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

Question

Quizplus · Accepted Answer

The total momentum of the system can be found by calculating the momentum of each ball and then combining these vectors. The momentum of the 4.00 kg ball moving east is $4.00 \, \text{kg} \cdot 4.00 \, \text{m/s} = 16.00 \, \text{kg} \cdot \text{m/s}$ eastward, and the momentum of the 6.00 kg ball moving north is $6.00 \, \text{kg} \cdot 3.00 \, \text{m/s} = 18.00 \, \text{kg} \cdot \text{m/s}$ northward. The resultant momentum vector can be found using the Pythagorean theorem: $\sqrt{16^2 + 18^2} = \sqrt{256 + 324} = \sqrt{580} \approx 24.1 \, \text{kg} \cdot \text{m/s}$. The angle $\theta$ with respect to the eastward direction can be found using the tangent function: $\tan(\theta) = \frac{18}{16}$, so $\theta \approx 48.4^\circ$ north of east.

A 4.00 kg4.00 \mathrm{~kg}4.00 kg Ball Is Moving At 4.00 m/s4.00 \mathrm{~m} / \mathrm{s}4.00 m/s To the EAST and A

A $4.00 \mathrm{~kg}$ Ball Is Moving At $4.00 \mathrm{~m} / \mathrm{s}$ To the EAST and A