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man the answer is -5.76 !!!
very simple: mv-mu = change in momentum. 1.2 ( -0.8 - 4 ) = -5.76 kg m/s , u shud notice the time of collision. the initial velocity being 4 m/s and final being -0.8 .
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Physics: Post your doubts here!
- Thread starter XPFMember
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very simple: mv-mu = change in momentum. 1.2 ( -0.8 - 4 ) = -5.76 kg m/s , u shud notice the time of collision. the initial velocity being 4 m/s and final being -0.8 .
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have some doubts in paper 1 m/j 05 q14, o/n 05 q5,7,16, o/n 09 q9,14,28, m/j 10(P11) q22,26,27, o/n 10(P11) q18,24,37, m/j 11(P11) q15,34
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this is so messy. can u please upload the links and write clearly. thanks
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Sure, here..........
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s05_qp_1.pdf Q14
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w05_qp_1.pdf Q5,7,16
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w09_qp_11.pdf Q9,14,28
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_11.pdf Q22.26,27
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_11.pdf Q18,24,37
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf Q15,34
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But the marking scheme states it to be +57.6
Also it states the working to be 1.2(4+0.8)
Here:http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s09_ms_2.pdf
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o
this is what marking scheme says:-
(change in momentum = 1.2 (4.0 + 0.8) ....................................... C2
(correct values, 1 mark; correct sign {values added}, 1 mark )
= 5.76 N s …(allow 5.8) .....)
now the question is why positive sign so i guess the answer to this is that 4 m/s is the initial velocity and -0.8 is the final velocity so it will be 1.2( 4 - (-0.8) = 5.76 Nm
ooooooo brother its 5.76 Nm
this is what marking scheme says:-
(change in momentum = 1.2 (4.0 + 0.8) ....................................... C2
(correct values, 1 mark; correct sign {values added}, 1 mark )
= 5.76 N s …(allow 5.8) .....)
now the question is why positive sign so i guess the answer to this is that 4 m/s is the initial velocity and -0.8 is the final velocity so it will be 1.2( 4 - (-0.8) = 5.76 Nm
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If it's C then its vice versa. means the steeper the I/V gradient the lesser the resistance. The resistance is equal to = 1/ gradient.. the larger the value of gradient the lower the resistance.
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but the gradient is higher at B :/
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http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w10_qp_51.pdf
PLz someone know how to do this Uncertaintiy Q2b
PLz someone know how to do this Uncertaintiy Q2b
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November 2005, Paper 1, Q.16;
Q.16) The barrel moves up the slope with a constant speed. Hence forces are observed to be in equilibrium. The resultant force on the barrel is zero since the barrel moves up the slope with constant speed.
Work done = P.E + K.E + Wf
Work done in the equation above this line represents the work done on the barrel. P.E represents the work done against gravity on the barrel. K.E represents the work done against the kinetic energy of the barrel and Wf represents the work done against friction in the movement of the barrel. In this case, K.E and Wf are zero and hence the equation becomes;
Work done = P.E
Work done = mgsintheta
Note we are resolving forces horizontally. Since there's no vertical component of the applied force, there's just horizontal component of it there too.
Work done = (1.0 * 10^3) ( 5 ) sin30
therefore work done = 5 * 10^3(0.5 )
therefore work done = 2.5 * 10^3 Ans B
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November 2005, Paper 1, Q.5;
Q.5) let rogue be p
p = m/v
therefore p = m/(l * b * h)
delta p/p = delta m/m + delta l/l + delta b/b + delta h/h
therefore delta p/p = 0.1/25 + 0.01/2 + 0.01/2 + 0.01/1
therefore delta p/p = 0.004 + 0.002 + 0.005 + 0.01 Calculation part is a little bit tricky so I suggest you redo this question too.
The value of density is given. And the question is asking for the uncertainty in the result of the calculated density. And what you've calculated now is the fractional uncertainty.
therefore delta p/2.5 = 0.021
therefore delta p = 0.05 gcm^-3 Ans C
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s05 q. 14) see the resultant and all other forces initiate from the same point. the resultant is coming outwards from the paper u gotta imagine it in 3D. in 2d most likely the right direction is D
w05 qp 5) uncertainty in density / density = uncertainity over mass + Uncer. / length + Uncer / breath + Uncert./ height. equate the values and you will get 0.05 g/ cm^3
7) s= ut + 1/2 gt^2 . ut is zero as initial velocity is xero. g = 2s / t^2 right so now we need to subtract both times in order to get one expression giving acceleration : 2s / t2^2- t1^2 D option.
16 ) calculate the weight component along the slope which will be 1 into 10^3 into sin 30. multiply the force now with distance i.e 5 m. u will get B 2.5 into 10^3 Nm
oct/ nov o9 ) q.9 i dnt knw q.14 dnt knw, q. 28 ) the polarity has to be negative as the weight is acting downwards the electric force must be upwards in order to balance it so it stays at equilibrium position. E = F / q , q = F/E i.e B
s 10) q.22 its c , Infra-red radiation. u need to remember the wavelengths and frequencies of visible spectrum..plus the size of slit should always be very close to the wavelength in order to see diffraction.
26) the force acts on every charge present in the available field region, the direction of force is parallel to the field. fact.
27) zero A because work done = force into distance travelled in the direction of force. the direction of force is in the middle due to centripetal force but the charge is moving elsewhere. so work done zer0
w10) q 18) Power = Fv , upwards resultant force > F= mg , so (m1-m2) gv
q 24)
you need to identify water waves as transverse waves. For transverse waves, point particles move up and down and NOT left and right.
So each of the points A, B , C and D will move up and down ONLY, so do not get tricked that they will move to the right.
First of All, B and D are wrong because they are the maximum displacement , so obviously, the acceleration is max and speed is zero.
To get the right answer you have to image the wave moving. If you just shift this wave to the right abit, you will notice that A will be at the bottom of the shifted wave and C will be at the top. Thus, C is the point with the maximum speed going upwards.
q. 37) u can see that the voltmeter is place across the power supply, the voltmeter will give us the e.m.f. not voltage against any of the two resistors.-----------------------------------------------------------------------------------------------------------------------------------------------------------------------
s 11 paper 11) q. 15 ) loss in g.pe = gain in kinetic energy plus work done against friction
2 . 9.8 . 3 = 1/2 . 2 . v^2 + 5 into 7
58.86 - 35 = v^2 v = 4.9 m/s
34 I don't know sorry.
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you have to draw out several lines from the origin to each of the points A/B/C/D.
So you will notice that the line is steepest i.e highest gradient when it connects the origin and point C.
A common error is drawing tangents to the points on the graph so be careful.
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BUT THAT IS EXACTLY WHAT I DON'T UNDERSTAND!
The change in momentum is supposed to be m(v-u), isn't it?
Then shouldn't it be 1.2(-0.8-4)?
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OK MAN TAKE IT EASY HERE IS THE TRICK !!! at least u realised dat u were quoting 58.76 again and again
this is what i got:-
yes it is m(v-u)
the final velocity is 4 m/s NOT 0.8 m/s. after the collision mass B will have a increase in its speed not decrease as the ball S is much heavier. when the ball B is travelling towards right, its direction is taken to be positive, when it rebounds after colliding with ball S, now its in negative direction but 1.2 ( 4 - ( -0.8) end up with positive 0.8 ( as minus into minus = plus) plus 4 into 1.2 i.e 5.76 N m/s
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If I were you, I wouldn't get too attached to the mark scheme and I highly doubt the examiners would have marked you down for that question. Your working is correct and the change in momentum is negative. Cambridge just seemed to give the magnitude of the change in momentum, ignoring their own sign convention.
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so u m
so u mean dat v is -1 and u is 4 m/s in this case?
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Check out what the examiner report has to say:http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_er.pdf
The marking scheme and the examiner report agree with each other: the velocities must be added, not subtracted
OK MAN TAKE IT EASY HERE IS THE TRICK !!! at least u realised dat u were quoting 58.76 again and again
this is what i got:-
yes it is m(v-u)
the final velocity is 4 m/s NOT 0.8 m/s. after the collision mass B will have a increase in its speed not decrease as the ball S is much heavier. when the ball B is travelling towards right, its direction is taken to be positive, when it rebounds after colliding with ball S, now its in negative direction but 1.2 ( 4 - ( -0.8) end up with positive 0.8 ( as minus into minus = plus) plus 4 into 1.2 i.e 5.76 N m/s
First of all : DO YOU ACTUALLY THINK THAT QUOTING THE DECIMAL FIGURE INCORRECTLY IS MORE IMPORTANT THAN SOLVING THE QUESTION???????????????
lol just kidding
I don't get you point about the final velocity being 4m/s :the question states the final velocity to be -0.8 and the initial to be +4m/s, how can you change that?
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HAHAHA Its very important. one decimal can change ur life.Check out what the examiner report has to say:http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_er.pdf
The marking scheme and the examiner report agree with each other: the velocities must be added, not subtracted
First of all : DO YOU ACTUALLY THINK THAT QUOTING THE DECIMAL FIGURE INCORRECTLY IS MORE IMPORTANT THAN SOLVING THE QUESTION???????????????
lol just kidding
I don't get you point about the final velocity being 4m/s :the question states the final velocity to be -0.8 and the initial to be +4m/s, how can you change that?
man i also don't get what i said. i m happy to loose marks in this question.
i dnt knw what it is. sorry