Question

Starting with an initial speed of 4.49 m/s at a height of 0.148 m, a 2.76-kg ball swings downward and strikes a 5.25-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.76-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.76-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.25-kg ball just after the collision. (d) How high does the 2.76-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.25-kg ball swing after the collision, ignoring air resistance?

Answer 1

let
m1 = 2.76 kg
u1 = 4.49 m/s
h1 = 0.148 m/s
m2 = 5.25 kg

a) (1/2)*m1*v1^2 = m1*g*h1 + (1/2)*m1*u1^2

v1^2 = 2*g*h1 + u1^2

v1 = sqrt(2*g*h1 + u1^2)

= sqrt(2*9.8*0.148 + 4.49^2)

= 4.80 m/s

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

= (2.76 - 5.25)*4.80/(2.76 + 5.25)

= -1.49 m/s

|v1f| = 1.49 m/s

direction : opposite to initial direction

c) v2f = 2*m1*v1/(m1 + m2)

= 2*2.76*4.8/(2.76 + 5.25)

= 3.31 m/s

direction : apposite to m1

d) m1*g*h1 = (1/2)*m1*v1f^2

h1 = v1f^2/(2*g)

= 1.49^2/(2*9.8)

= 0.113 m

e)

m2*g*h2 = (1/2)*m2*v1f^2

h2 = v2f^2/(2*g)

= 3.31^2/(2*9.8)

= 0.559 m