Question

particle 1 of mass 3 m initially moving with a speed vo in the positive x direction collides with particle 2 of mass m moving in opposite xirection with unknown speed v. Determine v and v'.

Answer 1

Answer on Question #55879-Physics-Mechanics-Relativity

particle 1 of mass $3\mathrm{m}$ initially moving with a speed $v_{o}$ in the positive x direction collides with particle 2 of mass m moving in opposite x direction with unknown speed v. After collision particle 1 moves along the negative y direction with speed $\frac{v_{o}}{2}$ and particle 2 moves with v' in a direction making angle of 45deg with positive x direction. Determine v and v'.

Solution

From the conservation of momentum on the x-axis:

$$3 m v _ {0} - m v = m v ^ {\prime} \cos 4 5 = \frac {\sqrt {2}}{2} m v ^ {\prime}.$$

From the conservation of momentum on the y-axis:

$$0 = m v ^ {\prime} \sin 4 5 - 3 m \frac {v _ {0}}{2} = m v ^ {\prime} \frac {\sqrt {2}}{2} - 3 m \frac {v _ {0}}{2}.$$

So,

$$v ^ {\prime} = 3 \frac {\sqrt {2}}{2} v _ {o}.$$

Substitute it in the first equation

$$3 m v _ {0} - m v = \frac {\sqrt {2}}{2} m \cdot 3 \frac {\sqrt {2}}{2} v _ {o} = \frac {3}{2} m v _ {0}.$$

Then,

$$v = \frac {3}{2} v _ {0}.$$

Answer: $\pmb{v} = \frac{3}{2}\pmb{v}_0$ ; $\pmb{v}' = 3\frac{\sqrt{2}}{2}\pmb{v}_0$ .

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