Starting with an initial speed of 5.64 m/s at a height of 0.130 m, a 2.25-kg bal

Question

Starting with an initial speed of 5.64 m/s at a height of 0.130 m, a 2.25-kg ball swings downward and strikes a 4.56-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.25-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.25-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.56-kg ball just after the collision. (d) How high does the 2.25-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.56-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

GIVEN: initial speed = 5.64 m/s
height =0.130 m
masses m1= 2.25-kg and m2= 4.56-kg

a).speed of the 2.25-kg ball just before impact

By using the conservation if mechanical energy m1gh=(1/2)m1V01 ^2

V01 =sqrt(2gh)=sqrt(2*9.81*0.130)=1.59m/s

b). find the velocity (magnitude and direction) of the 2.25-kg ball just after the collision.

since the collision is elastic,

Vf1 =((m1-m2)/(m1+m2))*V01 =((2.25-4.56)/(4.56+2.25))*(1.59)=-0.54m/s

c) Vf2 =((2m1)/(m1+m2))*V01 =((2*2.25)/(4.56+2.25))*(1.59)=1.05m/s

d)By using the conservation of energy

mgh=(1/2)mV^2

h1=(V^2)/2g=(0.54*0.54)/(2*9.81)=0.015m

e)h2=(V^2)/2g=(1.05*1.05)/(2*9.81)=0.056m

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