Question

A 0.40 kg pendulum bob starts with an initial velocity of 2.0 m/s is released from a height of
0.275 m. It collides at the bottom of its swing with a 1.3 kg ball which is at initially at rest.
(a) Find the speed of the pendulum bob just before it strikes more massive ball at rest.
(b) Assume the collision is elastic. Determine the magnitude and direction of the velocities following
the collision.
(c) Using conservation of energy, determine the maximum height to which ball travels following the
collision.

Answer 1

here,

mass of pendulum bob , m1 = 0.4 kg

initial velocity , u1 = 2 m/s

height , h0 = 0.275 m

mass of ball , m2 = 1.3 kg

a)

let the speed of pendulum bob just before collison be v1

using conservation of energy

0.5 * m1 * u1^2 + m * g * h0 = 0.5 * m1 * v1^2

0.5 * 0.4 * 2^2 + 0.4 * 9.81 * 0.275 = 0.5 * 0.4 * v1^2

v1 = 3.07 m/s

the speed of pendulum bob before collison is 3.07 m/s

b)

let the final velocity of the bob be v1' and ball be v2'

using consevation of momentum

m1 * v1 = m1 * v1' + m2 * v2'

0.4 * 3.07 = 0.4 * v1' + 1.3 * v2' ....(1)

and

using conservation of kinetic energy

0.5 * m1 * v1^2 = 0.5 * m1 * v1'^2 + 0.5 * m2 * v2'^2

0.4 * 3.07^2 = 0.4 * v1'^2 + 1.3 * v2'^2 ....(2)

from (1) and (2)

v1' = - 1.63 m/s

v2' = 1.44 m/s

b)

the velocity of pendulum after the collison is 1.63 m/s to the left

the velocity of block after the collison is 1.44 m/s to the right

c)

the maximum height of the ball , hm = v2'^2 / 2g

hm = 1.44^2 /(2*9.81) m

hm = 0.11 m