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Solutions of Exam# 1 Q1 (a) A particle of mass m moves in a Yukawa potential k V r e r / a ; where k , a are consts. r Find the equation of a bound orbit of the particle (having an orbital angular momentum ( ℓ ) about the center of the potential) to the first order of r/a. (b) A particle of mass m moves in a central force field that has a constant magnitude F0 but always points toward the origin. (i) Find the angular velocity ω φ required for the particle to move in a circular orbit of radius ro . (ii) Find the frequency of ω r of small radial oscillations about the circular orbit. Both answers should be in terms of m, r0, F0. (a) The bound motion means that E m v2 k V 0 where V e r a . 2 r The orbit of particle moving in this central force potential is given by r / r 2 dr r rmin 2 2 E V 2 2 r r 2 rmin 1 r2 dr E 2 ke r / a r 2 r 2 In first order of (r/ a), that is (e-r/a ~ 1-r/a) (r) 2 dr r2 E 2 k k 2 r 2r a 2 dr 2 k k r2 E a r 2r2 k Now effectively, this is the orbit of particle of total energy E moving in potential a k . It is well known that this orbit is given by (see Chapter 8 Eqs 8.38, 8.40, 8.41) r r Where 2 k and 1 22 k2 1 cos k E a If 0 1, the orbit is ellipsoid; if 0 , the orbit is circular. (b) (i) In equilibrium, for a circular orbit of radius r0, F0 m 2r0 (ii) F0 m r0 The angular momentum (which is conserved) of a particle in circular orbit is L m r02 mF0 r03 The force acting on a particle, which is placed a distance r (r is very close to equilibrium position ro ) from the center of force is F m 2r F0 L3 F0 m r3 L3 3L2 3L2 r r F 0 0 r r0 k r r0 m r03 m r02 m r04 where k 3L2 /m r04 . So the frequency of oscillation is r 3F0 k 3L2 2 4 m m r0 m r0 Q2 For the elastic scattering of T0= 5.0 MeV α-particles ( 24 He ) from gold ( 197 79 Au )(initially at rest) at a scattering angle in the LAB of ψ = 90◦. Find the following: (i) The LAB kinetic energies of the scattered α-particle and gold. (ii) The recoil scattered angle of gold ζ. (iii) The scattering angles of α-particle (θ1) and gold (θ2) in the CM system. (iv) The impact parameter b and the distance of closest approach between the particles. (v) The differential cross section (LAB) of scattering at ψ = 90◦. (vi) The ratio of the probabilities of scattering at ψ = 90◦ to that at ψ = 30◦. (i) m 1 mass of ( 24 He) particle, m 2 mass of gold ( 197 79 Au ) u1 ,u1 : velocity of particle in LAB and CM before collision v1 ,v1 : velocity of particle in LAB and CM after collision u2 ,u2 : velocity of 197 79 Au in LAB and CM before collision v2 ,v2 : velocity of 197 79 Au in LAB and CM after collision u2 0 , T0 m1u 12 5.0 MeV 2 90 is angle through which particle is deflected in LAB 1 and 2 are the angles through which particle and is recoil angle of 197 79 197 79 Au are deflected in CM Au in LAB (i) Conservation of momentum in LAB: m 1 u1 m 2 v2 cos v1 u1 tan m 1 v1 m 2 v2 sin Conservation of energy in LAB: (1) m 1u12 m 1v12 m 2v22 2 2 2 m m1 2 v 12 2 u1 m 2 m1 Thus, and (2) 2 2m12 v 22 u 1 (3) m m m 2 1 2 The kinetic energy of particle after collision in LAB is T1 The kinetic energy of particle 197 79 m1v 12 m 2 m1 T 0 4.80 MeV 2 m1 m 2 Au after collision in LAB is T2 m 2v 22 2m1 T 0 0.20 MeV 2 m1 m 2 Evidently, conservation of energy is satisfied. (ii) From Eq (1) , (2) and (3) we obtain the recoil scattering angle of 197 79 Au m m1 tan 2 0.9799 44.42 m 2 m1 (iii) The velocity of CM of system is vCM The velocity of 197 79 m 1u1 m1 m2 Au in CM after collision is v2' v2 vCM . From the above figure we can obtain the scattering angle of particle tan 2 197 79 Au in CM to be (use Eqs 1, 3) v 2 sin m 22 m12 49.24 2 88.84 v 2 cos v cm m12 In CM, clearly after collision, particle moves in opposite direction of that of 197 79 Au i.e. θ1 =180 – 88.84 = 91.16◦. (iv) The impact parameter in CM is given by Eq (9.139). b where k T 0 CM, k cot 1 2T 0 2 q1q 2 z 1z 2e 2 2 79 1.44 MeV .fm 227.52 MeV .fm and 4 0 4 0 m 1m 2 2 1 m1u12 m 2u 22 u1 4.90 MeV 2 2 m1 m 2 is the total energy of system in b so k cot 1 2.28 1014 m 2T '0 2 We note that b is the impact parameter of particle with respect to CM, so the impact m1 m 2 b 2.32 1014 m. parameter of particle with respect to 197 79 Au is m2 To find the closest distance from particle to the center of mass: In CM system, the orbit equation of particle is r 1 cos rm in 2 m 1k 1 1 2 and 1 rmin 1 2 where 0 corresponds to r rm in is closest distance from particle to the center of mass, and m1u1b 2 m 1k 2T 0'b 2 2.24 1014 m k m u b E 2 1 2T 0' 1 1 m 1k m 1k 2 4T 0'2b 2 1 k 1.12 1014 m But the actual minimum distance between particles is rmin (v) m1 m 2 1.14 1014 m. rmin m2 Using formula LAB CM where x m1 , 90 , m2 CM 1 x cos 1 k2 4T 0 2 1 x 2 sin 2 1 x 2 sin 2 1 5.18 1028 m 2 ; 4 1 sin 2 2 LAB CM 1 1 x 2 CM 1 m 1 1 0.9998 CM 1 m2 where T0’ = 4.90 MeV; θ1= 91.16◦, k = 227.52 MeV.fm We find this differential cross section in LAB at = 90°: 2 LAB 90 5.18 1028 m2 (vi) Since the probability of scattering at an angle ψ = that the ratio of probability is LAB sin LAB ' sin For ψ’=30◦ the corresponding θ’ is related by tan ' LAB ' CM where x x cos ' ' 1 m1 , ' 30 , m2 1 x 2 sin 2 ' dN LAB ( ) sin d d we see N sin 1 ' 1 ' 30.58 m cos 1 ' 1 m2 2 1 x 2 sin 2 ' CM '1 k2 4T 0 2 1 1028 m 2 ; 4 '1 sin 2 Q3 (a) Find the center of mass of a uniformly solid cone of base diameter (2a) and height (h) and a solid hemisphere (of the same material as the cone)of radius (a) where the two basis are touching.. (b) A 104 kg spherical probe of radius R = 20 cm is launched vertically upward from the surface of Earth with an initial speed 6000 m/s. If the air resistance F = -(1/2) cwρAv2; where the constant cw = 0.2, ρ = air density ( =1.3 kg/m3 ), A = πR2 and v is the speed of the probe, determine the maximum height reached and the time taken to reach this height. Solve the problem taking g = constant first then solve it again taking g = G M /r2; where M is the mass of Earth. (a) For the hemisphere: Put the shell in the z > 0 region, with the base in the x-y plane. By symmetry, x y 0 . 2 2 z 0 2 0 r2 0 r r1 2 r2 zr2 drsin d d 0 r r1 r2 drsin d d Using z = r cos and doing the integrals gives z For the cone: By symmetry, x y 0 . 8 r r 3 r24 r14 3 2 3 1 Use cylindrical coordinates , , z. 0 m ass density h h a 0 0 0 z 0 h 2 a h 4 0 0 0 z 0 2 z a z d d dz d d dz h 4 The center of mass is on the axis 3 of the cone h from the vertex. 4 For the compound problem: (see Fig) By symmetry, x y 0 . The center of mass of the cone is at z 1 h. 4 3 The center of mass of the hemisphere is at z a r2 a,r1 0 8 So the problem reduces to z1 1 1 h; m1 1 a 2 h 4 3 3 2 z2 a; m2 2 a3 8 3 z m 1z1 m 2z2 1h2 32a2 m1 m2 4 1h 22a for 1 2 z h2 3a3 4 2a h (b1) When we add the expression for air resistance, the differential equation that describes the projectile’s ascent is dv 1 v Fm m g cW A v2 m g 1 dt 2 vt 2 (1) where vt 2m g cW A 25 km s1 would be the terminal velocity if the object were falling from a sufficient height (using 13 kg m differential equation gives 3 as the density of air). Solution of this v0 gt vt vt v t vt tan tan 1 This gives v = 0 at time vt g tan1 v0 vt (2) 300 s. The velocity can in turn be integrated to give the y-coordinate of the projectile on the ascent. The height it reaches is the y-coordinate at time : h vt2 v ln 1 0 2g vt 2 (3) which is 600 km. (b2) G M Changing the acceleration due to gravity from –g to e R e y 2 g R e R e y changes our differential equation for y to 2 y 2 R 2 e y g vt R e y (5) Q4 A pojectile is fired due East from a point on the surface of Earth at a northern latitude λ with an initial velocity V0 and at an angle α to the horizontal. Take South to be x-axis and East to be y-axis and z-axis to be vertically upward from the surface of Earth. (a) Assume that the z motion to be unaffected by the Coriolis force, show that the side deflection when the projectile strikes Earth is 4V03 x T sin cos sin 2 2 g (b) Now take the effect of the Coriolis force in the z-motion find the range of the projectile (Range = y at z=0). (a) The coordinates x, y, z as in the diagram: South is x-axis and East is y-axis, z is vertically upward. Then, the velocity of the particle and the rotation frequency of the Earth are expressed as v 0, y, z x ,0, z cos , 0, sin (1) so that the acceleration due to the Coriolis force is a 2 r 2 yz , zx , yx (2) Since the motion is in y-z plane, then we see that the lateral (side) deflection of the projectile is in the x direction and that the acceleration is ax x 2 z y 2 sin V0 cos (1) Integrating this expression twice and using the initial conditions, x 0 0 and x 0 0 , we obtain xt V0t2 cos sin Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In this approximation, we have (2) zt V0tsin 1 2 gt 2 (3) from which the time T of impact is obtained by setting z = 0: T 2V0 sin g (4) Substituting this value for T into (2), we find the lateral deflection at impact to be x T 4V03 sin cos sin2 g2 (5) (b) In the previous part we assumed the z motion to be unaffected by the Coriolis force. Recall that the Coriolis acceleration is: a 2 r 2 yz , zx , yx With v 0, y, z 0,V0 cos ,V0 sin gt x ,0, z cos , 0, sin The upward acceleration is given by 2 x y so that z 2 V0 cos cos g (6) from which the time of flight is obtained by integrating twice, using the initial conditions, and then setting z = 0: T 2V0 sin g 2V0 cos cos (7) Now, the acceleration in the y direction is a y y 2x z 2 cos V0 sin gt (8) Integrating twice and using the initial conditions, y 0 V0 cos and y 0 0 , we have yt 1 gt3 cos V0t2 cos sin V0tcos 3 (9) Substituting (7) into (9), the range R is R 4V 03 sin3 cos 2V 02 cos cos 8 V 03 g sin3 cos 3 g 2V 0 cos cos 3 g 2V 0 cos cos 2 g 2V 0 cos cos (10) We can stop here. However, if we want to compare R’ with the range for the case ω=0, we have the following: We now expand each of these three terms, retaining quantities up to order but neglecting all quantities proportional to 2 and higher powers of . In the first two terms, this amounts to neglecting 2V0 cos cos compared to g in the denominator. But in the third term we must use 2V 02 cos sin 2V 0 g 1 cos cos g 2V 02 cos sin g R 0 2V 0 cos cos 1 g 4V 03 sin cos2 cos g2 (11) where R0 is the range when Coriolis effects are neglected [see Example 2.7]: R 0 2V 02 cos sin g (12) The range difference, R R R0 , now becomes R 4V03 1 cos sin cos2 sin3 2 g 3 (13) Substituting for V 0 in terms of R0 from (12), we have, finally, R 2R 0 1 cos cot1 2 tan3 2 g 3 (14)