Consider two particles of equal mass m. If one particle is stationary while the other collides with this particle obliquely with velocity v. Now, what is the angle between the velocity vectors of the two particles after the collision?

Options

A. `45^@`
B. `30^@`
C. `90^@`
D. none of these

Answer

C

Explanation

Given that mass of each particle = m
initial velocity of one moving particle = v
initial velocity of other particle = 0 (because it is stationary)
Let after collision,
One particle moves with velocity `v_1` with angle `theta_1` w.r.t. -axis, and other particle moves with velocity `v_2` with angel `theta_2` w.r.t. x axis.
Now, by momentum conservation in x-direction (initial momentum) along x-axis.
= (final momentum) along y-axis `(p_i)_(x-"axis") = (p_f)_(x-"axis")`
`implies mv + m(0) = mv_(1) cos theta_1 + mv_(2) cos theta_2`
`implies v= v_1 cos theta_1 + v_2 cos theta_2 ... (1)`
Now, by momentum conservation in y-direction
`(p_i)_("y-axis") = (p_f)_("y-axis")`
`implies m(0) + m(0) = mv_(1) sin`
`theta_1 – mv_2 sin theta_2`
`implies 0 = v_1 sin theta_(1) – v_2 sin theta_2 ... (2)`
and by law of conservation of kinetic energy
`("Kinetic energy")_("before collision") = ("Kinetic energy")_("after collision")`
`implies1/2mv^2+1/2m(0)^2=1/2mv_1^2+1/2mv_2^`
`implies v^2=v_1^2+v_2^2" "...(3)`
Addition of square of eq. (1) and eq. (2)
`implies v^2 + 0^2 = (v_1 cos theta_1 + v_2 cos theta_2)^(2) + (v_1 sin theta_1 - v_2 sin theta_2)^2`
`impliesv^2=v_1^2costheta_1^2+v_2^2cos theta_2^2+2v_1v_2cos theta_1cos theta_2+v_1^2sin theta_(1)^2 +v_2^2 sin theta_2^(2)- 2v_1v_2 sin theta_(1) sin theta_2`
`impliesv^2=v_1^2(cos theta_1^2+sintheta_2^2)+v_2^2(costheta_2^2+sintheta_2^2)`
`2v_1v_2(cos theta_1cos theta_2-sintheta_2sintheta_2)`
`impliesv^2=v_1^2+v_2^2+2v_1v_2`
`(cos theta_1cos theta_2-sintheta_1sintheta_2)`
Using eq.(3)
`implies v_1^2+v_2^2=v_1^2+v_2^2+ 2v_1v_2(cos theta_1cos theta_2-sin theta_1 sin theta_2)`
`implies 2v_1v_2cos(theta_1+theta_2)=0`
`implies cos (theta_1 +theta_2) = 0`
`implies theta_1 +theta_2 = cos^(-1)(0) = 90`
So, angle between the velocity vectors of the two particles after the collision is `90^@` .