University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2008
Views:

108

Pages:

19
Author:

Vasientep9m

29-5 Relativistic Momentum and Mass When a object approaches the light speed, the classical momentum express p = mv, is not valid. For examples, if a large mass with a speed v collides with a small mass at rest, the small mass can get a speed 2v; This is not valid if the large mass has a speed v larger than 0.5c, since the speed of the small mass cannot be greater than c. It can be shown that the correct relativistic momentum for the magnitude : Relativistic Momentum p= mv 2 v 1− 2 c SI unit: kg.m/s 29 − 5 The difference of the relativistic and the classical momentum Figure 29-13 Relativistic Momentum Exercise 29-3 Find (a) the classical and (b) the relativistic momentum of a 2.4 kg mass moving with a speed of 0.81c. Solution (a) For classical momentum, p=mv = (2.4kg)(0.81x3.00x10 (2 4k )(0 81 3 00 108 m/s) / ) = 5.8 5 8 x 108 kg.m/s k / (b) For relativistic momentum momentum, p= mv v2 1− 2 c = (2.4kg )(0.81× 3.00 × 108 m / s ) (0.81c) 2 1− c2 = 9.9 × 108 kg.m / s The relativistic momentum is always larger than that of the classical ! Example 29-5 The Missing Mass A satellite, t llit initially i iti ll att restt iin space, explodes l d iinto t ttwo pieces. i O One piece has a mass of 150kg and moves away from the explosion with a speed of 0.76c. The other piece moves away in a opposite direction with a speed of 0 0.88c. 88c Find the mass of the second piece of the satellite. Solution). The magnitude of the momentum for the piece 1 with m1=150kg: 1). The magnitude of the momentum for the piece 1 with m1=150kg: p1 = m1v2 v12 1− 2 c = (150kg )(0.76 × 3.00 × 108 m / s ) (0.76c) 2 1− c2 = 5.3 × 1010 kg.m / s 2) The magnitude of the momentum of the piece 2: 2). p2 = m2 v2 v 1− c 2 2 2 = (m2 )(0.88 × 3.00 × 108 m / s ) (0.88c) 2 1− c2 3)) p2 = p1 : (m2 )(0.88 × 3.00 × 108 m / s ) (0.88c) 2 1− c2 So m2 = 95 kg So, = 5.3 × 1010 kg.m / s The mass increasing In Equation 29-5, we have p= m0 v v2 1− 2 c =( m0 v2 1− 2 c )v = mv The mass increasing with speed v as m= m0 v2 1− 2 c 29 − 6 Note: 1) When v = 0, m = m0; 2) When v approaches c, m approaches infinite. 29-6 Relativistic Energy and E= mc2 Since mass increases at high speed, when work is done on an object: 1) part of the work is used to increase the speed ; 2) and d partt is i used d tto iincrease itits mass ! Considering an object with mass m0 at rest. When an object moves with ith a speed d v, itits ttotal t l energy iis given i as: Relativistic Energy E= m0 c 2 v2 1− 2 c = mc 2 SI unit: J Need infinite energy to achieve the speed of light ! 29 − 7 I t d the Instead, th energy off an object bj t att rest, t the th restt energy E0 is: i Rest Energy with rest mass m0 E = m0 c 2 29 − 9 SI unit: J This is why y material can be converted into nuclear energy! gy Exercise 29-4 Find the rest energy of a 0.12-kg apple. S l ti Solution: E0 = m0c2 = (0.12kg)(3.00x108m/s) = 1.1 x 1016 J It could supply the energy needs of the entire United State for about one hour ! Example 29-6 The Energy of the Sun Energy is radiated by the Sun at the rate of about 3.92x1026 W. Find the corresponding decrease in the Sun’s mass for every second that it radiates. Solution: 1) Calculate the energy (power) radiated by the Sun in 1.00 s: p= 3.92x1026 W = 3.93x1026 J/s. So, ΔE = p Δt = (3.92x1026 J/s)(1.00s) = 3.92x1026 J 2) Calculate the rest mass: Δm= ΔE E / (c ( 2) =(3.92x10 (3 92 1026 J)/(3.00x10 J)/(3 00 1018 m/s) / )2 = 4.36 4 36 x10 109 kg k This is only a small amount of the total mass of the Sun! The mass loss of the Sun in 1,500 years is only 10-10 of the Sun. Relativistic Kinetic Energy When work is done on a rest object, its speed increases, and thus total energy increases (Equation 29-7). The increase in the energy because of the speed, compared with the rest energy, is call (relativistic) Kinetic energy K: E= m0 c 2 v2 1− 2 c = m0 c 2 + K Relativistic Kinetic Energy K= SI unit: J m0 c 2 v2 1− 2 c − m0 c 2 29 − 9 Compared with the classic Kinetic energy Figure 29-16 1 m0 v 2 2 Relativistic and Classical Kinetic Energies Example 29-7 Relativistic Kinetic Energy An observer watching a high-speed spaceship passing by notices that a clock on board runs slow by a factor of 1.50. If the rest mass of the clock is 0.320kg, what is its kinetic energy. Example 29-7 Relativistic Kinetic Energy Solution 1) Using time dilation to calculate the speed v: Δ t = Δ t v c Δ t Δ t0 1 − i .e . 0 2 , 2 1 = 1 .5 = 1 − ( v ) c 2 So, v = 0.745 c, that is v/c = 0.745 2) Calculate kinetic energy K: K= m0c2 v2 1− 2 c − m0c2 = (0.320kg)(3.00×108 m / s)2 1− (0.745)2 − (0.320km)(3.00×108 m / s)2 K= (0.320kg)(3.00×108 m / s)2 1− (0.745)2 = 1.44×1016 − (0.320km)(3.00×108 m / s)2 J Comparison, the classical kinetic energy is 7.99x1015 J , always less than that of the relativistic !