How much kinetic energy is lost in the collision? (See full question below)

Question

A snooker ball, having a mass of #0.04# kg and initially moving with a speed of
#2ms^-1# strikes a stationary ball of the same mass. After the collision, the two balls are both moving with equal speeds at equal angles of #30°# to the original direction of the incident ball. How much kinetic energy is lost in the collision?

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Answer 1 · 2018-05-13T17:22:06

0.026J

Start with a diagram.

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We need to find the final speed of the balls; let this be #v# for now.

By the conservation of momentum
Momentum before = momentum after

#2xx0.04+0=0.04vcos30+0.04vcos30#

#0.08=0.08vcos30#
#1=vcos30#
#v=1/cos30#
#v=2/sqrt3#
#v=1.15ms^-1#

Now we can work out the kinetic energies.

#KE_"before"=1/2xx0.04xx2^2#
#=0.08J#

#KE_"after"=1/2xx0.04xx(2/sqrt3)^2+1/2xx0.04xx(2/sqrt3)^2#
#=0.04xx(2/sqrt3)^2#
#=0.04xx4/3#
#=4/75#

#DeltaKE=0.08-4/75#
#=0.026J#