A-Level Physics Paper 1 doubts (MCQ)

[NidZ- Hero]

http://www.xtremepapers.me/forums/viewtopic.php?f=26&t=9216&start=10

How about this


Conditions for an object to be in equilbrium ?

[Ghost Of Highbury]

Well, you gotta measure the lengths, and check if R+U = W

[NidZ- Hero]

Kinda figure that out so the answer is A ..



N that momentum question can yu pls show the working

[Ghost Of Highbury]

Sketch the wave on an axis as it would appear a few seconds later in time. Compare to the original... VOILA !!!

Could you please elaborate a li'l more? I didn't get it..

[elemis]

Could you please elaborate a li'l more? I didn't get it..

Give me some time.

[Ghost Of Highbury]

Give me some time.

Sure, thanks a lot

[elemis]

I've done the sketch below. The dotted curve represents the wave a few seconds later in time.

Notice all the points except C have moved downwards.

[Ghost Of Highbury]

Kinda figure that out so the answer is A ..



N that momentum question can yu pls show the working

Total Ke b4 collision => 0.5*2m*u^2 + 0.5*m*(-u)^2 = mu^2 + mu^2/2 = 3/2 mu^2
Total momentum b4 => 2mu - mu = mu

A. KE after collision -> 0.5*2m*(u/3)^2 + 0.5*m*(5u/3)^2 = mu^2 /9 + 25mu^2/18  = 3/2 mu^2
    Momentum = 2m*(-u/3) + m*(5u/3) = 5mu/3 - 2mu/3 = 3mu/3 = mu

[Ghost Of Highbury]

I've done the sketch below. The dotted curve represents the wave a few seconds later in time.

Notice all the points except C have moved downwards.

Okay, thanks, but how does it show that at that point the wave is moving UPWARDS with MAXIMUM speed?

[elemis]

Okay, thanks, but how does it show that at that point the wave is moving UPWARDS with MAXIMUM speed?

If you draw a straight vertical line through point A, the point where the line intersects with the dotted curve is the point at which A is later on in time.

Notice, point A will have moved downwards.

Do the same with the rest. Only C moves upwards.

Additionally, C moves the greatest distance in the same time as the rest. Velocity = distance/time

[NidZ- Hero]

So yu mean to say for the above explanation we read the graph from opp direction lke from D

[NidZ- Hero]

http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov/

Question 13

[elemis]

http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FA+Level%2FPhysics%2FCIE%2F2009+Nov/

Question 13

Which variant ?

[Ghost Of Highbury]

Im not sure about variant 1. The torque is obviously 4.5Nm
{Tension = Torque/diameter = 3/0.1 = 30
  Torque = tension*diameter = 30*0.15=4.5}

Although, i'm not sure why the tension is 60. Maybe, because the lower belt has no tension, and for a torque two forces acting in opposite directions
are required, I guess, the tension in the lower belt is transferred to the upper belt which adds up to 60 (30+30).

The projectile one is easy
Let's say the initial velocity of the projectile is u
The velocity at highest point is horizontal with velocity ucos 45

KE = 0.5m*(ucos45)^2 = 0.5m*u^2*0.5
0.5E

[NidZ- Hero]

Which variant ?

Variant 11