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Chapter 10: Q10Q (page 410)

What is it about analyzing collisions in the center-of-mass frame that simplifies the calculations?

Short Answer

Expert verified

The velocity of the system does not change while analyzing collisions in the center of the mass frame which also simplifies the calculation.

Step by step solution

01

Significance of the law of conservation of momentum of a system

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

The law of conservation of momentum is helpful for analyzing the collisions in the center-of-mass frame.

02

Analysing the collisions in the center-of-mass frame

From the law of the conservation of momentum, due to the occurrence of the collisions in the center of the mass frame, the momentum gets affected. On the other hand, the velocity of the system does not change if the system is a closed system. Moreover, the system moves by concentrating the masses between the system at a particular point. Hence, as the velocity of the system does not change, it helps in analyzing the collisions in the center of the mass frame which simplifies the calculation.

Thus, the velocity of the system does not change while analyzing collisions in the center of the mass frame which also simplifies the calculation.

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Most popular questions from this chapter

Object A has mass mA=7kgand initial momentumP→Aj=(17,-5,0)kg.m/s2, just before it strikes object B , which has mass mA=11kg. Object B has initial momentum p→Bj=(4,6,0)kh.m/s2. After the collision, object A is observed to have final momentum P→Af=(13,3,0)kg.m/s2. 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)?

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

A charged pion ( mpc2= 139.6 MeV) at rest decays into a muon (mμ= 105.7 MeV) and a neutrino (whose mass is very nearly zero). Find the kinetic energy of the muon and the kinetic energy of the neutrino, in MeV (1 MeV = 1×106eV).

In outer space, rock 1 whose mass is5Kg and whose velocity was(3300,-3100,3400)m/s struck rock 2, which was at rest. After the collision, rock 1’s velocity is(2800,-2400,3700)m/s . (a) What is the final momentum of rock 2? (b) Before the collision, what was the kinetic energy of rock 1? (c) Before the collision, what was the kinetic energy of rock 2? (d) After the collision, what is the kinetic energy of rock 1? (e) Suppose that the collision was elastic (that is, there was no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2? (f) On the other hand, suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where∆Ethermal,1+∆Ethermal,2=7.16×106J . What is the final kinetic energy of rock 2? (g) In this case (some of the kinetic energy being converted to thermal energy), what was the transfer of energy Q (microscopic work) from the surroundings into the two-rock system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

In a reference frame where a ∆+ particle is at rest it decays into a proton and a high-energy photon (a “gamma ray”): ∆+→P++γ+. The mass of the ∆+ particle is 1232 MeV/c2and the mass of the proton is 938 MeV/c2(1 MeV = 106eV). Calculate the energy of the gamma ray and the speed of the proton.
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