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Chapter 10: Q9Q (page 410)

Give an example of what we can learn about matter through the use of momentum and energy conservation applied to scattering experiments. Explain what it is that we cannot learn this way, for which we need to measure the distribution of scattering angles.

Short Answer

Expert verified

A matter can be analyzed by scattering experiments like the structure of atoms

Step by step solution

01

Given information

The momentum and energy conservation applied to scattering experiments.

02

The concept of conservation of momentum and energy

The law of conservation of momentum asserts that if a system of bodies has no net external forces acting on it, the total momentum remains constant at all times (it is conserved). If all forces, external or internal, can be given a potential, then the total energy remains constant as well.

03

The merits and demerits for what we can learn about matter through the use of momentum and energy conservation applied to scattering experiment

Scattering tests can be used to identify subatomic particles and disclose the structure and behavior of atoms, nuclei, and other tiny particles in matter. (For example, Rutherford identified the nucleus by measuring alpha particle scattering against a very thin gold foil, using the momentum conservation principle).

Scattering angle distributions provide a broader view of the situation (for example, the relative sizes, the incident flux and angular size)

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Most popular questions from this chapter

Object A has mass mA=7kgand initial momentumP鈫�Aj=(17,-5,0)kg.m/s2, just before it strikes object B , which has mass mA=11kg. Object B has initial momentum p鈫�Bj=(4,6,0)kh.m/s2. After the collision, object A is observed to have 铿乶al momentum P鈫�Af=(13,3,0)kg.m/s2. In the following questions, 鈥渋nitial鈥� refers to values before the collisions, and 鈥滐瑏nal鈥� 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 铿乶al momentum of object B. (c) The initial kinetic energy of object A. (d) The initial kinetic energy of object B. (e) The 铿乶al kinetic energy of object A. (f) The 铿乶al kinetic energy of object B. (g) The total initial kinetic energy of the system. (h) The total 铿乶al kinetic energy of the system. (i) The increase of internal energy of the two objects. (j) What assumption did you make about Q (energy 铿俹w from surroundings into the system due to a temperature difference)?

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.

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.

A uranium atom traveling at speed4脳104m/scollides elastically with a stationary hydrogen molecule, head-on. What is the approximate final speed of the hydrogen molecule?

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