Starting with an initial speed of 4.22 m/s at a height of 0.353 m, a 2.45-kg bal

Question

Starting with an initial speed of 4.22 m/s at a height of 0.353 m, a 2.45-kg ball swings downward and strikes a 5.65-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.45-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.45-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.65-kg ball just after the collision. (d) How high does the 2.45-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.65-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

Given that

m1 =2.45kg

m2 =5.65kg

From the conservation of energy to calculate the velocity of the ball we get

mgh =(1/2)mV2

V =Sqrt(2gh) =Sqrt(2*9.81*0.353) =2.631m/s

b)

Since the collision is elastic then

Vf1 =(m1-m2/m1+m2)V =(2.45-5.65/2.45+5.65)(2.631m/s) =-8.4192/8.1=-1.039m/s

Vf2 =(2m1/m1+m2)V =2(2.45)/2.45+5.65)*2.631 =4.9*2.631/8.1=1.591m/s

c)

Using the conservation of enegy to get the heights

mgh =(1/2)mv2 ===> h =v2/2g

Height h1 =(-1.039)2/2*9.81 =0.0550m

Height h2 =(1.591)2/2*9.81 =0.1290m

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