A ball falls straight down onto a wedge that is sitting on frictionless ice. The

Question

A ball falls straight down onto a wedge that is sitting on frictionless ice. The ball has a mass of 3.21 kg, and the wedge has a mass of 5.31 kg. The ball is moving a speed of v = 4.67 m/s when it strikes the wedge, which is initially at rest (see the figure).

a) Assuming that the collision is instantaneous and perfectly elastic, what is the velocity of the wedge after the collision?


b) Assuming that the collision is instantaneous and perfectly elastic, what is the velocity of the ball after the collision?

Explanation / Answer

conservation of energy: 0.5*mb*V^2 = 0.5*mb*Vb'^2 + 0.5*mw*Vw'^2
horitanzal momentum should be conserved: mb*Vb' + mw*Vw' =0

>>>> Vb' = -mw*Vw'/mb

>>>> mb*V^2 = mb*(mw*Vw'/mb)^2 + mw*Vw'^2

>>>> Vw'^2 = (mb^2/(mw^2+mb*mw))*V^2

>>> Vw' = -2.23m/s (Vw is negative)

-----------------------------------------------------------------

b)

Vb' = -mw*Vw'/mb

     = -5.31*2.23/3.21 = +3.69m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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