Answer to Question #157151 in Mechanics | Relativity for Vanessa

Question

A smooth sphere A of mass 3m moving on a smooth horizontal table with speed 4u, impiges directly on another smooth sphere B of mass 2m, moving with speed u in the opposite direction to A. The coefficient of restitution between A and B is e.
(a) Find the impulse exerted on A by the impact.
At the moment of the impact, the line of centres of the spheres is perpendicular to a vertical wall which is at a distance x from the point of collision and nearer to B than to A, and B subsequently collides with the wall.
(b) Find in terms of x, the distance of A from the wall at the instant B hits the wall.

Answer 1

The coefficient of restitution (COR), also denoted by (e), is the ratio of the final to initial relative velocity between two objects after they collide.

$P_1 + P_2 = P_1' + P_2'$

$E_1 + E_2 = E_1' + E_2'$

$3m*4u-2m*u = 3m * x + 2m*y$

$\large\frac{3m*16u^2}{2} + \large\frac{2m*u^2}{2} = \large\frac{3m*x^2}{2}+\large\frac{2m*y^2}{2}$

${\displaystyle {\text{Coefficient of restitution }}(e)={\frac {\left|{\text{Relative velocity after collision}}\right|}{\left|{\text{Relative velocity before collision}}\right|}}}$

$e = \large\frac{y - x}{4u - u}$ $\to y - x = 3ue$

$10mu = m (3x+2y) \to 3x + 2y = 10u$

$48u^2 + 2u^2 = 3x^2 + 2y^2$

$9x^2 + 12xy+4y^2 = 100u^2$

$50u^2 = 3x^2 + 2y^2$ / * 2

$9x^2 + 12xy+4y^2 = 6x^2 + 4y^2$

$3x^2 + 12xy = 0$

$3x(x+4y) = 0$

$x = -4y$

$y - x = 5y = 3ue \to y = 0.6ue$

$x = - 4y = -2.4ue$ the minus consider the sphere direction

$P_1' = 3mx = 3m * 2.4ue = 7.2mue$