Detail please A particle of mass m moving with a spired has a collision with a p

Question

Detail please

A particle of mass m moving with a spired has a collision with a particle of mass 2 m that is At rest before the collision. After the collision each particle is observed to be moving at an angle theta measured from the original direction of the incoming particle a) What is the fractional change in the kinetic energy of the particles after the collision as compared to the kinetic energy before the collision? b) What value of theta would correspond to an elastic collision? For part a one wants fractional change =

Explanation / Answer

let

m1 = m

m2 = 2*m

let v1 and v2 are the speeds of m1 and m2 after the collision.

Apply consrvation of momentum in x-direction

m*vo = m*v1*cos(theta) + 2*m*v2*cos(theta)

vo = (v1+2*v2)*cos(theta) ---(1)

Apply consrvation of momentum in y-direction

0 = m*v1*sin(theta) - 2*m*v2*sin(theta)

m*v1*sin(theta) = 2*m*v2*sin(theta)

v1 = 2*v2

substitute above value in the equation 1

vo = (2*v2+2*v2)*cos(theta)

= 4*v2*cos(theta)

v2 = vo/(4*cos(theta))

so, v1 = vo/(2*cos(theta))

Eo = 0.5*m*vo^2

Efinal = 0.5*m*v1^2 + 0.5*2*m*v2^2

= 0.5*m*(vo/(2*cos(theta)) )^2 + 0.5*2*m*(vo/(4*cos(theta)) )^2

= 0.1875*m*vo^2/(cos^2(theta))

a) fractional change in the KE = Eo - Efinal / Eo

= (0.5*m*vo^2 - 0.1875*m*vo^2/(cos^2(theta)) )/(0.5*m*vo^2)

= (1 - 0.375/cos^2(theta) )<<<<<<<<<<--------------Answer

b) for Elastic collsion, Efinal -Eo = 0

(0.5*m*vo^2 - 0.1875*m*vo^2/(cos^2(theta)) ) = 0

(1 - 0.375/cos^2(theta) ) = 0

1 = 0.375/cos^2(theta)

cos(theta) = sqrt(0.375)

theta = tan^-1(0.612)

= 31.5 degrees <<<<<<<<<<--------------Answer

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