天同師兄,岩岩考完AL,有幾題想請教你

My suggested answers are given below.

4. option DGraph (1) is impossible, P and Q are initially moving in the same direction with P faster than Q (notice that the slope of the line in a s-t graph represents velocity). Hence, during collision, P exerts a "pushing force" to Q, and we would expect to have Q gains in velocity. But the graph shows that the velocity of Q decreases after collision.

7. option C

Moment of inertia I of the rod and beads = mL^2 + m(2L)^2 = 5mL^2
where m is the mass of each bead, and 2L is the length of the rigid rod

Loss of potential energy when the rod is vertical
= mgL + mg(2L) = 3 mgL
Gain in kinetic energy = (1/2).I.w^2 = (1/2).(5mL^2).w^2 = (5/2)(mL^2w^2)
where w is the angular speed when the rod is vertical
Hence, 3mgL = (5/2)(mL^2w^2)
w^2 = 6g/(5L)
Centripetal acceleration of Q = (2L)w^2 = (2L).((6g/5L) = 12g/5

13. option B
From the given figure, wavelength = 4 x 10^-7 m = 400 nm
since 400 nm, corresponds to a frequency of 7.5x10^14 Hz, is the wavelqngth of violet light.

21. option B
The strongest eelctric field occurs when the equipotential lines are closest apart.

30. option D
The path traversed by the particle is an arc of a circle, of radius R say, subtentded at the circle centre with angle 2(theta).
Length of path = R[2(theta)]
Time of travel t = R[2(theta)]/v
But mv^2/R = qvB (m is mass of particle, q is its charge)
R = mv/(qB)
hence, t = 2m(theta)/qB

31. option A
Output = 220 x (100/1100) v = 20 v
notice that the two secondary coils supply current to the output alternatively, one at each half cycle (+ve and -ve cycle) of the ac.