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Chapter 10: Q8Q (page 410)
What properties of the alpha particle and the gold nucleus in the original Rutherford experiment were responsible for the collisions being elastic collisions?
The property that both the nuclei remained in their ground states
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Object:has mass and initial momentum, just before it strikes object , which has mass . Object has initial momentum. After the collision, object is observed to have final momentum . In the following questions, “initial” refers to values before the collisions, and “final” refers to values after the collision. Consider a system consisting of both objects and. Calculate the following quantities: (a) The total initial momentum of this system. (b) The final momentum of object B. (c) The initial kinetic energy of object A. (d) The initial kinetic energy of object B. (e) The final kinetic energy of object A. (f) The final kinetic energy of object B. (g) The total initial kinetic energy of the system. (h) The total final kinetic energy of the system. (i) The increase of internal energy of the two objects. (j) What assumption did you make about Q (energy flow from surroundings into the system due to a temperature difference)?
In a reference frame where a ∆+ particle is at rest it decays into a proton and a high-energy photon (a “gamma ray”): . The mass of the ∆+ particle is 1232 and the mass of the proton is 938 (1 MeV = eV). Calculate the energy of the gamma ray and the speed of the proton.
A Fe-57 nucleus is at rest and in its first excited state, 14.4 keV above the ground state (14.4 × 103 eV, where 1 eV = 1.6×10−19 J). The nucleus then decays to the ground state with the emission of a gamma ray (a high-energy photon). (a) Wthe recoil speed of the nucleus? (b) Calculate the slight difference in eV between the gamma-ray energy and the 14.4 keV difference between the initial and final nuclear states. (c) The “Mössbauer effect” is the name given to a related phenomenon discovered by Rudolf Mössbauer in 1957, for which he received the 1961 Nobel Prize for physics. If the Fe-57 nucleus is in a solid block of iron, occasionally when the nucleus emits a gamma ray the entire solid recoils as one object. This can happen due to the fact that neighbouring atoms and nuclei are connected by the electric interatomic force. In this case, repeat the calculation of part (a) and compare with your previous result. Explain briefly
Redo the analysis of the Rutherford experiment, this time using the concept of the centre-of-momentum reference frame. Let m = the mass of the alpha particle and M = the mass of the gold nucleus. Consider the specific case of the alpha particle rebounding straight back. The incoming alpha particle has a momentum p1 , the outgoing alpha particle has a momentum p3 , and the gold nucleus picks up a momentum p4 . (a) Determine the velocity of the centre of momentum of the system. (b) Transform the initial momenta to that frame (by subtracting the centre-of-momentum velocity from the original velocities). (c) Show that if the momenta in the centre-of-momentum frame simply turn around (180◦), with no change in their magnitudes, both momentum and energy conservation are satisfied, whereas no other possibilitysatisfies both conservation principles. (Try drawing some other
momentum diagrams.) (d) After the collision, transform back to
the original reference frame (by adding the center-of-momentum
velocity to the velocities of the particles in the center-of-mass
frame). Although using the center-of-momentum frame may be
conceptually more difficult, the algebra for solving for the final
speeds is much simpler
A spring has an unstretched length of 0.32 m. a block with mass 0.2 kg is hung at rest from the spring, and the spring becomes 0.4 mlong.Next the spring is stretched to a length of 0.43 mand the block is released from rest. Air resistance is negligible.
(a) How long does it take for the block to return to where it was released? (b) Next the block is again positioned at rest, hanging from the spring (0.4 m long) as shown in Figure 10.43. A bullet of mass 0.003 kg traveling at a speed of 200 m/s straight upward buries itself in the block, which then reaches a maximum height above its original position. What is the speed of the block immediately after the bullet hits? (c) Now write an equation that could be used to determine how high the block goes after being hit by the bullet (a height h), but you need not actually solve for h.
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