Question

2. A 50 gram steel ball hangs from a 15 cm long string. The ball is brought back to an angle of 15 degrees from vertical at the bottom of the swing it contacts and bounces off the 25 gram ball on a 15cm string. Find the angle of both balls.

Answer 1

here,

the mass of first ball , m1 = 50 g = 0.05 kg

mass of second ball , m2 = 25 g = 0.025 kg

theta = 15 degree

length of string , l = 15 cm = 0.15 m

the initial speed of the first ball, u = sqrt(2 * g * l * ( 1 - cos(theta)))

u = sqrt(2 * 9.81 * 0.15 * ( 1 - cos(15)))

u = 0.32 m/s

let the final speed of each ball be v1 and v2

using conservation of momentum

m1 * u = m1 * v1 + m2 * v2

0.05 * 0.32 = 0.05 * v1 + 0.025 * v2 .....(1)

and

using conservation of kinetic energy

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

0.05 * 0.32^2 = 0.05 * v1^2 + 0.025 * v2^2 .....(2)

from (1) and (2)

v1 = 0.1067 m/s

v2 = 0.427 m/s

for first ball

let the angle for first ball be theta1

using conservation of energy

0.5 * m1 * v1^2 = m * g * l * ( 1 - cos(theta1))

0.5 * 0.1067^2 = 9.81 * 0.15 * ( 1 - cos(theta1))

solving for theta1

theta1 = 5.04 degree

for second ball

let the angle for first ball be theta2

using conservation of energy

0.5 * m2 * v2^2 = m2 * g * l * ( 1 - cos(theta2))

0.5 * 0.427^2 = 9.81 * 0.15 * ( 1 - cos(theta2))

solving for theta2

theta2 = 20.3 degree