As shown in the figure below, object m 1 = 1.65 kg starts at an initial height h

Question

As shown in the figure below, object m1 = 1.65 kg starts at an initial height h1i = 0.305 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.

(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

by conservation of energy

initial energy = final energy

m1 * g * h + 0.5 * m1 * v1^2 = 0.5 * m1 * v^2

g * h + 0.5 * v1^2 = 0.5 * v^2

9.8 * 0.305 + 0.5 * 4^2 = 0.5 * v^2

v = 4.688 m/s

speed of m1 just before collision = 4.688 m/s

by conservation of momentum

initial momentum = final momentum

1.65 * 4.688 = 1.65 * v1 + 4.6 * v2 --------------- (1)

by conservation of energy

initialenergy = final energy

0.5 * 1.65 * 4.688^2 = 0.5 * 1.65 * v1^2 + 0.5 * 4.6 * v2^2 ------------(2)

on solving 1 and 2 we'll get

velocity of m1 or v1 = -2.212 m/s

velocity of m2 or v2 = 2.475 m/s

height of m1 after collision = (-2.212)^2 / (2 * 9.8)

height of m1 after collision = 0.249 m

height of m2 after collision = 2.475^2 / (2 * 9.8)

height of m2 after collision = 0.312 m

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