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Chapter 10: Q4Q (page 410)

You know that a collision must be 鈥渆lastic鈥� if: (1) The colliding objects stick together. (2) The colliding objects are stretchy or squishy. (3) The sum of the final kinetic energies equals the sum of the initial kinetic energies. (4) There is no change in the internal energies of the objects (thermal energy, vibrational energy, etc.). (5) The momentum of the two-object system doesn鈥檛 change.

Short Answer

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3) The sum of the final kinetic energies equals the sum of the initial kinetic energies and 4) there is no change in the internal energies of the objects (thermal energy, vibrational energy, etc.) and 5) the momentum of the two-object system doesn鈥檛 change.

Step by step solution

01

Significance of the law of conservation of momentum and elastic collision of a system

This law states that the momentum of a particular system before and after collision is constant if no external force acts on the system.

The total momentum of a system is conserved in an elastic collision and the mechanical energy is also conserved.

02

Determination of the elasticity of a collision

From the law of conservation of momentum, the momentum of a system gets conserved. Hence, all the bodies should be included while defining a system of bodies. However, when the objects collide and also bounce back between them, a moment redistribution amongst the bodies occurs. Moreover, in the elastic collision, the initial and the final kinetic energy of an object does not change.

Thus, 3) the sum of the final kinetic energies equals the sum of the initial kinetic energies and 4) there is no change in the internal energies of the objects (thermal energy, vibrational energy, etc.) and 5) the momentum of the two-object system doesn鈥檛 change.

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Most popular questions from this chapter

A gold nucleus contains 197nucleons (79protons and 118neutrons) packed tightly against each other. A single nucleon (proton or neutron) has a radius of about1脳10-15m. Remember that the volume of a sphere is43蟺谤3. (a) Calculate the approximate radius of the gold nucleus. (b) Calculate the approximate radius of the alpha particle, which consists of 4nucleons, 2protons and 2neutrons. (c) What kinetic energy must alpha particles have in order to make contact with a gold nucleus?

Rutherford correctly predicted the angular distribution for 10 MeV(kinetic energy) alpha particles colliding with gold nuclei. He was lucky: if the alpha particle had been able to touch the gold nucleus, the strong interaction would have been involved and the angular distribution would have deviated from that predicted by Rutherford, which was based solely on electric interactions.

In a nuclear 铿乻sion reactor, each 铿乻sion of a uranium nucleus is accompanied by the emission of one or more high-speed neutrons, which travel through the surrounding material. If one of theneutrons is captured in another uranium nucleus, it can trigger 铿乻sion, which produces more fast neutrons, which could make possible a chain reaction

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鈼�), 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

In an elastic collision involving known masses and initial momenta, how many unknown quantities are there after the collision? How many equations are there? In a sticking collision involving known masses and initial momenta, how many unknown quantities are there after the collision? Explain how you can determine the amount of kinetic energy change.

In outer space a rock whose mass is 3kg and whose velocity was(3900,-2900,3000)m/sstruck a rock with mass 13kg and velocity(220,-260,300)m/s. After the collision, the 3kg rock鈥檚 velocity is(3500,-2300,3500)m/s. (a) What is the 铿乶al velocity of the 13kg rock? (b) What is the change in the internal energy of the rocks? (c) Which of the following statements about Q (transfer of energy into the system because of a temperature difference between system and surroundings) are correct? (1)Q鈮�0 because the duration of the collision was very short. (2)Q=Ethermal of the rocks. (3)Q鈮�0 because there are no signi铿乧ant objects in the surroundings. (4)Q=鈭�k of the rocks.
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