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for s12 qp11Thank u !
Can u help me with q15 of the same paper.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
And Q14 ,18 and 21 0f http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
#15
L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it.
note the system is in equilibrium so there must be no net force and no net torque
immediately eliminate A and B because there is a horizontal H force that is unbalanced
now the logic comes for C, D.
for D, the reaction is shifted closer to the right (you can consider centre of mass shifted closer to right), so clockwise moment decreases, anticlockwise moment increases. Thus balances the system. so D is correct
for W12 qp 11
#14
initial momentum = final momentum
momentum = mv
0.005 x 200 = (0.005 + 0.095) v
v = 10 m/s
use kinematics equation
v^2 = u^2 + 2gs
0 = 10^2 - 2 x 9.81 x s
s = 100/(2x9.81) = 5.1 answer = A
#18
Ek = (mv^2)/2 so Ek is proportional to v^2
Ek = cv^2 where c is a constant
4Ek = c(new v^2)
so new v must be 2v because 2^2 = 4
#21
efficiency = output/input
output power = VI = 200 x 6000 = 1200000
input = E(potential)/t = mgh/t = 500x300x 9.85 = 1477500
efficiency = 1200000/1477500 = 82%