In a perfectly elastic collision, both momentum and mechanical energy are conser

Question

In a perfectly elastic collision, both momentum and mechanical energy are conserved. Two balls with masses mA and mB are moving toward each other with speeds vA and vB, respectively. Select the appropriate equations you would use to find the speeds of the two balls after the collision. There are more than one answer

A.vA2 = ((mA + mB) / (mA - mB))vA1 + ((mA + mB) / 2mB)vB1

B.vB2 = (2mA / (mA + mB))vA1 + ((mB - mA) / (mA + mB))vB1

C.vA2 = ((mA - mB) / (mA + mB))vA1 + (2mB / (mA + mB))vB1

D. vB2 = ((mA + mB) / 2mA)vA1 + ((mA + mB) / (mB - mA))vB1

E. vA2 = (2mB / (mA + mB))vA1 + ((mA - mB) / (mA + mB))vB1

F. vA2 = (2mA / (mA + mB))vA1 + ((mB - mA) / (mA + mB))vB1

G. vB2 = ((mB - mA) / (mA + mB))vA1 + (2mA / (mA + mB))vB1

H. vB2 = ((mA - mB) / (mA + mB))vA1 + (2mB / (mA + mB))vB1

Explanation / Answer

from conservation of momentum

mAvA1 +mBvB1 = mAvA2 +mBvB2 .. (1)

from conservation of energy

(1/2)mAvA1^2 +(1/2)mBvB1^2 = (1/2)mAvA2^2 +(1/2)mBvB2^2 ..(2)

From (1) and (2) we get

vA2 = ((mA-mB)/(mA+mB))vA1 +(2mBvB1/(mA+mB))

vB2 = (2mAvA1/(mA+mB)) +((mB - mA)/(mA+mB))vB1

Correct options are (B) and (C)

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