A ball of mass 0.540 kg moving east (+x direction) with a speed of 3.30 m/s coll

Question

A ball of mass 0.540 kg moving east (+x direction) with a speed of 3.30 m/s collides head-on with a 0.640 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

ball originally at rest ___3.02________ m/s   East

Ball originally moving east _________________ m/s   East

I got the first part correct but I got -0.279 for the second part an it was wrong. I also tried to make it positive but that was wrong too.

Please explain in detail the answer and why.

Thank you!

Explanation / Answer

Equation for perfectly elastic collision:

u1 - u2 = -(v1 - v2)

u2 = 0

v2 = u1 + v1

and momentum conservation:

m1u1 + m2u2 = m1v1 + m2v2

m1u1 = m1v1 + m2(u1+v1)

v1 = (m1 - m2)*v1/(m1 + m2) = (0.540 - 0.640)*3.3/(0.540 + 0.640) = -0.279 m/sec

you should put the answer +0.279 m/sec West

because negative sign means velocity is in opposite direction.

v2 = u1 + v1 = -0.279 + 3.3 = 3.021

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.