As shown in the figure below, object m 1 = 1.55 kg starts at an initial height h

Question

As shown in the figure below, object m1 = 1.55 kg starts at an initial height h1i = 0.330 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.60 kg which is initially at rest. Determine the following. PLEASE SHOW YOUR WORK! No work, no credit!

(a) speed of m1 just before the collision.

___________m/s

(b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.)
m/s (m1)
m/s (m2)

(c) height to which each ball swings after the collision (ignoring air resistance)
m (m1)
m (m2)

Explanation / Answer

Part A)

Apply mgh + .5mv^2= .5mv^2

(9.8)(.33) + (.5)(4)^2 = (.5)(v^2)

v = 4.74 m/s

Part B)

Apply conservation of momentum and conservation of KE

mv = mv' + Mv''

(1.55)(4.74) = (1.55)(v') + (4.6)v''

v' = 4.74 - 2.97v''

.5mv^2 = .5mv'^2 + .5Mv''^2

(.5)(1.55)(4.74)^2 = (.5)(1.55)(v'^2) + .5(4.6)(v''^2)

17.4 = .775v'^2 + 2.3v''^2

17.4 = (.775)(4.74 - 2.97v'')^2 + 2.3v''^2

17.4= .775(22.5 - 28.2v'' + 8.82v''^2) + 2.3v''^2

17.4 = 17.4 -21.85v'' + 9.14v''^2

0 = v''(9.14v'' - 21.85)

v'' = 2.39 m/s

v' = 4.74 - 2.97(2.39)

v' = -2.36 m/s

Summary for part B - velocity for m1 = -2.36 m/s (negative is sign convention since it will be going backwards - ignore the negative if needed)

velocity for m2 = 2.39 m/s

Part C)

For m1

mgh = .5mv^2

(9.8)(h) = (.5)(2.36)^2

h = .284m (28.4 cm)

For m2

mgh = .5mv^2

(9.8)(h) = (.5)(2.39)^2

h = .291 m (29.1 cm)

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