Respuesta :
Answer:
[tex]\Delta KE=8.67958\ J[/tex]
Explanation:
Given:
- mass of the first ball, [tex]m_1=2\ kg[/tex]
- initial velocity of the first ball, [tex]u_1=9\ m.s^{-1}[/tex]
- mass of the second ball, [tex]m_2=6.2\ kg[/tex]
- initial velocity of the second ball, [tex]u_2= 4\ m.s^{-1}[/tex]
- Final velocity of the first ball, [tex]v_1=8\ m.s^{-1}[/tex]
Using the law of conservation of linear momentum:
[tex]m_1.u_1+m_2.u_2=m_1.v_1+m_2.v_2[/tex]
where: [tex]v_2=[/tex] final velocity of the second ball
[tex]2\times 9+6.2\times 4=2\times 8+6.2\times v_2[/tex]
[tex]v_2=4.3225\ m.s^{-1}[/tex]
Now the kinetic energy lost in the collision:
Δ KE = (Sum of initial individual kinetic energy) - (Sum of individual final kinetic energy)
[tex]\Delta KE=\frac{1}{2} m_1.u_1^2+\frac{1}{2} m_2.u_2^2-(\frac{1}{2} m_1.v_1^2+\frac{1}{2} m_1.v_2^2)[/tex]
[tex]\Delta KE=0.5\times 2\times 9^2+0.5\times 6.2\times 4^2-(0.5\times 2\times 8^2+0.5\times 6.2\times 4.3225^2)[/tex]
[tex]\Delta KE=8.67958\ J[/tex]
