A ball of mass 0.540 kg moving east (+x direction) with a speed of 4.00 m/s coll

Question

A ball of mass 0.540 kg moving east (+x direction) with a speed of 4.00 m/s collides head-on with a 0.380 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision? a)Ball originally at rest (m/s) b) ball originally moving east (m/s)

Explanation / Answer

m1 = 0.54, u1 = 4 m/s m2 = 0.38, u2 = 0 m/s a)Collision is perfectly elastic =>v2-v1 = u1-u2 =>v2 = v1+u1 = v1+4 Conservation of momentum =>m1u1 + m2u2 = m1v1 + m2v2 =>(0.54*4) = (0.54v1) + (0.38*v1) + (0.38*4) =>0.92v1 = (0.16*4) =>0.92*v1 = 0.64 =>v1 = 0.695652174 m/s v2 = 4.695652174 m/s So a)Ball originally at rest, speed = v2 = 4.695652174 m/s b)ball originally moving east, speed = v2 = 0.695652174 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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