A 4.00 kg ball is moving at 4.00 m/s to the EAST and a 6.00 kg ball is moving at 3.00 m/s to the NORTH. The total momentum of the system is
A) 14.2 kg m/s at an angle of 48.4 degrees SOUTH of EAST.
B) 48.2 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
C) 48.2 kg m/s at an angle of 48.4 degrees NORTH of EAST.
D) 24.1 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
E) 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST.

Question

Quizplus · Accepted Answer

The total momentum of the system can be found by vector addition of the individual momenta. The momentum of the 4.00 kg ball is $4.00 \, \text{kg} \times 4.00 \, \text{m/s} = 16.00 \, \text{kg m/s}$ to the EAST, and the momentum of the 6.00 kg ball is $6.00 \, \text{kg} \times 3.00 \, \text{m/s} = 18.00 \, \text{kg m/s}$ to the NORTH. Using Pythagoras' theorem, the magnitude of the total momentum is $\sqrt{16^2 + 18^2} = \sqrt{256 + 324} = \sqrt{580} = 24.1 \, \text{kg m/s}$. The angle $\theta$ with the EAST can be found using $\tan(\theta) = \frac{18}{16}$, which gives $\theta = \arctan\left(\frac{18}{16}\right) = 48.4^\circ$. Therefore, the total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST.

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