# A 12,500 kg railroad car is rolling along a level track with a speed of 4.5m/s when it collides with a second stationary railroad car with a mass of 24,700 kg. The two cars lock together after the collision. Help? Questions in details

## (a) Determine the final speed of the two cars after the collision. (b) Determine the change in kinetic energy of the system. Where does this kinetic energy go?

Apr 5, 2018

I tried this but check my maths anyway....

#### Explanation:

This is an inelastic collision where at the end the objects get stuck together. Kinetic Energy will not be conserved so that at the end, after the impact, part of the initial Kinetic Energy will be "lost" or better changed into heat inside the bumpers (increasing the internal energy of the two objects....they get hotter!!!) and sound.

Linear Momentum is conserved though because no matter enters or leaves the system and no external forces are acting.
So we use conservation of linear momentum between the before (two separate masses) and after (sum of the two masses) collision conditions:

${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = \left({m}_{1} + {m}_{2}\right) V$

with our data:

$12 , 500 \cdot 4.5 + \cancel{24 , 700 \cdot 0} = \left(12 , 500 + 24 , 700\right) V$

$V = 1.5 \frac{m}{s}$

The initial Kinetic Energy was:

${K}_{i} = \frac{1}{2} {m}_{1} {v}_{1}^{2} = \frac{1}{2} \left(12 , 500\right) {\left(4.5\right)}^{2} = 126 , 562.5 J$

The final Kinetic Energy is:

${K}_{f} = \frac{1}{2} \left({m}_{1} + {m}_{2}\right) {V}^{2} = \frac{1}{2} \left(12 , 500 + 24 , 700\right) {\left(1.5\right)}^{2}$
giving:

${K}_{f} = 41 , 850 J$

Giving: a change in Kinetic Energy of:

$126 , 562.5 - 41 , 850 = 84 , 712.5 J$