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Chapter 2: Q119CE (page 71)

Exercise 117 gives the speed u of an object accelerated under a constant force. Show that the distance it travels is given by
$x = \frac{m c^{2}}{f^{2}} [\sqrt{1 + {(\frac{F t}{m c})}^{2}} - 1]$

Short Answer

Expert verified

Distance travelled by an object under constant force is determined by integrating the expression of relativistic speed.

Step by step solution

01

Determine the expression for the differential change in distance dx

Consider an object moving under a constant force, the relativistic speed is given by

$u = \frac{1}{\sqrt{1 + {(\frac{F t}{i t c})}^{2}}} \frac{F}{m} t$

Here, velocity $u = \frac{d x}{d t}$ and let $k = \frac{F}{m c}$ .

$\frac{d x}{d t} - \frac{c}{\sqrt{1 + (k t)^{3}}} d t$

$d x = \frac{t k i}{\sqrt{1 + (k t)^{2}}} d t$

02

Integrate the above expression

Let鈥檚 say

m = 1 + k^{2} t^{2} 鈬� d m = k^{2} (2 t d t) 鈬� t d t = \frac{d m}{2 k^{2}}

.

Therefore,

$d x = \frac{c}{2 k \sqrt{m}} d m$

$鈭� d x = \frac{c}{2 k} 鈭� \frac{1}{\sqrt{m}} d m$

$x = \frac{c}{2 k} [\frac{\sqrt{m}}{1 / 2}] + C$

$x = \frac{c}{k} {(1 + k^{2} t^{2})}^{1 / 2} + C$

${At}^{t = 0}, x = 0$ Therefore, the integration constant

$C = - \frac{c}{k}$

$x = \frac{c}{k} {(1 + k^{2} t^{2})}^{1 / 2} - \frac{c}{k}$

$x = \frac{c}{k} (\sqrt{1 + (k t)^{2}} - 1)$

$x = \frac{m c^{2}}{F} [\sqrt{1 + {(\frac{F t}{m c})}^{2}} - 1]$

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Most popular questions from this chapter

Question: Two objects isolated from the rest of the universe collide and stick together. Does the system鈥檚 final kinetic energy depend on the frame of reference in which it is viewed? Does the system鈥檚 change in kinetic energy depend on the frame in which it is viewed? Explain your answer.

Question: Equation (2-38) show that four-momentum of a particle obeys a Lorentz transformation. If we sum momentum and energy over all particles in a system, we see that the total momentum and energy also constitute a four-vector. It follows that ${(E_{t o t a l} / c)}^{2} - P_{t o t a l}^{2}$ is the same quantity in any frame of reference. Depending on what is known, this can be a quicker route to solving problems than writing out momentum and energy conservation equations. In the laboratory frame, a particle of mass m and energy $E_{i}$ collide with another particle of mass initially stationary, forming a single object of mass . (a) Determine the frame of reference where the after-collision situation is as simple as possible, then determine the invariant in that frame. (b) Calculate the invariant before the collision in the laboratory frame in terms of M and $E_{i}$ . (You will need to use $E_{i}^{2} / c^{2} - p^{2} = m^{2} c^{2}$ for the initially moving particle to eliminate its momentum.) Obtain an expression for M in terms of m and E_i . (c) Write out momentum and energy conservation in the laboratory frame, using $u_{f}$ for the speed of the initially moving particle and for the speed of the final combined particle. Show that they give the same result for M in terms of m and $E_{i}$ . (Note: The identity $纬_{u}^{2} u^{2} = 纬_{u}^{2} c^{2} - c^{2}$ will be very handy.)

Particles of light have no mass. Does the Sun鈥檚 mass change as a result of all the light it emits? Explain.

A light beam moves at an angle $胃$ with the x-axis as seen from frame S. Using the relativistic velocity transformation, find the components of its velocity when viewed from frame $S'$ . From these, verify explicitly that its speed is c.

Derive the following expressions for the components of acceleration of an object , $a_{x}'$ and $a_{y}'$ in the frame $S'$ in terms of its components of acceleration and velocity in the frame $S$ .

${a_{x}}^{'} = \frac{a_{x}}{{纬_{v}}^{3} {(1 - \frac{u_{x} v}{c^{2}})}^{3}} {a_{y}}^{'} = \frac{a_{y}}{{纬_{v}}^{2} {(1 - \frac{u_{x} v}{c^{2}})}^{2}} + \frac{a_{x} \frac{u_{y} v}{c^{2}}}{{纬_{v}}^{2} {(1 - \frac{u_{x} v}{c^{2}})}^{3}}$

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