※ 引述《dinks (丁克思)》之銘言： : 題目及我解到後面卡住的內容如下。 : 本題正解是(D)，但可憐的我算不出來 QQ : 【出處】普物 : 【題目】 : https://i.imgur.com/GF4PuRQ.jpg : 【瓶頸】下面是我的解法，算到後面發現M的動能居然是負的，我就知道我一定有什麼觀 : 念是錯的，所以整個卡死。還望先進指點，感恩。 : https://i.imgur.com/7M72kVE.jpg You should deal with this problem more carefully due to the fact that the mechanical energy is not conserved since the normal force would do the work during the process. Before entering the derivation, the mass is used to label these two objects. Some notations are introdueced and defined as follows: t: time. N(t): the normal force between these two objects. v_x(t): the horizontal component for the obejct m's velocity. v_y(t): the vertical component for the object m's velocity. \theta(t): the polar angle respect ot the vertical direction for the object m. V_x(t): the horizontal component for the object M's velocity. Above quantities are functions of time. Assume when the moment t=\tau, the object m will leave from the object M. At first, in the lab frame, the linear momentum is conserved along the horizontal direction. It gives the following relation: mv_x(t) = MV_x(t), for all 0<t<\tau -------------------------------(1) On the other hand, apply work-kinetic energy theorem on object M, it follows: \int^{\tau}_{0} N(t)sin(\theta(t))V_x(t)dt = M(V_x(\tau))^2/2------(2) where \int^{\tau}_{0} is the intetgration on the interval [0,\tau]. Similarly, apply work-kinetic energy theorem on object m, it arrives that: m[(v_x(\tau))^2+(v_y(\tau))^2]/2 = mgR(1-cos(\theta(\tau)) ) + \int^{\tau}_{0} N(t)sin(\theta(t))v_x(t)dt - \int^{\tau}_{0} N(t)cos(\theta(t))v_y(t)dt -------(3) where g is the gravitational accelaration and R is the radius of the object M. Now the rest frame in which the object M is in the rest is considered to connect v_x(t), v_y(t), \theta(t) and gR. In this frame, the object m moves along the surface of the object M. Hence, v_x(t), v_y(t), and \theta(t) have the following relation: v_y(t)/[v_x(t)+V_x(t)] = tan(\theta(t)) for all o<t<\tau----------(4) It is noted that the horizontal component for the object's velocity in this rest frame becomes v_x(t) + V_x(t). Finally, at the moment t = \tau, the normal force becomes zero and only the gravity provides the centripetal force which gives the following equality: (v_y(\tau))^2 + [v_x(\tau) + V_x(\tau)]^2 = gRcos(\theta(\tau))-------(5). In the end, Combing Equs. (1) to (5) and applying some caculation, it arrives: [(1+m/M)tan^2(\theta(\tau))+1]cos^3(\theta(\tau)) = 2[1 - cos(\theta(\tau))](1+m/M) -----(6) Substitute the fact cos(\theta(\tau)) = 0.7 into Equ. (6), the ratio M/m can be solved uniquely and it is about 2.43 Q.E.D. --

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