A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a perfectly inelastic (stick together) collision. The kinetic energy lost in the collision is
A) 34.8 J.
B) 28.7 J.
C) 20.4 J.
D) 18.7 J.
E) 14.9 J.

Question

Quizplus · Accepted Answer

The kinetic energy before the collision can be calculated using the formula for kinetic energy, $KE = \frac{1}{2}mv^2$, for each object and then summing them up. For the 4.00 kg object moving north at 5.00 m/s, $KE_1 = \frac{1}{2}(4.00\,kg)(5.00\,m/s)^2 = 50.0\,J$. For the 6.00 kg object moving west at 2.00 m/s, $KE_2 = \frac{1}{2}(6.00\,kg)(2.00\,m/s)^2 = 12.0\,J$. The total initial kinetic energy is $50.0\,J + 12.0\,J = 62.0\,J$.After the collision, the objects stick together, moving as a single mass. The velocity of the combined mass can be found using conservation of momentum. The initial momentum in the north direction is $p_{north} = (4.00\,kg)(5.00\,m/s) = 20.0\,kg\cdot m/s$, and in the west direction, it is $p_{west} = (6.00\,kg)(2.00\,m/s) = 12.0\,kg\cdot m/s$. The magnitude of the total momentum vector is $\sqrt{(20.0)^2 + (12.0)^2} = \sqrt{544} = 23.3\,kg\cdot m/s$.The combined mass is $4.00\,kg + 6.00\,kg = 10.00\,kg$. The speed of the combined mass is $v = \frac{p}{m} = \frac{23.3\,kg\cdot m/s}{10.00\,kg} = 2.33\,m/s$.The kinetic energy after the collision is $KE_{after} = \frac{1}{2}(10.00\,kg)(2.33\,m/s)^2 = 27.2\,J$.The kinetic energy lost in the collision is the initial kinetic energy minus the final kinetic energy: $62.0\,J - 27.2\,J = 34.8\,J$.

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