Question

Starting with an initial speed of 3.77 m/s at a height of 0.126 m, a 1.25-kg ball swings downward and strikes a 5.95-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 1.25-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.25-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.95-kg ball just after the collision. (d) How high does the 1.25-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.95-kg ball swing after the collision, ignoring air resistance?

Answer 1

Given that

m1 = 1.25 kg

m2 = 5.95 kg

u = 3.77 m/s

h1 = 0.126 m

A) using law of conservation of mechanical energy

m1*g*h1+(0.5*m1*u^2) = 0.5*m1*v1^2

(1.25*9.8*0.126)+(0.5*1.25*3.77^2) = (0.5*1.25*v1^2)

v1 = 4.08 m/s

b) v1f = (m1-m2)*v1/(m1+m2)

v1f = (1.25-5.95)*4.08/(1.25+5.95) = -2.66 m/s

negative sign shows that the diection is opposite to the original direction

c) V2f = 2*m1*V1/(m1+m2) = (2*1.25*4.08)/(1.25+5.95) = 1.42 m/s

d) m*g*h2 = (0.5*m*v1f^2)

(1.25*9.8*h2) = (0.5*1.25*2.66^2)

h2 = 0.361 m

e) h3 = V2f^2/(2*g) = 1.42^2/(2*9.8) = 0.102 m