Answer to Question #122436 in Mechanics | Relativity for Quanda

Question #122436
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.
1
Expert's answer
2020-06-21T20:08:35-0400

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

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

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

Therefore,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\theta & = \\small \\alpha_2 -\\alpha_1\\\\\n&= \\small 26.565^0\n\\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\n\\begin{aligned}\n\\small 2mU\\cos\\theta+0 &= \\small 2m V\\cos x + mV_2\\\\\n\\small 2U\\cos\\theta &= \\small 2 V \\cos x +V_2 \\cdots\\cdots(1)\\\\\n\\end{aligned}"


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


Adding (1) & (2),

3) "\\qquad\\qquad\n\\begin{aligned}\n\\small U\\cos \\,\\theta(2-e) &= \\small 3V \\cos x\\cdots \\cdots(3)\n\\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\n\\begin{aligned}\n\\small U \\sin \\theta &= \\small V \\sin x \\cdots\\cdots (4)\n\\end{aligned}"

Now (4) /(3),

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

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

  • Now compare "\\tan x" with e,

"\\qquad\\qquad\n\\begin{aligned}\n\\small tan x & = \\small \\frac{3}{(4-2e)}\n\\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\n\\begin{aligned}\n\\small \\bold{\\frac{3}{4} \\le \\tan x \\le \\frac{3}{2}}\n\\end{aligned}"


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

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

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

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

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

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

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



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