A $4.00 \mathrm{~kg}$ object is moving at $5.00 \mathrm{~m} / \mathrm{s}$ NORTHWEST. It strikes a $6.00 \mathrm{~kg}$ object that is moving SOUTHWEST at $2.00 \mathrm{~m} / \mathrm{s}$. The objects collide perfectly inelastically (stick together). The velocity of the $6.00 \mathrm{~kg}$ object after the collision is
A) $3.89 \mathrm{~m} / \mathrm{s}$ at an angle of 35.0 degrees NORTH of WEST.
B) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 35.0 degrees NORTH of WEST.
C) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 45.0 degrees NORTH of WEST.
D) $3.89 \mathrm{~m} / \mathrm{s}$ at an angle of 14.0 degrees NORTH of WEST.
E) $2.33 \mathrm{~m} / \mathrm{s}$ at an angle of 14.0 degrees NORTH of WEST.

Question

Quizplus · Accepted Answer

In a perfectly inelastic collision, the objects stick together, and momentum is conserved. The initial momentum in the x-direction (west-east) can be found by adding the x-components of each object's momentum, and similarly for the y-direction (north-south). The 4.00 kg object has an initial momentum of $4.00 \mathrm{~kg} \times 5.00 \cos(45^\circ) \mathrm{~m/s}$ in the x-direction and the same value in the y-direction due to the 45-degree angle for northwest direction. The 6.00 kg object has an initial momentum of $6.00 \mathrm{~kg} \times 2.00 \cos(45^\circ) \mathrm{~m/s}$ in the x-direction and a negative value in the y-direction because it's moving southwest. After the collision, the total mass is $4.00 \mathrm{~kg} + 6.00 \mathrm{~kg} = 10.00 \mathrm{~kg}$, and the final velocity can be found by dividing the total momentum in each direction by the total mass. The angle of the resultant velocity vector can be found using the arctan of the ratio of the y-component to the x-component of the final velocity. The correct answer must account for both the magnitude and direction accurately, reflecting the combined mass's movement and the initial directions of both objects.

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