Starting with an initial speed of 5.42 m/s at a height of 0.355 m, a 1.35-kg bal

Question

Starting with an initial speed of 5.42 m/s at a height of 0.355 m, a 1.35-kg ball swings downward and strikes a 5.26-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.35-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.35-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.26-kg ball just after the collision. (d) How high does the 1.35-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.26-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

a) mgh = (1/2)m*v^2
velocity just before collision = (2gh) = (2*9.81*0.355) = 2.64 m/s

b) V1 = ((m1-m2)v1 + 2m2v2)/(m1+m2)

        = ((1.35-5.26)*2.64 + 2*5.26*0)/(1.35+5.26)

        = -1.56 m/s  i.e 1.56 m/s in the negative x direction

c) V2 = (2m1u1 + u2(m2-m1))/(m1+m2)

        = (2*1.35*2.64 + 0)/(1.35+5.26)

       = 1.08 m/s in positive x direction

d) mgh = (1/2)mv^2

 => h = (1.08)^2/9.81 = 0.119 m

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