Question

A 200-g particle moving at 5.0 m/s on a frictionless horizontal surface collides with a 300-g particle initially at rest. After the collision the 300-g object has a velocity of 2.0 m/s at 45o below the direction of motion of the incoming particle. What is the velocity of the incoming particle after the collision? What percentage of the initial kinetic energy is lost in the collision?

Answer 1

here,

mass of 1 , m1 = 200 g = 0.2 kg

u1 = 5 i m/s

mass of 2 , m2 = 300 g = 0.3 kg

v2 = 2 * ( cos(45) i + sin(45) j) m/s

v2 = 1.41 i m/s + 1.41 j m/s

let the final velocity of particle be v1

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

0.2 * 5 i = 0.2 * v1 + 0.3 * ( 1.41 i + 1.41 j)

solving for v1

v1 = 2.89 i m/s - 2.12 j m/s

the magnitude of velocity , |v1| = sqrt(2.89^2 + 2.12^2) = 3.58 m/s

theta = arctan(2.12/2.89) = 36.3 degree from orignal line of motion

the percentage of the initial kinetic energy is lost in the collision , % = ( 0.5 * m1 * u1^2 - 0.5 * m1 * v1^2 - 0.5 * m2 * v2^2) /(0.5 * m1 * v1^2) * 100

% = ( 0.2 * 5^2 - 0.2 * 3.58^2 - 0.3 * 2^2)/(0.2 * 5^2) * 100

% = 24.7 %