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 = 0.270 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 in 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)

Explanation / Answer

using conservation of mechanical energy

m1gh1+1/2 m1v1i^2=1/2 m1vf^2

vf=4.614 m/sec

For an elastic collision with a stationary object we have the following

v1a = (m1 - m2)/(m1 + m2) *v1 where v1 is the velocity of the incoming ball and v1a is its velocity after the collsion

so for ball1 we have v1a = (1.65-4.6)/(1.65+4.6)*4.614 =-2.177 m/sec

(The negative indicates ball 1 is moving in the opposite direction)

For ball 2 = have v2 = 2*m1/(m1 + m2)*v1=(2*1.65/6.25)*4.614=2.436 m/sec

h1=v1^2/2g=2.177^2/2*9.8=0.242 m

h2=v2^2/2g=2.436^2/2*9.8=0.303 m

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