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 3.45 kg, and the wedge has a mass of 5.43 kg. The ball is moving a speed of v = 4.23 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

Considering that on the system horitanzal momentum should be conserved and must be 0.

mbvb' + mwvw' =0

3.45vb' + 5.43vw' =0

or (vb'/vw')2 = 2.48 ............ eq1

Considering the conservation of energy:
0.5mb x v2 = 0.5 x mb x vb'2 + 0.5 x mw x vw'2

3.45 x 4.232 = 3.45vb'2 + 5.43vw'2
dividing by vw'2   

63.7905/vw'2= 3.45(vb'/vw')2 +5.43 ..................from eq 1 substitute the value

63.7905/vw'2 = 3.45(2.48)+5.43

vw'2=4.56

or vw' = - 2.135 ( velocity of wedge..disregard +ve root as wedge is moving in -ve x direction)

vb'2=2.48vb'2 ....from eq1

vb'2=11.308

or vb'=3.362 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.