Two masses slide along the horizontal bar as shown A is kg

Two masses slide along the horizontal bar as shown. A is @ kg moving at 3m/sec while it is 0.75 kg moving at the left at 1m/sec the coefficient of restitution for the two masses is e=o.7 Determine the velocity of each mass after the collision. Also, what is the percentage of kinetic energy lost during the collision?

Solution

u1 = 3 m/s

u2 = -1 m/s

Let the velocity of mass 1 afger collision be v1

and for mass 2 be v2

So, coefficient of restitution,

e = (v2-v1)/(u2-u1)

So, 0.7 = (v2 - v1)/(-1 - 3)

So, v2- v1 = -2.8 <------- (1)

Now, conserving momentum,

2*3 + 0.75*(-1) = 2*v1 + 0.75*v2 -------(2)

Solving these two equations , we get :

v1 = 2.67 m/s <-------- velocity of mass 1 after collision

v2 = -0.13 m/s <---- velocity of mass 2 after collision

So, energy lost in collision:

E = 0.5*2*3^2 + 0.5*0.75*1^2 - (0.5*2*2.67^2 + 0.5*0.75*1^2)

= 1.87 J

So, percent of energy lost = 1.87/( 0.5*2*3^2 + 0.5*0.75*1^2)

= 0.199

= 19.9 percent