Answer to Question #195487 in Mechanics | Relativity for TAWANDA SIMON

Question

A particle of mass m, traveling at a velocity vi, strikes a stationary second object of mass

2m, and the two of them scatter at different angles to the initial direction. The incoming

particle scatters at an angle of cos 3 2 5 1

1

   . This collision is inelastic, with exactly

half of the incoming kinetic energy being lost in the collision. Determine the final velocities

of both particles in terms of vi only.

N/B Leave your answers in exact form. e.g. 2 rather than 1.414… [10]

Answer 1

Initial Velocity $=v_i$

Mass of moving body $=m$

Mass of stationary body $=2m$

Kinetic energy lost $=\dfrac{1}{4}mv_i^2$

Kinetic energy after collision $K_f=\dfrac{1}{4}mv_i^2$

$\dfrac{1}{2}mv_m^2+\dfrac{1}{2}(2m)(v_{2m})^2=\dfrac{1}{4}mv_i^2$

$\dfrac{v_m^2}{2}+v_{2m}^2=\dfrac{v_i^2}{4}$

Let the particle of mass 2m be deflected by an angle $\theta_1$ and particle of mass m be deflected by $\theta_2$

Momentum balance in x direction,

$mv_mcos\theta_2+2mv_{2m}cos\theta_1=mv_i$

Momentum balance in y direction,

$mv_msin\theta_2+2mv_{2m}sin\theta_1=0$

$v_m=\dfrac{-2v_{2m}sin\theta_1}{sin\theta_2}$

Substituting value of $v_m$ in the equation of momentum balance in x direction

$\dfrac{-2v_{2m}sin\theta_1}{sin\theta_2}cos\theta_2+2v_{2m}cos\theta_1=v_i$

$v_m=\dfrac{v_i\sin\theta_2}{2(\cos\theta_1\sin\theta_2-\sin\theta_1\cot\theta_2)}$

$v_{2m}=\dfrac{v_i\sin\theta_2^2}{-4\sin\theta_1(\cos\theta_1\sin\theta_2-\sin\theta_1\cot\theta_2)}$