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Shantou yimo mathematics grade three
The conservation of momentum in (1) system leads to mava=mbvb, ①.

Ma=2mb=2? Kg, va = 4.5m/s

Solution: vb=9.0m/s? ②

Let the speed of slider B when it reaches point B be vB, which is obtained by kinetic energy theorem. μmbgL= 12mbv2B? 12mbv2b③

When entering the circular orbit, let the supporting force of slider B be FN, which is obtained from Newton's second law, FN? mbg=mbv2BR④

By Newton's third law fn =? f’N⑤

The pressure f ′ n of slider B on the track is obtained from ③ ④ ⑤. 59N, vertically downward.

(2) If the small slider B can reach the highest point of the circular trajectory, the speed is vC.

Then by the conservation of mechanical energy,12mbv2b = mbg2r+12mbv2c6.

The solution is vc=3.0m/s⑦.

The speed at which the small piece B can just pass the highest point is V, so mbg=mbv2? R⑧

Solution, v = gr =10m/s pet-name ruby.

Because VC < V, the small slider B can't reach the highest point C of the circular orbit.

Answer: (1) When the small slider B passes through the point B of the circular track, the pressure on the track is 59N, and the direction is vertical downward.

(2) The small slider B cannot reach the highest point C of the circular orbit.