A billiard ball rolling across a table at 1.35 m/s makes a head-on elastic colli

Question

A billiard ball rolling across a table at 1.35 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs. (a) The second ball is initially at rest. first ball m/s second ball m/s (b) The second ball is moving toward the first at a speed of 1.05 m/s. first ball Your response differs from the correct answer by more than 10%. Double check your calculations. m/s second ball m/s (c) The second ball is moving away from the first at a speed of 1.00 m/s. first ball The response you submitted has the wrong sign. m/s second ball m/s

Explanation / Answer

Since this is an elastic collision, kinetic energy is conserved. The momentum before Pb is equal to the momentum after Pa. Masses are equivalent. The KE of the first ball is 1.35^2m/2 = KE1+KE2 after the collision =m(V1a^2+V2a^2)/2 Pb = Pa1+Pa2 =>mV1 = m(V1+V2) => 1.35=V1a+V2a => V1a=1.35-V2a Substitute this into the above equation 1.35^2=1.35^2-2.7V2a+2V2a^2 => V2a=1.35m/s which means the first ball stops and the second ball has the same velocity and direction as the first. V1a=0,V2a=1.35 b)Pb = m(1.35-1.1) = m/4=mV1a+mV2a =>V1a=0.25-V2a => V1a^2=0.0625-V2a/2+V2a^2 KEb= (m/2)(1.35^2+1.1^2) => 2=V1a^2+V2a^2= 0.0625-V2a/2+2V2a^2 => -1.9375-V2a/2+2Va2^2=0 Using quadratic formula V1a = 1.12 V2a= -0.87 c) Pb = m(1.35+0.9) = 1.25m => V1a=1.25-V2a => V1a^2 = 1.5625-2.5V2a+V2a^2 1.35^+0.9^2= 2.6325 = V1a^2+V2a^2 = 1.5625-2.5V2a+2V2a^2 Quadratic formula gives V1a= -0.34, V2a = 1.59

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