Answer:
a) [tex]v = 0.118 \frac{m}{s}[/tex], b) [tex]K_{A} = 0.016 J[/tex], c) [tex]K_{B} = 0 J[/tex], d) [tex]K = 0.010 J[/tex], e) No, f) Yes.
Explanation:
a) Final velocity of the two-cart system:
The final velocity can be derived of the momemtum equation associated with the system:
[tex]m_{A} \cdot v_{A} + m_{B} \cdot v_{B} = (m_{A} + m_{B})\cdot v[/tex]
[tex]v = \frac{m_{A}\cdot v_{A} + m_{B}\cdot v_{B}}{m_{A}+m_{B}}[/tex]
[tex]v = \frac{(0.9 kg)\cdot (0.19 \frac{m}{s} )+(0.550 kg)\cdot (0 \frac{m}{s} )}{0.9 kg + 0.55 kg}[/tex]
[tex]v = 0.118 \frac{m}{s}[/tex]
b) Initial kinetic energy of cart A:
The kinetic energy is:
[tex]K_{A} = \frac{1}{2} \cdot m_{A} \cdot v_{A}^{2}[/tex]
[tex]K_{A} = \frac{1}{2} \cdot (0.9 kg) \cdot (0.19 \frac{m}{s} )^{2}[/tex]
[tex]K_{A} = 0.016 J[/tex]
c) Initial kinetic energy of cart B:
As the cart B is initially at rest, the kinetic energy is:
[tex]K_{B} = 0 J[/tex]
d) Final kinetic energy of the system:
The kinetic energy is
[tex]K = \frac{1}{2}\cdot (m_{A} + m_{B})\cdot v^{2}[/tex]
[tex]K = \frac{1}{2} \cdot (0.9 kg + 0.55 kg)\cdot (0.118 \frac{m}{s} )^{2}[/tex]
[tex]K = 0.010 J[/tex]
e) No. The kinetic energy is not conserved for inelastic collisions since some of the energy is dissipated during the collision.
f) Yes. The momentum for inelastic collisions is conserved due to the absence of external forces.
