Starting with an initial speed of 6.12 m/s at a height of0.317 m, a 2.23-kg ball

Question

Starting with an initial speed of 6.12 m/s at a height of0.317 m, a 2.23-kg ball swings downward and strikes a 4.70-kg ballthat is at rest, as the drawing shows. (a) Using the principle ofconservation of mechanical energy, find the speed of the 2.23-kgball just before impact. (b) Assuming that the collision iselastic, find the velocity (magnitude and direction) of the 2.23-kgball just after the collision. (c) Assuming that the collision iselastic, find the velocity (magnitude and direction) of the 4.70-kgball just after the collision. (d) How high does the 2.23-kg ballswing after the collision, ignoring air resistance? (e) How highdoes the 4.70-kg ball swing after the collision, ignoring airresistance? I got a) is 6.61 m/s... but then things got pretty hopelessafter that. If anyone could please walk me through the stepsto figure this out I'd greatly appreciate it! Thanks!

Explanation / Answer

   m1 g h = (1/2)m1 V2
   V = [ 2 g h]1/2
   = [ 2(9.8)(0.317)]1/2  
   = 2.4926
   = 2.5m/s
According to consrvation of momentum
   m1V + 0 =m1V1f +m2V2f   
According to conservation of energy
   (1/2)m1V2 + 0 =(1/2)m1V1f2 +m2V2f2   
on solving the above equations
   V1f =(m1-m2)V / (m1+m2)
   = -0.89 m/s  
   = 0.89 m/stowards left.
   V2f = 2m2 V / (m1+m2)
   = 1.6m/stowards right.

Height from energy conservation
   h1 =V1f2 /2g
   = 0.04m
   h2 =V2f2 /2g
   = 0.1312m

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