Chapter 6: Q50P (page 1250)

The net force $\vec{F}$ on a particle of mass m is directed at $30.0 °$ counter-clockwise from the +x - axis. At one instant of time, the particle is travelling in the +x - direction with a speed (measured relative to the earth) of 0.700c . At this instant, what is the direction of the particle’s acceleration?

Expert verified

The direction of the particle’s acceleration is at $48.5 °$ counter clockwise from the +x - axis.

Step by step solution

Speed of the particle $v = 0.700 c$

Force on a particle of mass m is directed $θ_{F} = 30.0 °$

The rate of change of velocity with respect to time is known as acceleration.

$\vec{a} = \frac{d v}{d t}$

The force of particle is F = ma .

From the above figure:

$F_{l l} = y^{3} m a_{l l} F_{⊥} = y m a_{⊥} \frac{F_{⊥}}{F_{l l}} = \frac{a_{⊥}}{y^{2} a_{l l}} \frac{a_{⊥}}{a_{l l}} = \frac{y^{2} F_{⊥}}{F_{l l}}$

From the figure (a):

$F_{l l} = F \cos θ_{F} F_{⊥} = F \sin θ_{f}$

And from the figure (b):

$a_{l l} = a \cos θ_{a} a_{⊥} = a \sin θ_{a}$

So, $\frac{a \sin θ_{a}}{a \cos θ_{a}} = \frac{F \cos θ_{F}}{(1 - v^{2} / c^{2}) F \sin θ_{F}}$

Where, is speed of particle and $γ = 1 / \sqrt{1 - v^{2} / c^{2}}$

Now, $θ_{a} = \tan^{- 1} [\frac{\tan θ}{(1 - v^{2} / c^{2})}] = \tan^{- 1} [\frac{\tan θ}{(1) - {(0.700)}^{2}}] = 48.5 °$

Hence, the direction of the particle’s acceleration is at $48.5 °$ counter clockwise from the +x - axis.

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