetic energy=0.5(6)(2)^2
Kinetic energy=12J Ans B
The two trolleys move together after their collision in Q.12/.m' is the sum of trolleys and v' is velocity of
those trolleys after collision.

 

[Mustehssun Iqbal]

m' is the sum of mass of trolleys in Q.13 *.

 

[rijabukhari]

thankyou for the explanations,but i was only able to understand question no 12 and 13.
q8 is the most confusing .how did u know that this question needs both the fromulas?i was just solving it by s=ut + 1/2 at2 which was obviously wrong.
in q 11 they have asked us the total momentum but what you have calculated ,isnt it the initial momentum?shouldnt i also add the final momentum ? sorry for asking such dumb questions but physics is getting really tricky

 

[Mustehssun Iqbal]

no problem!
In Q.11, it's asked to calculate the total momentum, or you can call it initial momentum or the final momentum. They might be asking the total momentum of the system before collision or the total momentum of the system after the collision. However, the total momentum before collision and after collision of a system remains constant.
m1u1-m1v1(initial momentum of the system)=m2v2-m1v1(final momentum of the system)
Since those momenta are both equal, you can calculate either one of them, or in other words, either side of the equation.
If you add or solve the total momentum of a system before collision and after collision, you would probably get zero as an answer to that equation, almost whatever collision it might be.
In Q.8, the tricky part is the word problem in that question. If you solve it with one equation, you might calculate the time taken , but that would be the time taken to travel 40m, which isn't what the question demands. You might put 40m in place of s(distance) and get that answer. But the question asks the time taken to travel the last 10m to the ground. In that case, the initial speed ,u is not zero. The stone is already travelling. Distance is known, and so is the acceleration of free fall. We can take this motion in two parts, as two different motions, first to travel 30m and second to travel 10m to the ground.
So the final velocity of the first motion of this stone, is the initial velocity of the second motion of this stone. For the first motion, u=0, s=30 and a= 10 m/s^2 . You calculate the v of that equation, put that value of v as u in the second equation and calculate the time taken , t of the second equation.

 

[abcde]

9. One way of dealing with this question is realising that at the lowest point, the velocity will have a maximum magnitude (min G.P.E and max K.E). Secondly, it will be negative since downward motion is ascribed a negative value. Hence, C is correct.
12. Momentum before collision = momentum after collision
m(-2v) + 3m(v) = 4m V
=> V = mv/4m = v/4.
Hence, A is correct.

I was going through this question today when I realised that the explanation and the answer I gave to MCQ 9 above is utterly bogus. My apologies for the error! The correct answer is D. That's because the mass is moving up AND down on the end of a spring. Say the velocities are positive as it moves TO the lowest point (downward vertical motion). When it will move AWAY from the lowest point (upward vertical motion), the direction of motion is reversed so the velocities will attain negative values. Therefore, for the brief moment when the mass is AT the lowest point, the velocity would be zero.
Really sorry for the erroneous explanation @leosco1995! Hope you didn't make that mistake in your exam.

 

[Astellaz]

Trick Question for all who can answer!!!
A uniform metre rule of weight 0.9N is suspended horizontally by 2 vertical loops of thread A and B placed at 20cm and 30cm from its ends respectively. Find the distances from the centre of the rule at which a 2N weight must be suspended:
i) to make loop A become slack
ii) to make loop B become slack
Lets see whos gonna guess the answer

 

[Ashique]

june '07 q10) force = momentum*time
means area under graph
f = [p1 + (-p2)] / (t2 - t1)
f = (p1 - p2) / (t2 - t1)

I don't really get what you have done. How is area under the curve (p1 - p2) / (t2 - t1), when F=mt?

 

[omg]

phy P5 helppp??? ((((