Answer:
[tex]\Delta K=-160.89\ J[/tex]
Explanation:
It is given that,
Mass of object A, [tex]m_A = 2\ kg[/tex]
Initial speed of object A, [tex]u_A=15\ m/s[/tex]
Mass of object B, [tex]m_B = 5\ kg[/tex]
Initial speed of object A, [tex]u_B=0[/tex] (at rest)
Le V is the final speed when they lock and move with one common velocity. Using the conservation of momentum to find it.
[tex]m_Au_A+m_Bu_B=(m_A+m_B)V[/tex]
[tex]m_Au_A=(m_A+m_B)V[/tex]
[tex]V=\dfrac{m_Au_A}{(m_A+m_B)}[/tex]
[tex]V=\dfrac{2\times 15}{(2+5)}[/tex]
V = 4.28 m/s
Initial kinetic energy of the system is :
[tex]K_i=\dfrac{1}{2}m_Au_A^2[/tex]
[tex]K_i=\dfrac{1}{2}\times 2\times (15)^2[/tex]
[tex]K_i=225\ J[/tex]
Final kinetic energy of the system is :
[tex]K_f=\dfrac{1}{2}(m_A+m_B)V^2[/tex]
[tex]K_f=\dfrac{1}{2}\times (2+5)\times (4.28)^2[/tex]
[tex]K_f=225\ J[/tex]
[tex]K_f=64.11\ J[/tex]
Let [tex]\Delta K[/tex] is the change in kinetic energy of the system after the collision. It is given by :
[tex]\Delta K=K_f-K_i[/tex]
[tex]\Delta K=64.11-225[/tex]
[tex]\Delta K=-160.89\ J[/tex]
So, the change in kinetic energy of the system is 160.89 Joules.
