Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 42, Problem 14P
To determine
The speed of the atoms moving before collision.
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Before attempting any two-dimensional relativistic collisions, do this classical (i.e., non-
relativistic) two-dimensional collision problem:
Two balls collide elastically. Ball 1 has a mass of 0.70 kg and an initial velocity of 10 m/s
in the +x-direction. Ball 2 has a mass of 0.90 kg and is initially stationary. After the
collision, ball 1 is moving at a 30° angle above the horizontal. Find the speeds of each
ball and the direction of ball 2 after the collision.
An initially stationary rocket explodes in empty space. Out of the debris two pieces are recovered: m1=10kg traveling with speed of 500m/s and a second mass m2=20kg, traveling with speed 1500m/s.
What is the minimum amount of energy released in the explosion?
What is the minimum amount of energy released in the explosion?
Group of answer choices
K≥2.250×107J
K≥3.500×104J
K≥2.375×107J
K≥1.250×106J
A linear particle accelerator using beta particles collides electrons with their anti-matter counterparts, positrons. The accelerated electron hits the stationary positron with a velocity of 29 x 106 m/s, causing the two particles to annihilate.If two gamma photons are created as a result, calculate the energy of each of these two photons, giving your answer in MeV (mega electron volts), accurate to 1 decimal place. Take the mass of the electron to be 5.486 x 10-4 u, or 9.109 x 10-31 kg.
Chapter 42 Solutions
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Ch. 42.3 - Prob. 42.1QQCh. 42.3 - Prob. 42.2QQCh. 42.4 - Prob. 42.3QQCh. 42.4 - Prob. 42.4QQCh. 42.8 - Prob. 42.5QQCh. 42 - Prob. 1OQCh. 42 - Prob. 2OQCh. 42 - Prob. 3OQCh. 42 - Prob. 4OQCh. 42 - Prob. 5OQ
Ch. 42 - Prob. 6OQCh. 42 - Prob. 7OQCh. 42 - Prob. 8OQCh. 42 - Prob. 9OQCh. 42 - Prob. 10OQCh. 42 - Prob. 11OQCh. 42 - Prob. 12OQCh. 42 - Prob. 13OQCh. 42 - Prob. 14OQCh. 42 - Prob. 15OQCh. 42 - Prob. 1CQCh. 42 - Prob. 2CQCh. 42 - Prob. 3CQCh. 42 - Prob. 4CQCh. 42 - Prob. 5CQCh. 42 - Prob. 6CQCh. 42 - Prob. 7CQCh. 42 - Prob. 8CQCh. 42 - Prob. 9CQCh. 42 - Prob. 10CQCh. 42 - Prob. 11CQCh. 42 - Prob. 12CQCh. 42 - Prob. 1PCh. 42 - Prob. 2PCh. 42 - Prob. 3PCh. 42 - Prob. 4PCh. 42 - Prob. 5PCh. 42 - Prob. 6PCh. 42 - Prob. 7PCh. 42 - Prob. 8PCh. 42 - Prob. 9PCh. 42 - Prob. 10PCh. 42 - Prob. 11PCh. 42 - Prob. 12PCh. 42 - Prob. 13PCh. 42 - Prob. 14PCh. 42 - Prob. 15PCh. 42 - Prob. 16PCh. 42 - Prob. 17PCh. 42 - Prob. 18PCh. 42 - Prob. 19PCh. 42 - Prob. 20PCh. 42 - Prob. 21PCh. 42 - Prob. 23PCh. 42 - Prob. 24PCh. 42 - Prob. 25PCh. 42 - Prob. 26PCh. 42 - Prob. 27PCh. 42 - Prob. 28PCh. 42 - Prob. 29PCh. 42 - Prob. 30PCh. 42 - Prob. 31PCh. 42 - Prob. 32PCh. 42 - Prob. 33PCh. 42 - Prob. 34PCh. 42 - Prob. 35PCh. 42 - Prob. 36PCh. 42 - Prob. 37PCh. 42 - Prob. 38PCh. 42 - Prob. 39PCh. 42 - Prob. 40PCh. 42 - Prob. 41PCh. 42 - Prob. 43PCh. 42 - Prob. 44PCh. 42 - Prob. 45PCh. 42 - Prob. 46PCh. 42 - Prob. 47PCh. 42 - Prob. 48PCh. 42 - Prob. 49PCh. 42 - Prob. 50PCh. 42 - Prob. 51PCh. 42 - Prob. 52PCh. 42 - Prob. 53PCh. 42 - Prob. 54PCh. 42 - Prob. 55PCh. 42 - Prob. 56PCh. 42 - Prob. 57PCh. 42 - Prob. 58PCh. 42 - Prob. 59PCh. 42 - Prob. 60PCh. 42 - Prob. 61PCh. 42 - Prob. 62PCh. 42 - Prob. 63PCh. 42 - Prob. 64PCh. 42 - Prob. 65APCh. 42 - Prob. 66APCh. 42 - Prob. 67APCh. 42 - Prob. 68APCh. 42 - Prob. 69APCh. 42 - Prob. 70APCh. 42 - Prob. 71APCh. 42 - Prob. 72APCh. 42 - Prob. 73APCh. 42 - Prob. 74APCh. 42 - Prob. 75APCh. 42 - Prob. 76APCh. 42 - Prob. 77APCh. 42 - Prob. 78APCh. 42 - Prob. 79APCh. 42 - Prob. 80APCh. 42 - Prob. 81APCh. 42 - Prob. 82APCh. 42 - Prob. 83APCh. 42 - Prob. 84APCh. 42 - Prob. 85APCh. 42 - Prob. 86APCh. 42 - Prob. 87APCh. 42 - Prob. 88APCh. 42 - Prob. 89CPCh. 42 - Prob. 90CPCh. 42 - Prob. 91CP
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- An observer in a rocket moves toward a mirror at speed v relative to the reference frame labeled by S in Figure P1.30. The mirror is stationary with respect to S. A light pulse emitted by the rocket travels toward the mirror and is reflected back to the rocket. The front of the rocket is a distance d from the mirror (as measured by observers in S) at the moment the light pulse leaves the rocket. What is the total travel time of the pulse as measured by observers in (a) the S frame and (b) the front of the rocket? Figure P1.30arrow_forwardThe muon is an unstable particle that spontaneously decays into an electron and two neutrinos. If the number of muons at t = 0 is N0, the number at time t is given by , where τ is the mean lifetime, equal to 2.2 μs. Suppose the muons move at a speed of 0.95c and there are 5.0 × 104 muons at t = 0. (a) What is the observed lifetime of the muons? (b) How many muons remain after traveling a distance of 3.0 km?arrow_forward(a) What is the momentum of a 2000 kg satellite orbiting at 4.00 km/s? (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that =1+(1/2)v2/c2 at low velocities.)arrow_forward
- A particle of mass 'm' moves in a potential V(x) =mw? +muv², where 'x' is the position co-ordinate, v is the speed, and o and u are constants. The canonical conjugate momentum of the particle is the p = (1+ µ)mv O p = (1 – µ)mv. O p = µmv p = mvarrow_forwardElectron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e − + 7 B e + ⟶ 7 L i + ν The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: 7 B e ⟶ 7 L i + ν The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forwardElectron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e − + 7 B e + ⟶ 7 L i + ν The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: 7 B e ⟶ 7 L i + ν The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (You will make life much easier for yourself if you recognize that practically all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forward
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Length contraction: the real explanation; Author: Fermilab;https://www.youtube.com/watch?v=-Poz_95_0RA;License: Standard YouTube License, CC-BY