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Thank you.Q8
The time when the express train crosses the goods train will be the time when both of them have traveled the same distance.
Distance traveled by goods train = speed * time = 10t
Distance traveled by express train = ut + 1/2at^2 = 1/2at^2
u = 0 (express train starts from rest) and t cancels:
1/2at = 10
at = 20 (acceleration = 0.5)
t = 40s
Answer is D
Q11:
The collision is inelastic. Masses are added on the RHS:
m1v1 + m2v2 = (m1 +m2)v
v2 = 0, so:
2 * 3 = (2 + 1) v
v = 2
KE lost = Initial KE - Final KE = 0.5m1v1^2 - 0.5(m1 + m2)v^1/2
= 0.5*2*3^2 - 0.5*3*2^2 = 3 J
Answer is D
Q16:
Since the tension T in the cable alone is countering the other forces to keep the trapdoor in equilibrium, it is the largest. The only option with T being maximum is C, so the answer is C.
Q17:
v = (2gh)^1/2
HEIGHT IS 0.72 NOT 0.8! Measure from the base of the ball. Diameter is 0.08. We have to subtract that.
Use 1/2mv^2 = initial KE to calculate mass.
u = (2gh)^1/2
Again, height is 0.45-0.08 m.
KE after impact = 1/2mu^2 = 0.39 J
Answer is C
Q18:
P = E/t
P = (1/2mv^2)/t [m = density p * volume V]
P = (1/2pVv^2)/t [Vol = Area A * length s]
P = (1/2pAlv^2)/t [l/t = speed v]
P = 1/2Alv^3
Substitute values, Answer is B.
ummm. Question number 35 too please.
