Answer to Question #157279 in Mechanical Engineering for Tinotenda

Question

A particle of mass m moving at 5.0m/s in the positive x direction makes a glancing elastic collision with a particle of mass 2m that is at rest before the collision. After the collision, m moves off at an angle of 450 to the x axis and 2m moves off 600 to the x axis. Calculate the speed of m after the collision.  

Answer 1

Let $v_1 = 5 \frac{m}{s}$ be the initial speed of the particle of mass m.

Let $v_1'$ and $v_2'$ be the speed of particle of mass $m$ and $2m$ after the collision respectively.

Hence, the law of conservation of momentum $\bold p_1 = \bold p_1' + \bold p_2'$ , written in coordinate form is:

$(m v_1, 0) = (m v_1' \cos 45, m v_1' \sin 45) + (2 m v_2' \cos 60, -2 m v_2' \sin 60)$.

From y components of the last equation, obtain $v_2' = \frac{v_1' \sin 45}{2 \sin 60}$.

From x components of the last equation, obtain: $v_1' = \frac{v_1 - 2 v_2' \cos 60}{\cos 45}$ .

Substituting $v_2'$ from the first equation into the second, obtain:

$v_1' = \frac{v_1 - v_1' \frac{\sin 45 \cos 60}{\sin 60}}{\cos 45} = \frac{v_1}{\cos 45} - \tg 45 \ctg 60 v_1'$, from where $v_1' = \frac{v_1}{\cos 45 (1 + \tg 45 \ctg 60)} \approx 4.48 \frac{m}{s}$.