As shown in the figure below, object m 1 = 1.60 kg starts at an initial height h

Question

As shown in the figure below, objectm1=1.60kg starts at an initial heighth1i=0.275mand speedv1i= 4.00 m/s,swings downward and strikes (in an elastic collision) objectm2=4.45kgwhich is initially at rest. Determine the following.

(a) speed ofm1just before the collision.

(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.)

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

Explanation / Answer

A) So the first thing you need to find in the speed after collision. This can be found using SQRT Vo^2+2gHo. So square root of initial velocity squared plus two times gravity times height. Which for you would be. SQRT (4.00^2)+(2)(9.8)(0.275)=4.62m/s.

B) The next equation is ((m1-m2)/(m1+m2))*(Vo1) <---you just found
So for you it would be ((1.60-4.45)/(1.60+4.45))*(4.62)= -2.176m/s

The equation is ((2)m1)/(m1+m2)*(Vo1)
or ((2)(1.60)/(1.60+4.45))*(4.62)=2.44m/s

D) its asking for height after impact Vo2^2/2(g) is used
or -2.176^2/2(9.8)=0.241m

Same thing Vo2^2/2(g)
or 2.44^2/2(9.8)=0.303m

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