can somebody explain the answers for Q9,13,15 & 28 O/N 2009/11
Thx
9. otal 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
13.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).
15. First find out the velocity of the 1kg trolley
(2*-2)+(1*x) = 0
-4+x = 0
x = 4m/s
KE = 0.5mv^2 = 0.5*2*(-2)^2 + 0.5*1*(4)^2 = 12
D
28. Charge to mass
We know, that Force acting on it = Weight
E = F/q
F = Eq
Eq = mg
m = Eq/g
Charge to mass = q/(Eq/g) = g/E
Polarity should be negative because the positive plate will pull it upwards and its weight will pull it downwards. Hence they balance o ut