A billiard ball collides elastically with an identical stationary one. By lookin

Question

A billiard ball collides elastically with an identical stationary one. By looking at the collision in the Center of Mass frame, show that the angle between the resulting trajectories in the lab frame is 90 degrees.

Please show all steps and calculations.

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A billiard ball collides elastically with an identical stationary one. By looking at the collision in the Center of Mass frame, show that the angle between the resulting trajectories in the lab frame is 90 degree s.

Explanation / Answer

initia lvelocities

ux =u uy = 0

after collision v1 = v1xI+v1yj.........v1^2 = v1x^2+v1y^2

v2 = v2xi+v2yj.....v2^2 =v2x^2+v2y^2

v1 . v2 = (v1xI+v1yj) . (v2xi+v2yj) = v1x*v2x + v1y*v2y

momentum is conserved along x axis

mux = m*v1x+m*v2x

u = v1x+v2x

squaring

u^2 = v1x^2+v2x^2+2v1xv2x...........(1)

along y axis

uy = 0

m uy = mv1y + mv2y

0 = v1y+v2y

squaring

v1y^2+v2y^2+2v1yv2y = 0 ..........(2)

1+2

u^2 = v1x^2+v2x^2+2v1xv2x + v1y^2+v2y^2+2v1yv2y

u^2 = v1^2 + v2^2+ 2*v1x*v2x +2*v1y*v2y..............(3)

KE is conserved along

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

u^2 = v1^2+v2^2................(4)

3 - 4)

2*v1x*v2x +2*v1y*v2y = 0

v1 . v2 = 0

as the dot product is aero

v1 and v2 are orthogonal

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