Question

Two equal mass object experience a totally inelastic elastic collision. Mass 1 has an initial velocity of 10 m/s in the negative y-direction. Mass 2 has an initial velocity of 10 m/s in the positive x-direction. The collision occurs at the origin. What is the magnitude and direction of the velocity of the combined mass? What is the kinetic energy conserved in the collision? If not what fraction of kinetic energy was lost?

Answer 1

let m1 = m2 = m

v1 = 10 m/s (towards -y axis)
v2 = 10 m/s (towards +x axis)
so, v1x = 0, v1y = -10 m/s
v2x = 10 m/s, v2y = 0
let vf is the velocity of combined mass.

Apply conservation of momentum in x-direction

m1*v1x + m2*v2x = (m1 + m2)*vfx

==> vfx = (m1*v1x + m2*v2x)/(m1 + m2)

= (m*v1x + m*v2x)/(m + m)

= (v1x + v2x)/2

= (0 + 10)/2

= 5 m/s

Apply conservation of momentum in y-direction

m1*v1y + m2*v2y = (m1 + m2)*vfy

==> vfy = (m1*v1y + m2*v2y)/(m1 + m2)

= (m*v1y + m*v2y)/(m + m)

= (v1y + v2y)/2

= (-10 + )/2

= -5 m/s

magnitude of the velocity of the combined mass, vf = sqrt(vfx^2 + vfy^2)

= sqrt(5^2 + 5^2)

= 7.07 m/s <<<<<<---------------Answer

direction : theta = tan^-1(vfy/vfx)

= tan^-1(-5/5)

= 45 degrees below +x axis

The kinetic energy is not conserved.

KEf = (1/2)*(m1 + m2)*vf^2

= (1/2)*(2*m)*7.07^2

= 50*m

KEi = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= (1/2)*m*10^2 + (1/2)*m*10^2

= 100*m

fraction of kinetic energy lost = (KEi - KEf)/KEi

= (100*m - 50*m)/(100*m)

= 0.5 or 50%