A ping-pong ball and a bowling ball collide elastically on a frictionless surfac

Question

A ping-pong ball and a bowling ball collide elastically on a frictionless surface. The magnitude of the initial velocity of the ping-pong ball is v(p,0) and the direction of the velocity is in the positive x -direction. The magnitude of the initial velocity of the bowling ball is v(b,0) and the direction of the velocity is in the negative x -direction. You may assume that the mass of the bowling ball is much greater than the mass of the pingpong ball. After the collision what is the velocity of ping-pong?


I would really appreciate any help/explanation of this concept!! Thanks.
Please help :)

Explanation / Answer

the velocity of a body after elastic collision is given by
v1 = (u1(m1-m2) + 2*m2*u2 )/(m1+m2)

so here in case of pingpong ball m1 can be taken as zero since its mass is negligible compared to mass of bowling ball

so

v1 = 2u2-u1

whre u1 and u2 are velocities of bowling ball and pingpong ball respectively

the bowling ball will effectively have no impact

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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