Show that linear momentum is conserved for the collision just discussed and shown in Figure 2.29.

Strategy We use the relativistic momentum to modify the expressions obtained for the momentum of the balls thrown by Frank and Mary. We will then check to see whether momentum is conserved according to Frank. We leave to Problem 62 the question of whether momentum is conserved according to Mary’s system.

Question

Show that linear momentum is conserved for the collision just discussed and shown in Figure 2.29.

Strategy We use the relativistic momentum to modify the expressions obtained for the momentum of the balls thrown by Frank and Mary. We will then check to see whether momentum is conserved according to Frank. We leave to Problem 62 the question of whether momentum is conserved according to Mary’s system.

Accepted Answer

From Equation (2.39), the momentum of the ball thrown by Frank becomes

[latex]p_{F y}=m u_{0}[/latex] (2.39)

[latex]p_{F y}=\gamma m u_{0}=\frac{m u_{0}}{\sqrt{1-u_{0}^{2} / c^{2}}}[/latex]

For an elastic collision, the magnitude of the momentum for this ball is the same before and after the collision. After the collision, the momentum will be the negative of this value, so the change in momentum becomes, from Equation (2.40),

[latex]\Delta p_{F}=\Delta p_{F y}=-2 m u_{0}[/latex] (2.40)

[latex]\Delta p_{F}=\Delta p_{F y}=-2 \gamma m u_{0}=-\frac{2 m u_{0}}{\sqrt{1-u_{0}^{2} / c^{2}}}[/latex] (2.50)

Now we consider the momentum of Mary’s ball as measured by Frank. Even with the addition of the γ factor for the momentum in the x direction, we still have [latex]\Delta p_{M x}=0[/latex]. We must look more carefully at [latex]\Delta p_{M y}[/latex]. First, we find the speed of the ball thrown by Mary as measured by Frank. We use Equations (2.41) to determine

[latex]\begin{aligned}&u_{M x}=v \&u_{M y}=-u_{0} \sqrt{1-v^{2} / c^{2}}\end{aligned}[/latex] (2.41)

[latex]u_{M}=\sqrt{u_{M x}^{2}+u_{M y}^{2}}=\sqrt{v^{2}+u_{0}^{2}\left(1-v^{2} / c^{2} ight)}[/latex] (2.51)

The relativistic factor γ for the momentum for this situation is

[latex]\gamma=\frac{1}{\sqrt{1-u_{M}^{2} / c^{2}}}[/latex]

The value of [latex]p_{M y}[/latex] is now found by modifying Equation (2.42) with this value of γ.

[latex]\begin{aligned}&p_{M x}=m v \&p_{M y}=-m u_{0} \sqrt{1-v^{2} / c^{2}}\end{aligned}[/latex] (2.42)

[latex]p_{M y}=-\gamma m u_{0} \sqrt{1-v^{2} / c^{2}}=\frac{-m u_{0} \sqrt{1-v^{2} / c^{2}}}{\sqrt{1-u_{M}^{2} / c^{2}}}[/latex]

We insert the value of [latex]u_{M}[/latex] from Equation (2.51) into this equation to give

[latex]p_{M y}=\frac{-m u_{0} \sqrt{1-v^{2} / c^{2}}}{\sqrt{\left(1-u_{0}^{2} / c^{2} ight)\left(1-v^{2} / c^{2} ight)}}=\frac{-m u_{0}}{\sqrt{1-u_{0}^{2} / c^{2}}}[/latex] (2.52)

The momentum after the collision will still be the negative of this value, so the change in momentum becomes

[latex]\Delta p_{M}=\Delta p_{M y}=\frac{2 m u_{0}}{\sqrt{1-u_{0}^{2} / c^{2}}} [/latex] (2.53)

The change in the momentum of the two balls as measured by Frank is given by the sum of Equations (2.50) and (2.53):

[latex]\Delta p=\Delta p_{F}+\Delta p_{M}=0[/latex]

Thus Frank indeed finds that momentum is conserved. Mary should also determine that linear momentum is conserved (see Problem 62).

Question 2.9: Show that linear momentum is conserved for the collision jus...

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