Part A-type questions

Transcription

1 PHYS306: lecture 8 th February 008 Part A-type questions. You toss an apple horizontally at 8.7 m/s from a height of.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?. Consider an object of mass m moving at speed v at the surface of a gravitating body of mass M and radius r. If its gravitational potential energy is given by U = GMm/r, calculate the escape speed needed for it to escape from the gravitating body. 3. A force given by F = a x acts in the x direction, where a = 9.5 N/m /. Calculate the work done by this force acting on an object as it moves: (a) from x = 0 to x = 3 m; (b) from 3 m to 6 m. 4. Two mass-spring systems A and B oscillate so that their energies are equal. If their respective spring constants are related by k A = k B, how are the amplitudes of oscillation related? 5. A rod 0 m long moves with a velocity of 0.8c parallel and close to a window 5 m wide. A rod grabber is at rest in the inertial system of the window. Is the rod moving sufficiently fast that the grabber could grab the two ends of the rod and pull it through the window? And their answers. The vertical component of both motions is due to gravity. Therefore the equations along that direction are the same, and the objects take the same time to reach the ground.. The escape speed will be the one necessary to escape the attraction of the gravitating body. So we need to make the gravitational potential energy equal to the object s kinetic energy. This results in the equation GMm r = mv.

2 From this we conclude that v = GM/r. 3. The expression of the work done as a function of the force (one dimensional in this case) is W i f = xf x i F dx. (a) In the first case the work done by the force is W 0 3 = 3 0 a xdx = a 3 x3/ 3 0 = a 3 33/ m 3/ = N m / 33/ m 3/ = 3.9 J. (b) In the second one we have W 3 6 = 6 3 a xdx = a 3 x3/ 6 3 = a 3 (63/ 3 3/ )m 3/ = N m / 9.5m3/ = 60.7 J. Therefore there is more work done in the second case. 4. If we call the amplitudes of oscillation A and B respectively, then the associated energies are k BB [], k AA [] = k B A. And equating both energies, we obtain B = A. 5. The proper length of the rod is L p = 0 m. The contracted length, when the rod moves at 0.8c, is given by L = L p /γ [], where γ = / v /c. Therefore L = L p v /c = L p 0.8 = 0 m 0.6 = m. Therefore the rod grabber will be able to pull it through the window, which is 5 m wide. Longer questions on Relativity. A spaceship goes by at half the speed of light and you determine that it is 35 m long. What is its length as measured in its rest frame?

3 We are being asked about the proper length of the spaceship. According to the formula for length contraction the length L measured in any other frame but the proper one is related to L p, the proper length via where /γ = L = L p γ, v /c. Therefore L p = γl and we know that L = 35 m and v = 0.5c. Therefore γ = / 0.5 = / 3, and L p = L/ 3 = 40.4 m.. Find (a) the speed and (b) the momentum of a proton whose kinetic energy is 500 MeV. (a) The relativistic kinetic energy is given by K = mc. v c mc = mc v c We can rearrange the previous expression to obtain = + K v mc, c which we can invert and square to obtain the speed v c = ( + K mc ) v c = ( ). + K mc The proton mass is m = MeV/c, therefore v/c = /( + 500/938.73) = (b) The relativistic momentum is given by p = γmv. We have all ingredients, just have to calculate γ = / v /c = / =.533. Then p = c kg = kg.m/s 3

4 3. When the speed of an object increases by 5%, its momentum goes up by a factor of 5. What was the original speed? We have an object of mass m with relativistic momentum p = γmv. If we increase its speed by 5%, then the new speed is.05v. At this speed the momentum is 5p, with p the original momentum. Then mv 5p = 5 = m(.05v) v c (.05v) c. We cancel factors of mv in both numerators and, rearranging, we end up with 5.05v = v.05 c c. Now we square both sides of the equation and group factors of v /c together, ( ) 5 ) (.05v.05 c ( ) = v 5 ( ( ) 5 v c =.05 ) c which leads to.6757 = 4 v c v c = Electrons in a particle accelerator reach a speed of 0.999c relative to the laboratory. One collision of an electron with a target produces a muon that moves forward with a speed of 0.95c relative to the laboratory. The muon mass is kg. What is the muon s momentum in the laboratory frame and in the frame of the electron beam? Let the laboratory be the reference frame S. The reference frame S of the electron beam (i.e. a reference frame in which the electrons are at rest) moves in the direction of the electrons at v = 0.999c. The muon velocity in frame S is u = 0.95c. For the muon in the laboratory reference frame we have γ = u = /c 0.95 = 3.. Thus its momentum in that frame is p = γmu = (3.)( kg)( m/s) = kg.m/s. 4

5 The momentum is a factor of 3. larger than the Newtonian momentum mu. To find the momentum in the electron-beam reference frame we must use the velocity transformation equation to find the muon s velocity in frame S : u = u v uv/c = 0.95c 0.999c (0.95c)(0.999c)/c = 0.96c. In the laboratory frame the faster electrons are overtaking the slower muon. Hence the muon s velocity in the electron-beam frame is negative. γ for the muon in frame S is γ = u /c = = The muon s momentum in the electron-beam reference frame is p = γ mu = (3.66)( kg)( m/s) = kg.m/s. From the laboratory perspective the muon moves only slightly slower than the electron beam. But it turns out that the muon moves faster with respect to the electrons, although in the opposite direction, than it does with respect to the laboratory. 5. Calculate the rest energy and the kinetic energy of (a) a 00 g ball moving with speed of 00 m/s and (b) an electron with a speed of 0.999c. The ball, with u c, is a classical particle. We don t need to use the relativistic expression for its kinetic energy. The electron is highly relativistic. (a) For the ball, with m = 0. kg, E 0 = mc = J, K = mu = 500 J. (b) For the electron, we start by calculating γ = u /c =.4. Then, using m e = kg, we find E 0 = mc = J, 5

6 K = (γ )E 0 = J. The ball s kinetic energy is a typical kinetic energy. Its rest energy, by contrast, is a staggeringly large number. For a relativistic electron, on the other hand, the kinetic energy is more important than the rest energy. 6. It is often observed in proton-proton (p p) scattering at relativistic energies that more particles come out than went in: particles called pions (π +, π, π 0 ) can be created. Possible scenarios include p + p p + p + π 0. The neutral pion mass is 35 MeV, where we follow standard high energy practice in calling mc the mass, since this is the energy equivalent and, hence, the energy which, on creation of the particle in a collision, is taken from kinetic energy and stored in mass. (a) Figure out how much energy the incoming proton needs to create a neutral pion and calculate he corresponding incoming velocity. (Hint: consider the problem in the centre of mass frame and assume that the least possible energy is that necessary to produce the final particles at rest). (b) Compute these quantities in the lab frame. (a) In the centre of mass frame both incoming protons have equal and opposite velocities (and, therefore, momenta), i.e. there is no total momentum. Denoting by m the relativistic mass of the incoming protons in this frame we have that, by conservation of energy, E = mc = m p c + m π c, where m p and m π and the proton and pion rest masses respectively. By using m = m p / v /c we find from the previous equation that v = 0.36c. (b) Normally we would like, for practical purposes, to know these quantities in the lab frame, where one of the protons is initially at rest. Then the velocity of the moving proton would result of the relativistic addition of its centre of mass velocity, 0.36c to the relative velocity between frames, another 0.36c. This results in u = u v 0.36c c = = 0.64c. uv/c

7 Therefore γ = / 0.64 =.3, i.e. the incoming proton has a relativistic mass.3 times larger than its rest mass, and thus a kinetic energy of K = (γ )m p c = MeV 80 MeV. Thus to create a pion of of rest energy 35 MeV it is necessary to give the incoming proton at least 90 MeV of kinetic energy, which is the threshold energy for pion production (this is slightly higher than the result quoted above because there is a certain inefficiency due to the final particles emerging with some kinetic energy). 7