| Two ice skaters, of mass 30 kg and 80 kg, are skating across the surface of a frozen lake on a collision course, with respective velocities of 2.0 m/s in a general north direction, and 1.0 m/s generally west, as shown above. After they collide, the pair of skaters move off in a direction north of west with a momentum of approximately 100 kg m/s. How much kinetic energy was lost in the collision? |
 |
A) 0 J
B) 110 J
C) 55 J
D) 70 J
Correct Answer: C
Solution :
| The kinetic energy lost in the collision can be found by subtracting the skater?s final K from their initial K : |
| \[{{K}_{initial}}=\frac{1}{2}{{m}_{1}}{{v}_{1}}^{2}+\frac{1}{2}{{m}_{2}}{{v}_{2}}^{2}\] |
| \[{{K}_{final}}=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}_{f}}^{2}\] |
| \[\Delta K={{K}_{final}}-\,\,{{K}_{initial}}\] |
| \[\Delta K=\left( \frac{1}{2}({{m}_{1}}+{{m}_{2}})v_{f}^{2} \right)-\left( \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2} \right)\] |
| \[\Delta K=\left( \frac{1}{2}(30+80){{\left( \frac{100kg\,\centerdot \,m/s}{110kg} \right)}^{2}} \right)\]\[-\left( \frac{1}{2}(30kg){{(2m/s)}^{2}}+\frac{1}{2}(80kg)(1m/{{s}^{2}}) \right)\] |
| \[\Delta K=45J-(60+40)=-55J\] |
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