A $4.00 \mathrm{~kg}$ object is moving at $5.00 \mathrm{~m} / \mathrm{s}$ NORTH. It strikes a $6.00 \mathrm{~kg}$ object that is moving EAST at 2.00 $\mathrm{m} / \mathrm{s}$. The objects undergo a perfectly inelastic (stick together) collision. The velocity of the $6.00 \mathrm{~kg}$ object after the collision is
A) $2.00 \mathrm{~m} / \mathrm{s}$ at an angle of 59.0 degrees NORTH of EAST.
B) $2.00 \mathrm{~m} / \mathrm{s}$ at an angle of 45.0 degrees NORTH of EAST.
C) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 59.0 degrees NORTH of EAST.
D) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 35.0 degrees NORTH of EAST.
E) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 45.0 degrees NORTH of EAST.

Question

Quizplus · Accepted Answer

In a perfectly inelastic collision, the objects stick together, and momentum is conserved. The total momentum before the collision is equal to the total momentum after the collision. The momentum in the north direction before the collision is $4.00 \, \text{kg} \times 5.00 \, \text{m/s} = 20.00 \, \text{kg} \cdot \text{m/s}$, and in the east direction, it is $6.00 \, \text{kg} \times 2.00 \, \text{m/s} = 12.00 \, \text{kg} \cdot \text{m/s}$. After the collision, the combined mass is $4.00 \, \text{kg} + 6.00 \, \text{kg} = 10.00 \, \text{kg}$, and the velocity can be found using the Pythagorean theorem for the resultant velocity and the arctan function for the angle.The magnitude of the velocity is $\sqrt{(20.00^2 + 12.00^2) / 10.00^2} = \sqrt{4.00^2 + 1.44^2} = \sqrt{16.00 + 2.88} = \sqrt{18.88} \approx 4.34 \, \text{m/s}$. However, this calculation error led to an incorrect magnitude. The correct magnitude calculation should be based on the combined momentum and mass: $\sqrt{(20^2 + 12^2)} / 10 = \sqrt{400 + 144} / 10 = \sqrt{544} / 10 = 23.32 / 10 = 2.33 \, \text{m/s}$.The angle $\theta$ with respect to the east direction can be found using the arctan of the north component over the east component of the velocity: $\theta = \arctan(20.00 / 12.00) = \arctan(5/3)$, which gives an angle of approximately 59.0 degrees north of east. Therefore, the correct answer is $2.33 \, \text{m/s}$ at an angle of 59.0 degrees north of east.

A 4.00 kg4.00 \mathrm{~kg}4.00 kg Object Is Moving At 5.00 m/s5.00 \mathrm{~m} / \mathrm{s}5.00 m/s NORTH It Strikes A Is

A $4.00 \mathrm{~kg}$ Object Is Moving At $5.00 \mathrm{~m} / \mathrm{s}$ NORTH It Strikes A Is