oblique collision (only need part b to check answers, I\'ve done part a) (a) In

Question

oblique collision (only need part b to check answers, I've done part a)

(a) In a game of pool, the white ball, of mass 0.15 kg is travelling directly towards the stationary black ball without rotating (which has a mass of 0.15 kg), as shown in figure Q3a, determine the final velocity of the two balls if the black ball travels towards the pocket and the coefficient of restitution during the impact is 0.8. (b) The next shot requires that you hit the target ball at an oblique angle to cause it to move towards the pocket. The diameter of a standard pool ball is 57 mm and you aim the white ball so the line of the centres differs by 15 mm. The initial speed of the white ball is 1.5 m s^-1, it travels without rotating and the coefficient of restitution remains the same as part a) Determine: i) The angle of contact between the lines of the centres of impact and the initial velocity of the moving ball. ii) The final velocities of the two balls

Explanation / Answer

b) i)

let the angle of impact between the lines of centres of impact be theta

then sin(theta) = vertical distance between the centres/ sum of radius of the two balls

given vertical distance between the centres = 15 mm

radius of balls, r = 57/2 = 28.5 mm

sin(theta) = 15/28.5

theta = 31.75 degrees

ii)

the momentum of the ball will not change in tghe direction perpendicular to the direction of collision

initial velocity of ball 1, u = 1.5 m/s

initial velocity of ball 2 = 0

final velocity of ball 1 = v at angle theta to x axis

final velocity of ball 2 = w at angle phi to x axis

and mass of balls be m

then musin(31.75) = mv*sin(theta + 31.75)

considering ball 2

its final velocity has to be in the direction of conatact

so,

from conservation of momentum

m*1.5 = mvcos(theta) + m*w*cos(31.75)

mvsin(theta) = m*w*sin(31.75)

also coefficeint of restitution = 0.8

e = (w - vcos(theta + 31.75))/ucos(31.75)

solving

vcos(theta) + 0.85w = 1.5

vsin(theta) = 0.52w

1.020 = w - vcos(theta + 31.75) = w - v[cos(theta)cos(31.75) - sin(theta)sin(31.75)] = w - v[0.85(1.5 - 0.85w)/v - 0.276w/v]

1.020 = w - 1.275 + 0.7225w + 0.276w = 1.9985w - 1.275

w = 1.148 m/s

tan(theta) = 0.52w/(1.5 - 0.85w) = 1.139

theta = 48.729 degrees

v = 0.794 m/s

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