Question

A pendulum of length l with a ball of mass m is raised 90◦ clockwise from its hanging position and released. A block of mass 2m rests on a frictionless surface at the pendulum’s hanging position so that it is struck just as the pendulum reaches its lowest point. The two objects collide elastically and the pendulum rebounds, swinging back up to an angle θ clockwise from the vertical. What is the angle θ, and what is the speed of the block after the collision?

Answer 1

let
m1 = m
m2 = 2*m

let u1 is the speed of the pendulum just before the collision.

apply cosnervation of energy

(1/2)*m*u1^2 = m*g*l

u1^2 = 2*g*l

u1 = sqrt(2*g*l)

speed of the block just after the collision,

v1 = (m2 - m1)*u1/(m1 + m2)

= (2*m - m)*sqrt(2*g*l)/(m + 2*m)

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

let h is the height raised.

apply conservation of energy

m*g*h = (1/2)*m*v1^2

m*g*h = (1/2)*m*((2/3)*sqrt(2*g*l) )^2

g*h = (4/9)*g*l

h = (4/9)*l

let theta is angle raised.

now use, h = l*(1 - cos(theta))

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

4/9 = 1 - cos(theta)

cos(theta) = 1 - 4/9

cos(theta) = 5/9

theta = cos^-1(5/9)

= 56.3 degrees   <<<<<<<<<<<-------------------------------Answer

speed of the blocm after the collision,

v2 = 2*m1*u1/(m1 + m2)

= 2*m*sqrt(2*g*l)/(m + 2*m)

= (2/3)*sqrt(2*g*l) <<<<<<<<<<<-------------------------------Answer