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Two billiard balls of equal mass undergo a perfectly elastic head-on collision.If one...

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dongy | Student, Undergraduate | eNotes Newbie

Posted December 17, 2009 at 3:07 PM via web

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Two billiard balls of equal mass undergo a perfectly elastic head-on collision.

If one ball's initial speed was 2.00 m/s, and the other's was 3.00 m/s in the opposite direction, what will be their speeds after the collision?

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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted December 17, 2009 at 3:31 PM (Answer #2)

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As the balls are perfectly elastic there will be no loss of energy by the collision. Thus the sum of kinetic energies of the two balls before and after collision will be same.

Also, since the two ball are of the same size, the speeds and kinetic energy of the two balls will be same.

Given: Initial speed of the two ball are 2 m/s and 3 m/s.

We know kinetic energy of an object is given by formula (m*v^2)/2

Therefore sum of kinetic energy of the two ball before collision will be given by:

e = (m*2^2)/2 + (m*3^2)/2 = m*13/2

Where m = mass of each ball.

After collision: both the balls will share this energy (e) equally.

Therefore kinetic energy of each ball after collision:

m* (v1^2)/2 = e/2 = (m*13/2)/2

where v1 = common speed of balls after collision


v1^2 = 13/2

Therefore v1 = 2.5495 m/s


Speeds of ball after collision will be 2.5495 m/s in opposite directions.

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neela | High School Teacher | (Level 3) Valedictorian

Posted December 18, 2009 at 3:42 AM (Answer #3)

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When there is an  elastic collision, there conservation of   momentum and kinetic energy. If the initial velocitie of the balls of masses m each are u and v and the final velocities after elastic collision are x and y, respectively, then the required equations are : mu+mv= mx+my for the conservation of momentum and mu^2+mv^2 = mx^2+my^2 for the consrvation of energy. Solving these equations, for x and y , i.e, the ball's respective velocities after impact,  we get :

x = [(m-m)u +2mv ]/(m+m) and y = [2mu+(m-m)v]/(m+m)

Given that u=2 meter/s and v = 3 meter/s in opposite direction of u = -3 meter/s, and substituting  to find the x and y the speed of the first and second balls, we get :

x=[(0+2m(-3)]/(m+m) = -3  meter per second and

y=[2m(2)+0](m+m)  = 2 meter per second.

After the elastic collision, the first ball, which had its velocity +2m/s before impact, imparts its speed  and direction also to the second ball and the second ball retraces back with a speed of 2m/s .

The second ball whose speed before impact was -3 meter per second imparts its speed and direction to the first ball and the first ball rebounds back with a speed , -3m/s as contrasted  to its velocity of  +2m/s before impact.

In other words, the balls exchanged their velocities after impact.

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