Solution - Career Point

Assuming the two-particle system reaches aminimum of kinetic energy at a certain moment andat a certain finite distance between the particles (inthe laboratory frame), find the distance between theparticles at that moment and the value of thatminimal kinetic energy.Sol. For the sake of convenience, we will first solve theproblem in the frame of the centre of mass. Then, wewill transform the results into the laboratory frame.We will determine the velocity of the center of massby :v r mv0 2mvcm = − 0 1xˆ = – v0 xˆ ...(1)2m 2This velocity remains constant since there are noexternal forces acting on the whole system. In thevelocity of the centre of mass, the velocities of theparticles are :⎧ r r r 3⎪ u1= v1− vcm= v0xˆ⎨2...(2)r r r 3⎪u2 = v2− vcm= − v0xˆ⎩2By definition of a centre of mass frame, the totalmomentum of the particles is zero. The kineticenergy in this system at t = 0 is :K´ = 21 mu1 2 + 21 mu2 2 = 49 mv02...(3)Therefore, the total energy in the center of massframe at t = 0 isE´ = K´ + U´ = 49 mv0 2 – Rα...(4)where we define R ≡ r(t = 0). Note that since theforce is conservative, we have F r = – ∇ u. The scalarfunction is u = – α/r.The advantage of using the center of mass frame isevident when one inspects the moment of arrival at aminimal distance, t 0 . At that moment, in this system,the two particles stop and reverse their directions.The kinetic energy, therefore, vanishes at t 0 in thecenter of mass frame, or,'K´(t 0 ) = K min = 0 ...(5)αHence, E´(t 0 ) = – = E´ ...(6)R minPlugging in the value of E´, we find :4αRR max =...(7)24α − 9mv0RWe now transform the centre of mass frame to thelaboratory frame. Since R max is the relative distancebetween the two particles, it is unchanged by thetransformation. Recall that distance is an invariantquantity of the transformations of displacement and /or rotation. Therefore,α1K min – = E(0) = E = 2 1 mv0 + m(2v0 ) 2 α– 2 2 RR max....(8)The kinetic energy in the laboratory frame is,therefore :5K min = 2 α 4α − 9mv Rmv0 – +0α2 R 4αR⎛ 5 9 ⎞= ⎜ − ⎟ mv 2 1 20 = mv0 ...(9)⎝ 2 4 ⎠ 4Another way of finding the minimal kinetic energy isby using the following formula :1 2K = K´ + Mv cm...(10)2where K´ is the kinetic energy in the centre of massframe, K is the energy in the laboratory frame, and Mis the total mass of the system. In our case,12 1 2K = 0 + (2m)Mv cm = Mv 0 ...(11)24Note : Generally, in transforming from system S tosystem S´ with relative velocity V r , the kineticenergy is transformed as :p 2 2M ⎛ rrp ⎞K´ = K – + ⎜V− ⎟ ..(12)2M 2 ⎝ M ⎠where M is the total mass and p r is the totalmomentum in S. K´ is minimal in the center of massframe if we choose V r =K´ = K =2v r p rcm = . We obtain :Mp 2 ...(13)2M4. A hot body is being cooled in air according toNewton's law of cooling, the rate of fall oftemperature being K times the difference of itstemperature with respect to that of surroundings.Calculate the time after which the body will lose halfthe maximum heat it can lose. The time is to becounted from the instant t = 0.Sol. According to Newton's law of cooling, we havedθ = – K(θ – θ0 )dtwhere θ 0 is the temperature of the surrounding and θis the temperature of the body at time t. Supposeθ = θ 1 at time t = 0.Then,∫ θ dθ= –Kθ1 θ − θ ∫ t θ − θ0dt or, log = – Kt00 θ1− θ0or, θ – θ 0 = (θ 1 – θ 0 )e –Kt ...(1)The body continues to lose heat till its temperaturebecomes equal to that of the surroundings. The lossof heat in this entire period is dQ m = ms(θ 1 – θ 0 ).This is the maximum heat the body can lose. If thebody loses half this heat, the decreases in itstemperature will bedQ m θ1− θ= 02ms 2XtraEdge for IIT-JEE 21 APRIL 2010