Starting with an initial speed of 5.00 m/s at a height of 0.290 m, a 1.70-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.290 m, a 1.70-kg ball swings downward and strikes a 4.65-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.70-kg ball just before impact. m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.

= m/s (1.70 kg-ball)

= m/s (4.65 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?

= m (1.70 kg-ball)

= m (4.65 kg-ball)

Explanation / Answer

Here is the correct solution

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 (5.00^2)+(2)(9.8)(0.290)=5.53m/s.

B) The next equation is ((m1-m2)/(m1+m2))*(Vo1) <---you just found
So for you it would be ((1.70-4.65)/(1.70+4.65))*(5.53)= -2.569m/s

C) The equation is ((2)m1)/(m1+m2)*(Vo1)
or ((2)(1.70)/(1.70+4.65))*(5.53)=2.96m/s

D) its asking for height after impact Vo2^2/2(g) is used
or -2.569^2/2(9.8)=0.336m

E) Same thing Vo2^2/2(g)
or 2.96^2/2(9.8)=0.447m

Check my math if you so please but that should do it for you

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