Question

A billiard ball rolling across a table at 1.50 m/s makes a head-on collision with an identical ball. Find the speed of each ball after the collision (a) when thesecond ball is initially at rest, (b) when the second ball is moving toward the first at a speed of 1.00 m/s, and (c) when the second ball is moving away from thefirst at a speed of 1.00 m/s.

Answer 1

Applying the law of conservation of momentumwe have as,

m1u1+m2u2=m1v1+m2v2As m1=m21.50+0=v1+v2Applying the law of conservation ofKE we have as,1/2m1u1²+1/2m2u2²=1/2m1v1²+1/2m2v2²1.50²+0=v1²+v2²Substituting the above in this we get the eqn in quadraticwhich by solving we get v1 and v2 of it.When the second ball is moving with speed 1m/s thenu2=-1m/sWhen the second ball is moving awayby it then whave1m/s then u2=1m/sFor the rest we have to use the same way of it.

Answer 2

a) in first case the second ball is at rest and first ball collideshead on with 1.5 m/s
after collision first ball comes to rest and thekinetic energy of first ball is transferred to second ball
∴1/2 m1v1=1/2 m2v2
m1=m2
∴v1=v2
∴v2=1.5m/s
b) in second case the speed of secondball=v1-v2
=1.5-1
=0.5m/s
the first ball will come to rest
c) the speed of first ball after collision will be equal to itsinitial speed
the speed of second ball =v1+v2
=1.5+1
=2.5m/s