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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

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

An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10 MeV (10 × 106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Assuming that all speeds are small compared to the speed of light, answer the following questions about the collision. (a) What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? (b) What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? (c) What is the final kinetic energy of the alpha particle? (d) What is the final kinetic energy of the gold nucleus? (e) Assuming that the movement of the gold nucleus is negligible, calculate how close the alpha particle will get to the gold nucleus in this head-on collision.

A projectile of mass $m_{1}$ moving with speed $v_{1}$ in the +xdirection strikes a stationary target of mass $m_{2}$ head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. After obtaining your results, see what your equations would predict if $m_{1} ≫ m_{2}$ , or if $m_{2} ≫ m_{1}$ . Verify that these predictions are in agreement with the analysis in this chapter of the Ping-Pong ball hitting the bowling ball, and of the bowling ball hitting the Ping-Pong ball.

Consider a head-on collision between two objects. Object 1, which has mass $m_{1}$ , is initially in motion, and collides head-on with object 2, which has mass $m_{2}$ and is initially at rest. Which of the following statements about the collision are true?

(1) ${\vec{p}}_{1,initial} = {\vec{p}}_{1, final} + {\vec{p}}_{2, final}$ .

(2) $| {\vec{p}}_{1, final} | < | {\vec{p}}_{1, initial} |$ .

(3) If $m_{2} ≫ m_{1}$ , then $| Δ {\vec{p}}_{1} | > | Δ {\vec{p}}_{2} |$ .

(4) If $m_{1} ≫ m_{2}$ , then the final speed of object 2 is less than the initial speed of object 1.

(5) If $m_{2} ≫ m_{1}$ , then the final speed of object 1 is greater than the final speed of object 2.

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

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