In Fig. 10-41, ball 1 with an initial speed of 8 m/s collides elastically with s
ID: 1918608 • Letter: I
Question
In Fig. 10-41, ball 1 with an initial speed of 8 m/s collides elastically with stationary balls 2 and 3 that are initially in contact with each other. The centers of balls 2 and 3 are on a line perpendicular to the initial velocity of ball 1. The three balls are identical. Ball 1 is aimed directly at the contact point, and all motion is frictionless.(Hint: With friction absent, each impulse is directed along the line connecting the centers of the colliding balls, normal to the colliding surfaces.)Explanation / Answer
Energy: E = (1/2)*m*v^2 Momentum: p = m*v Where: E = energy p = momentum m = mass v = velocity so: initial values are E = (1/2)*m*8^2 = 32m p = m*8 = 8m Balls 2 and three move at an angle of 45 degrees from the direction ball 1 is moving so the final energy is: E = (1/2)*m*v1^2 + (1/2)*m*v2^2 + (1/2)*m*v3^2 and the final momentum is: p = m*v1 + m*(v2+v3)/sqrt(2) where v1= velocity of ball 1 v2 = velocity of ball 2 v3 = velocity of ball 3 the vertical momentum of ball 2 and 3 cancel and the horizontal is 1/sqrt(2) of the total. Since ball 2 and 3 are identical there final velocities are the same so: E = (1/2)*m*v1^2 + (1/2)*m*v2^2 + (1/2)*m*v2^2 = (m/2) * (v1^2 + v2^2) p = m*v1 + m*(v2+v2)/sqrt(2) = m*v1 + sqrt(2)*v2*m then from the initial values of energy and momentum: (m/2) * (v1^2 + v2^2) = 32m m*v1 + sqrt(2)*v2*m = 8m then simplifying: v1^2 + v2^2 = 64 v1 + sqrt(2)*v2 = 8 from the second: v1 = 8 - sqrt(2)*v2 and plugging into the first: ( 8 - sqrt(2)*v2 )^2 + v2^2 = 64 so: 64 - v2^2 + v2^2 = 64