Question

One object is traveling northeast with a momentum of 12 kg-m/s, while a second object is traveling southwest on the same path as the first object with a momentum of 4 kg-m/s. When the two objects collide, they stick together. What is the momentum after their collision?(1 point)
Responses

16 kg-m/s southwest
16 kg-m/s southwest

8 kg-m/s southwest
8 kg-m/s southwest

16 kg-m/s northeast
16 kg-m/s northeast

8 kg-m/s northeast

Answer 1

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.

1. **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)

2. **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)

3. **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:
\[
p_{y,\text{total}} = p_{1y} + p_{2y} = 6\sqrt{2} - 2\sqrt{2} = 4\sqrt{2} \text{ kg-m/s (north)}
\]

4. **Magnitude of total momentum:**
\[
\text{Magnitude} = \sqrt{(p_{x,\text{total}}^2 + p_{y,\text{total}}^2)} = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \text{ kg-m/s}
\]

5. **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**.