※ 引述《dinks (丁克思)》之铭言： : 题目及我解到後面卡住的内容如下。 : 本题正解是(D)，但可怜的我算不出来 QQ : 【出处】普物 : 【题目】 : https://i.imgur.com/GF4PuRQ.jpg : 【瓶颈】下面是我的解法，算到後面发现M的动能居然是负的，我就知道我一定有什麽观 : 念是错的，所以整个卡死。还望先进指点，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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