Answer to Question #122436 in Mechanics | Relativity for Quanda

Question

A smooth sphere A of mass 2m moving on a smooth horizontal floor with velocity (10 i+ 5 j) m/s impinges obliquely on a smooth stationary sphere B of mass m. At the instant of collision, the line joining the centres of the spheres is parallel to the vector a=3i+4j. Given that the coefficient of restitution between the two spheres is e and that the direction of motion of sphere A after impact makes an angle x with line of centres of the spheres. Show that :
a) (4-2e)tan x= 3
b) 3/4≤tanx≤3/2
c) Given that tan x ≤ 1, find e and hence find the velocities of A and B after impact.

Answer 1

Explanations & Calculations

• Consider the sketch attached & assume the angle the velocity vector creates with the line of centers to be $\theta$ & the velocities after the collision be V & V₂ as shown.
• To find this angle,

first consider the angle created by the velocity vector U and the x axis.

$\qquad\qquad \begin{aligned} \small \alpha_1 &= \small \tan^{-1}(\frac{5}{10}) = 26.565^0 \end{aligned}$

Secondly the line of centers, if it makes an angle of $\alpha_2$ with x axis,

$\qquad\qquad \begin{aligned} \small \alpha_2&= \small \tan^{-1}(\frac{4}{3})= 53.130^0 \end{aligned}$

Therefore,

$\qquad\qquad \begin{aligned} \small \theta & = \small \alpha_2 -\alpha_1\\ &= \small 26.565^0 \end{aligned}$

• Magnitude of the velocity |U| = $\small \sqrt{10^2+5^2} = 5\sqrt5 ms^{-1}$

• For the rest, apply the theories of conservation of linear momentum & Newton's experimental law along the line of centers in the moving direction.

a) ..................................................................................................................................

1) $\qquad\qquad \begin{aligned} \small 2mU\cos\theta+0 &= \small 2m V\cos x + mV_2\\ \small 2U\cos\theta &= \small 2 V \cos x +V_2 \cdots\cdots(1)\\ \end{aligned}$

2) $\qquad\qquad \begin{aligned} \small V\cos x -V_2 &= \small-e\,( U \cos \theta -0) \\ \small -e\,U \cos \theta &= \small V\cos x -V_2 \cdots\cdots(2) \end{aligned}$

Adding (1) & (2),

3) $\qquad\qquad \begin{aligned} \small U\cos \,\theta(2-e) &= \small 3V \cos x\cdots \cdots(3) \end{aligned}$

4) Due to the absence of any external force perpendicular to the line of centers, velocities along that direction, is kept constant during the collision. Therefore,

$\qquad\qquad \begin{aligned} \small U \sin \theta &= \small V \sin x \cdots\cdots (4) \end{aligned}$

Now (4) /(3),

$\qquad\qquad \begin{aligned} \small \frac{\tan \theta}{(2-e)} &= \small \frac{\tan x}{3}\\ \small \frac{\tan(26.565)}{(2-e)} &= \small \frac{\tan x}{3}\\ \small \frac{1}{2(2-e)} &= \small \frac{\tan x}{3}\\ \small \bold{ (4-2e)\tan x} &= \small \bold{3} \cdots\cdots(5) \end{aligned}$

b)................................................................................................................................

• Now compare $\tan x$ with e,

$\qquad\qquad \begin{aligned} \small tan x & = \small \frac{3}{(4-2e)} \end{aligned}$ & $\small 0 \le e \le 1$

• $\tan x$ maximize as (4-2e) minimize; when e @ maximum (e = 1)

Therefore $\tan x$ maximum = $\frac{3}{2}$

• $\tan x$ minimize as (4-2e) maximize; when e @ minimum (e =0)

Therefore, $\tan x$ minimum = $\frac{3}{4}$

• Therefore,

$\qquad\qquad \begin{aligned} \small \bold{\frac{3}{4} \le \tan x \le \frac{3}{2}} \end{aligned}$

c).................................................................................................................................

• Now it should be corrected as $\color{blue}\tan x = 1$ ; Therefore, $x =45^0$
• Considering equation (5),

$\qquad\qquad \begin{aligned} \small 4 -2e &= 3\\ \small e &= \small \frac{1}{2} =0.5 \end{aligned}$

• Therefore, velocity of A after the impact (V) ; use the equation (4),

$\qquad\qquad \begin{aligned} \small 5\sqrt5 \times \sin(26.565) &= \small V \sin (45) \\ \small V &= \small \bold{7.07 ms^{-1}} \end{aligned}$

• Considering equation (1), Velocity of B after impact can be found

$\qquad\qquad \begin{aligned} \small 2\times 5\sqrt5\cos(26.565) &= \small 2\times 7.07 \cos (45) +V_2\\ \small V_2 &= \small \bold{10.00 ms^{-1}} \end{aligned}$