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

Question

Quizplus · Accepted Answer

In a perfectly inelastic collision, the objects stick together and move with a common velocity after the collision. The total momentum before the collision is equal to the total momentum after the collision. To solve this, we first break down the velocities into their x (west-east) and y (south-north) components. The 4.00 kg object moving northwest at 5.00 m/s has components: $V_{x1} = -5/\sqrt{2}$ m/s (west) and $V_{y1} = 5/\sqrt{2}$ m/s (north). The 6.00 kg object moving southwest at 2.00 m/s has components: $V_{x2} = -2/\sqrt{2}$ m/s (west) and $V_{y2} = -2/\sqrt{2}$ m/s (south).The total momentum in the x-direction before the collision is $P_{x, \text{before}} = m_1V_{x1} + m_2V_{x2}$ and in the y-direction is $P_{y, \text{before}} = m_1V_{y1} + m_2V_{y2}$.Substituting the given values:- $P_{x, \text{before}} = 4 \cdot (-5/\sqrt{2}) + 6 \cdot (-2/\sqrt{2}) = -20/\sqrt{2} - 12/\sqrt{2} = -32/\sqrt{2}$ kg·m/s- $P_{y, \text{before}} = 4 \cdot 5/\sqrt{2} + 6 \cdot (-2/\sqrt{2}) = 20/\sqrt{2} - 12/\sqrt{2} = 8/\sqrt{2}$ kg·m/sAfter the collision, the objects stick together, so their total mass is $4 + 6 = 10$ kg. The velocity components of the combined mass can be found by dividing the total momentum in each direction by the total mass.- $V_{x, \text{after}} = P_{x, \text{before}} / (m_1 + m_2) = -32/\sqrt{2} / 10$- $V_{y, \text{after}} = P_{y, \text{before}} / (m_1 + m_2) = 8/\sqrt{2} / 10$The magnitude of the velocity is found using the Pythagorean theorem: $V = \sqrt{V_{x, \text{after}}^2 + V_{y, \text{after}}^2}$, and the angle $\theta$ with respect to the west direction can be found using $\tan(\theta) = V_{y, \text{after}} / V_{x, \text{after}}$.Substituting the values:- $V = \sqrt{(-32/\sqrt{2} / 10)^2 + (8/\sqrt{2} / 10)^2} = \sqrt{1024/20 + 64/20} = \sqrt{1088/20} = \sqrt{54.4}$ m/s- $\theta = \tan^{-1}(8/\sqrt{2} / -32/\sqrt{2}) = \tan^{-1}(1/4)$The magnitude calculation simplifies to approximately 2.33 m/s, and the angle calculation gives an angle significantly less than 45 degrees, closer to 14 degrees north of west, matching option D.

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