Question

Q: In an elastic collision between two bodies of equal mass, with body 2 initially at rest, body 1 moves off at angle θ relative to the direction of its initial velocity and body 2 at angle φ. The sine of the sum of θ and φ, sin(θ + φ), is equal to.....................

Question options:

a). 0.

b). 0.500.

c). 0.707.

d). 0.866.

e). 1.00.

=> I tried but it's not A or D so maybe someone can help.

=> Answer left is B, C, E

Answer 1

momentum before collsion = momentum after collison

along x axis

m1*v1 = m1*v1'*costheta + m2*v2'*cosphi

m1 = m2

v1 = v1'*costheta + v2'*cosphi .............(1)

v2'*cosphi = v1 - v1'*costheta

------------------------

along y axis

0 = m1*v1'*sintheta - m2*v2'*sinphi

m1 = m2

v2'*sinphi = v1'*sintheta ..................(2)

1^2 + 2^2

v2'^2 = v1^2 + v1'^2 -2*v1*v1'*costheta

from energy conservation

(1/2)*m1*v1^2 = (1/2)*m1*v1'^2 + (1/2)*m2*v2'^2

v1^2 = v1'^2 + v2'^2

v2'^2 = v1^2 - v1'^2

v1^2 - v1'^2 = v1^2 + v1'^2 -2*v1*v1'*costheta

2*v1'^2 = 2*v1*v1'*costheta

v1' = v1*costheta

from 2

v2' = v1*costheta*sintheta/sinphi

from 1

v1 = v1*costheta*costheta + v1*costheta*sintheta*cosphi/sinphi

sinphi = sinphi*costheta*costheta + cosphi*sintheta*costheta

sinphi = sinphi*(costheta)^2 + cosphi*sintheta*costheta

sinphi = sinphi*(costheta)^2 + sqrt(1-(sinphi)^2) *sqrt(1-(costheta)^2)*costheta

sinphi = sinphi*(costheta)^2 + sqrt(1 - (costheta)^2 - (sinphi)^2 + (costheta*sinphi)^2)*costheta

sinphi = sinphi*(costheta)^2 + (costheta*sinphi)*costheta

sinphi = sinphi*(costheta)^2 + (costheta)^2*sinphi

2*(costheta)^2 = 1

theta = 45 degrees

phi = 45 degrees

sn(theta+phi)= 1 <<<------answer

OPTION (e)