Question

A mass ‘m’ moves with a velocity v and collides
inelastically with another identical mass. After
collision the 1st mass moves with velocity v/√3 in a
direction perpendicular to the initial direction of
motion. Find the speed of the 2nd mass after
collision.

Answer 1

1. A mass 'm' moves with a velocity $v$ and collides inelastically with another identical mass. After collision the 1st mass moves with velocity $v / \sqrt{3}$ in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision.

We have to use the law impulse conservation during inelasticity of the collision.

$$\dot {m} v = m v _ {1} + m v _ {2}.$$

Let write the last vector equation in projectives onto the $X$ - and $Y$ -axes.

$$\left\{ \begin{array}{l} m v = m v _ {1} \cos \alpha \\ 0 = m v _ {1} \sin \alpha - m v _ {2} \end{array} . \right.$$

From the Eq. (2), we can find that

$$v _ {1} = \frac {v _ {2}}{\sin \alpha} = \frac {v}{\sqrt {3} \sin \alpha}.$$

Then, let write the Eq. (1):

$$m v = m \cdot \frac {v}{\sqrt {3} \sin \alpha} \cdot \cos \alpha , \quad t g \alpha = \frac {1}{\sqrt {3}}, \quad \alpha = 3 0 ^ {0}.$$

So, the speed of the second mass after collision:

$$v _ {1} = \frac {v}{\sqrt {3} \sin 3 0 ^ {0}} = \frac {v}{\sqrt {3} \cdot \frac {1}{2}} = v \cdot \frac {2 \sqrt {3}}{3}.$$

Answer: $v \cdot \frac{2\sqrt{3}}{3}$ .