Answer:
ΔE = 61.2 kJ
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved.
- We can express this mathematically as follows:
- p₀ = pf (1)
- The initial momentum (p₀) is just the product of the mass of the boxcar times its speed, as follows:
[tex]p_{o} = m_{bc} * v_{bc} = \frac{(200*10^{3})N}{9.8m/s2} * 3m/s = 61225 kg*m/s (2)[/tex]
- Assuming that the collision is inelastic, we can express the final momentum as the product of the combined mass of the boxcar and the caboose, times the final speed of both objects, as follows:
[tex]p_{f} = (m_{bc} + m_{cab}) * v_{f} = \frac{(600*10^{3})N}{9.8m/s2} * v_{f} = 61225 kg*m/s (3)[/tex]
- From (2) and (3) we can easily find that vf = 1 m/s.
- We can find the initial kinetic energy as follows:
[tex]K_{o} = \frac{1}{2} * m_{bc} * v_{bc} ^{2} (4)[/tex]
- Replacing mbc and vbc from (2), we get:
[tex]K_{o} = \frac{1}{2} * m_{bc} * v_{bc} ^{2} = \frac{1}{2} * 20408 kg* (3m/s)^{2} = 91837 J (5)[/tex]
- By the same token, we can express the final kinetic energy as follows:
[tex]K_{f} = \frac{1}{2} * (m_{bc} + m_{cab}) * v_{f} ^{2} (6)[/tex]
- Replacing masses and the final speed vf from (5) we get:
[tex]K_{f} = \frac{1}{2} * (m_{bc} + m_{cab}) * v_{f} ^{2} = \frac{1}{2} * 61225 kg* (1m/s)^{2} = 30612 J (7)[/tex]
- From (5) and (7) we found that the energy transferred from kinetic energy to other forms of energy, is just the difference between them, as follows:
- [tex]K_{transf} = K_{o} - K_{f} = 91837 J - 30612 J = 61.2 kJ (8)[/tex]
