A 4.00 kg object is moving at 5.00 m/s SOUTH. It strikes a 3.00 kg object that is moving WEST at 2.00 m/s. The collision is perfectly inelastic. The kinetic energy lost in the collision is
A) 24.9 J.
B) 18.4 J.
C) 10.5 J.
D) 8.33 J.
E) 2.61 J.

Question

Quizplus · Accepted Answer

The kinetic energy before the collision is the sum of the kinetic energies of both objects. For the 4.00 kg object moving south at 5.00 m/s, $KE_1 = \frac{1}{2}mv^2 = \frac{1}{2}(4.00)(5.00)^2 = 50.0 \,J$. For the 3.00 kg object moving west at 2.00 m/s, $KE_2 = \frac{1}{2}mv^2 = \frac{1}{2}(3.00)(2.00)^2 = 6.00 \,J$. The total initial kinetic energy is $50.0 \,J + 6.00 \,J = 56.0 \,J$.After the collision, the objects stick together (perfectly inelastic collision), moving as a single object with a mass of $4.00 \,kg + 3.00 \,kg = 7.00 \,kg$. The velocity of the combined object can be found using conservation of momentum. The initial momentum in the south direction is $p_{south} = (4.00 \,kg)(5.00 \,m/s) = 20.0 \,kg\cdot m/s$, and in the west direction is $p_{west} = (3.00 \,kg)(2.00 \,m/s) = 6.00 \,kg\cdot m/s$. The magnitude of the total initial momentum is $\sqrt{(20.0)^2 + (6.00)^2} = \sqrt{400 + 36} = \sqrt{436} = 20.9 \,kg\cdot m/s$.The velocity of the combined mass after the collision is $v = \frac{p}{m} = \frac{20.9 \,kg\cdot m/s}{7.00 \,kg} = 2.99 \,m/s$. The kinetic energy after the collision is $KE_{after} = \frac{1}{2}(7.00 \,kg)(2.99 \,m/s)^2 = 31.1 \,J$.The kinetic energy lost in the collision is the initial kinetic energy minus the final kinetic energy: $56.0 \,J - 31.1 \,J = 24.9 \,J$.

A 400 Kg Object Is Moving at 5