Solution: va = 2gh = 2×10× 3 = 60m/s.
At point a, according to Newton's second law, the elastic force FN = MVA2R = 1× 600.2 = 300 N, and the direction is horizontal to the right.
(2) Conservation of mechanical energy? mg(R+H)= 12mvB2
Solution: VB = 2×10× (3+0.2)1= 8m/s.
Newton's second law μ g = Ma1+0 for slider and trolley.
Solution: a1= μ g = 0.5×10 = 5m/S2.
According to μ g = ma2
Solution: A2 = μ GM = 0.5×105 =1m/S2.
The slider decelerates uniformly, and the sliding speed is v1= VB-a1t = 8-5 = 3m/s.
Displacement s 1 = vbt? 12a 1t2=8× 1？ 12× 5×1= 5.5m.
The trolley accelerates uniformly, and when the slider slides out, the trolley displacement S2 =12a2t2 =12×1×1= 0.5m.
Speed V2 = A2t =/kloc-0 /×1=1m/s
The length of the skateboard L=s 1-s2=5.5-5=5m.
(3) After the slider slides out of the trolley, do flat throwing, and the trolley moves at a uniform speed, h = 12GT02.
Solution: t0 = 2hg = 2×1.810 = 0.6s.
The distance from the landing point to the right end of the trolley board is s = (v1-v2) t0 = (3-1) × 0.6 =1.2m.
Answer: (1) When the slider passes through point A, the elastic force on the slider is 300N and the direction is horizontal to the right.
(2) The length of the trolley is 5m.
(3) When the slider lands, its horizontal distance from the right end of the trolley is1.2m. 。