Two solid spheres hung by thin threads from a horizontal support (Figure 1) are

Question

Two solid spheres hung by thin threads from a horizontal support (Figure 1) are initially in contact with each other. Sphere 1 has inertia m1 = 0.060 kg , and sphere 2 has inertia m2 = 0.10 kg. When pulled to the left and released, sphere 1 collides elastically with sphere 2. At the instant just before the collision takes place, sphere 1 has kinetic energy K1 = 0.086 J .

What is the speed of sphere 1 after the collision?

What is the speed of sphere 2 after the collision?

Calculate the coefficient of restitution of the collision.

I saw someone elese post this question but I have no idea where they even got the equation for number 3 in the picture. What does U1 stand for? I have not used that unit in class.

Explanation / Answer

part 1:

initial kinetic energy of the sphere 1=0.086 J

if its initial speed before collision is v,

then 0.5*0.06*v^2=0.086

==>v=sqrt(0.086/(0.5*0.06))=1.693 m/s

this veloicty is towards right direction

as sphere 2 is at rest, its kinetic energy=0

and its momentum=0

hence total initial energy of the system =0.086 J
and total initial momentum=m1*v=0.06*1.693=0.10158 kg.m/s

now, let after collison sphere 1 moves in left direction and

sphere 2 moves in right direction

magnitude of sphere 1 speed=v1

magnitude of sphere 2 speed=v2

then just after collision, total momentum=-0.06*v1+0.1*v2(assuming right direction is positive and left direction is negative)

as momentum is conserved,

initial momentum=final momentum

==>0.10158=-0.06*v1+0.1*v2

==>v2=0.6*v1+1.0158..(1)

as the collision is elastic, total energy will be conserved

hence energy after collision=energy before collison

==>0.5*0.06*v1^2+0.5*0.1*v2^2=0.086

==>0.03*v1^2+0.05*v2^2=0.086

using equation 1 as expression for v2,

0.03*v1^2+0.05*(0.6*v1+1.0158)^2=0.086

==>0.03*v1^2+0.05*(0.36*v1^2+1.21896*v1+1.0158^2)=0.086

==>0.03*v1^2+0.018*v1^2+0.061*v1+0.0516=0.086

==>0.048*v1^2+0.061*v1-0.0344=0

solving for v1, we get

v1=0.423 m/s (ignoring the negative solution)

then from equation 1, v2=0.6*v1+1.0158=1.2996 m/s

hence speed of sphere 1 after collison is 0.423 m/s, to the left

part 2:

speed of sphere 2 after collison=v2=1.2996 m/s, to the right

part 3:

coefficient of restitution is defined as relative speed after collison/relative speed before collision

relative speed before collison=v=1.693 m/s

relative speed after collision=v1+v2=0.423+1.2996=1.7226 m/s(added because they are in opposite direction)

hence coefficient of restitution=1.7226/1.693=1.0175

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