Starting with an initial speed of 5.47 m/s at a height of 0.250 m, a 1.52-kg bal

Question

Starting with an initial speed of 5.47 m/s at a height of 0.250 m, a 1.52-kg ball swings downward and strikes a 4.83-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.52-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.52-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.83-kg ball just after the collision. (d) How high does the 1.52-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.83-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

m1 = 1.52 kg

m2 = 4.83 kg

Use conservation of energy to calculate the velocity of ball

m1gh = 0.5 m1 Vo1^2

Vo1 = sqrt(2gh) = sqrt(2*9.81*0.25) = 2.21 m/s

collision is elastic so

Vf1 = (m1-m2)/m1+m2) Vo1 = (1.52 - 4.83)(1.52 +4.83)*2.21 = -1.15 m/s

Vf2 = 2m1/(m1+m2) Vo1 = 2(1.52)/(6.35) *2.21= 1.05 m/s

use conservation of energy to get height

mgh = 0.5 m Vf1^2

h = 1.15^2/2*9,81 = 0.067 m

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