Question

Two particles, m₁ = 2m and m₂ = m, are suspended on massless cords of the same length l = 1.00 m so that they touch each other (see Fig. below). Particle 1 is displaced to an angle of 30

Answer 1

given,

m1 = 2m

m2 = m

length of chord = 1m

first particle displaced by 30 degree

so its height = 1-L * cos(30)

height = 1 - cos(30)

height = 0.134 m

potential energy of particle = mgh

potential energy of particle = 2m * 9.8 * 0.134

potential energy of particle = 2.6264m

kinetic energy of first particle just before collision = 0.5 * m * v^2

by conservation of energy

2.6264m = 0.5 * m * v^2

2.6264 = 0.5 * v^2

v = 2.2919 m/s

speed of mass m₁ before the collision = 2.2919 m/s

second particle is displaced by 45 degree so

so its height = 1-L * cos(45)

height = 1- cos(45)

height = 0.293 m

potential energy of particle = mgh

potential energy of particle = m * 9.8 * 0.293

potential energy of particle = 2.871m

kinetic energy of first particle just after collision = 0.5 * m * v^2

by conservation of energy

2.871m = 0.5 * m * v^2

2.871 = 0.5 * v^2

v = 2.396 m/s

speed of m2 after collision = 2.396 m/s

by conservation of momentum

m2 * vi = m1v1 + m2v2

2m * 2.2919 = m * 2.396 + 2mv2

2 * 2.2919 = 2.396 + 2v2

v2 = 1.094 m/s

speed of m1 after collision = 1.094 m/s

relative change in KE = 0.5 * 2m * 2.2919^2 - (0.5 * m * 2.396^2 + 0.5 * 2m1.094^2) / 0.5 * 2m * 2.2919^2

relative change in KE = 0.5 * 2 * 2.2919^2 - (0.5 * 2.396^2 + 0.5 * 2*1.094^2) / 0.5 * 2 * 2.2919^2

relative change in KE = 0.2257