I\'m having trouble. Help me out with this. Thank you!! A uniform density disc [

Question

I'm having trouble. Help me out with this. Thank you!!

A uniform density disc [density rho, radius R] is placed upon a horizontal, frictionless surface. A smaller, uniform density disc of the same material [density rho, radius r] slides across the surface [velocity v] and collides in a perfectly inelastic event as indicated in the diagram. R=3r Find the (center of mass) velocity and the angular velocity (around the center of mass)of the system immediately following the collision. Check your solutions for the case that theta approaches zero and pi /2. Derive an expression for the ratio of the final system kinetic energy to the pre-collision system kinetic energy. Check your solutions for the case that theta approaches zero and pi /2

Explanation / Answer

The velocity of center of mass,

Vcm=[m1v1+m2v2] /(m1+m2)

(a) m1=65*10^3 kg

m2=92*10^3 kg

v1=+0.79 m/s

v2=+1.2 m/s

Vcm=[m1v1+m2v2] /(m1+m2)

Vcm=[ 65*10^3*0.79 +92*10^3*1.2 ] / ( 65*10^3 +92*10^3*)

Vcm=161.75 / 157 =1.03 m/s
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(b)The momentum of the system remains conserved.

The coupled boxcars move with same velocity V.

(m1+m2)V=m1v1+m2v2

V = [ m1v1+m2v2] / (m1+m2)

m1=65*10^3 kg

m2=92*10^3 kg

v1=+0.79 m/s

v2=+1.2 m/s

V=[m1v1+m2v2] /(m1+m2)

V=[ 65*10^3*0.79 +92*10^3*1.2 ] / ( 65*10^3 +92*10^3*)

V=161.75 / 157 =1.03 m/s

As the two boxcars are coupled, the velocity of center of mass Vcm =V=1.03 m/s
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(c)The momentum remains conserved in perfectly INELASTIC collision

As no external force has acted,the velocity of center of mass will not change.

Vcm before collision IS EQUAL TO Vcm after the collision

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