Two clay balls (with masses m and 2m) hang vertically down from two light string

Question

Two clay balls (with masses m and 2m) hang vertically down from two light strings of length l. (a) The mass m is raised so that its string is horizontal and then the mass is released from rest. It swings down, collides and sticks with mass 2m. Find the maximum angle to which they swing. How much mechanical energy was lost in the collision? b) Now raise mass 2m instead and let it swing down and collide with mass m. Find the maximum angle to which they swing. How much mechanical energy was lost in the collision?

Explanation / Answer

a) speed of m just beofre it hits the the seond mass,

v = sqrt(2*g*L)

now apply conservation of momentum

m*v = (m + 2*m)*v'

==> v' = v/3

= sqrt(2*g*l)/3

let h is the maximum height raised above its lowest point.

h = v'^2/(2*g)

= (2*g*l/9)/(2*g)

= (1/9)*l

if theta is the maximum angle made by rope with vertical

h = l*(1 - cos(theta))

h/l = 1 - cos(theta)

cos(theta) = 1 - h/l

cos(theta) = 1 - (1/9)*l/l

cos(theta) = 8/9

theta = cos^-1(8/9)

= 27.2 degrees

b)

speed of 2*m just beofre it hits the the second mass,

v = sqrt(2*g*L)

now apply conservation of momentum

2*m*v = (m + 2*m)*v'

==> v' = 2*v/3

= 2*sqrt(2*g*l)/3

let h is the maximum height raised above its lowest point.

h = v'^2/(2*g)

= (8*g*l/9)/(2*g)

= (4/9)*l

if theta is the maximum angle made by rope with vertical

h = l*(1 - cos(theta))

h/l = 1 - cos(theta)

cos(theta) = 1 - h/l

cos(theta) = 1 - (4/9)*l/l

cos(theta) = 5/9

theta = cos^-1(5/9)

= 56.2 degrees

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