To find the momentum after the collision, we need to first determine the momentum of each object in component form, then find the total momentum and its direction.
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Object 1:
- Momentum = \(12\) kg-m/s (northeast)
- In component form:
- \(p_{1x} = 12 \cos(45^\circ) = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}\) kg-m/s (east)
- \(p_{1y} = 12 \sin(45^\circ) = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}\) kg-m/s (north)
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Object 2:
- Momentum = \(4\) kg-m/s (southwest)
- In component form:
- \(p_{2x} = 4 \cos(225^\circ) = 4 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -2\sqrt{2}\) kg-m/s (west)
- \(p_{2y} = 4 \sin(225^\circ) = 4 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -2\sqrt{2}\) kg-m/s (south)
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Total momentum components:
- Total \(x\)-component of momentum: \[ p_{x,\text{total}} = p_{1x} + p_{2x} = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2} \text{ kg-m/s (east)} \]
- Total \(y\)-component of momentum: \[ p_{y,\text{total}} = p_{1y} + p_{2y} = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2} \text{ kg-m/s (north)} \]
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Magnitude of total momentum: \[ \text{Magnitude} = \sqrt{(p_{x,\text{total}}^2 + p_{y,\text{total}}^2)} = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \text{ kg-m/s} \]
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Direction of total momentum:
- This total momentum is in the direction of northeast because both components (east and north) are positive.
Therefore, the total momentum after the collision is \(8 \text{ kg-m/s}\) in the northeast direction.
The final answer is: 8 kg-m/s northeast.