**Problem Statement:**Consider the two pucks shown in the figure. As they move towards each other, the momentum of each puck is equal in magnitude and opposite in direction. Given that \( v_{i, \text{green}} = 11.0 \, \text{m/s} \), and \( m_{\text{blue}} \) is 20.0% greater than \( m_{\text{green}} \), what are the final speeds of each puck (in m/s), if \(\frac{1}{2}\) the kinetic energy of the system is converted to internal energy?**Diagram Explanation:**- The diagram shows two pucks: one green and one blue.- The green puck is initially moving to the right at an angle of 30.0°.- The blue puck is initially moving to the left at an angle of 30.0°, opposite to the direction of the green puck.- Both velocities are vectors, shown with red arrows indicating their directions. **Equations:**- \( v_{\text{green, final}} = \, \_\_\_\_ \, \text{m/s} \)- \( v_{\text{blue, final}} = \, \_\_\_\_ \, \text{m/s} \)This problem focuses on momentum conservation and energy conversion.

Solution: Given Values, vg=11.0 m/smb=20 % of mgHere,Let Mass of green is 'm' thenMass of green(mg)=mMass of blue(mb)=20% of mg =0.2 m+m=1.2 m KEF=12KEi(given) To determine: The final speeds of each pluck are calculated as below: Here, The momentum of both balls is the same but opposite. So, it is given as: We know that, Momentum (p)=mv Here, (1.2 m)vb=m vgvb=m vg1.2 m =m ×111.2 m=9.16 m/svb=9.16 m/s Let the final speed of each pluck is VBlue and Vgreen is given as: By using the law of momentum of inertia: we have, 1.2 m Vblue=m VgreenVgreen=1.2 Vblue By using the Law of Equation of Kinetic Energy: It is given that: KEF=12KEi(given)So,12mV2green+12(1.2 m)V2blue=1212mvg2+12(1.2 m)vb21.2 Vblue2+(1.2 )V2blue=12112+12×(1.2 )9.1621.44V2blue+1.2V2blue=60.5+50.342.64 V2blue=110.84V2blue=110.842.64V2blue =41.98 m/sVblue =6.47 m/sWe have, Vgreen=1.2×VblueSo,Vgreen=1.2×6.47 m/s=7.76 m/s Thus,The final speed of each puck is:Vblue =6.47 m/sVgreen=7.76 m/s

Answered: Consider the two pucks shown in the…