a 60 kg object moves north at 1200 m/s and collides with an 1800 kg object moving west at 15 m/s. what is the velocity of the combined masses just after impact?
Answer 

This problem can be solved using the principle of conservation of momentum. Since it's a collision where both masses stick together after impact, we apply the conservation of momentum in both the north (y) and west (x) directions separately.

### Step 1: Calculate the momentum before the collision in both directions.

Momentum is given by:
\[
p = mv
\]
where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

#### In the north direction (y-axis):
- Mass \( m_1 = 60 \, \text{kg} \)
- Velocity \( v_1 = 1200 \, \text{m/s} \)

So, the initial momentum in the north direction is:
\[
p_{1y} = m_1 v_1 = 60 \times 1200 = 72,000 \, \text{kg} \cdot \text{m/s}
\]

#### In the west direction (x-axis):
- Mass \( m_2 = 1800 \, \text{kg} \)
- Velocity \( v_2 = 15 \, \text{m/s} \)

So, the initial momentum in the west direction is:
\[
p_{2x} = m_2 v_2 = 1800 \times 15 = 27,000 \, \text{kg} \cdot \text{m/s}
\]

### Step 2: Calculate the total momentum after the collision.

Since the two objects combine after the collision, their total mass is:
\[
m_{\text{total}} = m_1 + m_2 = 60 + 1800 = 1860 \, \text{kg}
\]

Now, we calculate the velocity in both directions:

\[
v_{x} = \frac{p_{2x}}{m_{\text{total}}} = \frac{27,000}{1860} \approx 14.52 \, \text{m/s}
\]
\[
v_{y} = \frac{p_{1y}}{m_{\text{total}}} = \frac{72,000}{1860} \approx 38.71 \, \text{m/s}
\]

### Step 3: Calculate the magnitude and direction of the combined velocity.

To find the resultant velocity \( v_{\text{resultant}} \), we use the Pythagorean theorem:
\[
v_{\text{resultant}} = \sqrt{v_x^2 + v_y^2} = \sqrt{(14.52)^2 + (38.71)^2}
\]
\[
v_{\text{resultant}} \approx \sqrt{210.85 + 1498.18} = \sqrt{1709.03} \approx 41.35 \, \text{m/s}
\]

### Step 4: Calculate the direction (angle).

The direction of the velocity (θ) with respect to the west can be found using the tangent function:
\[
\tan(\theta) = \frac{v_y}{v_x} = \frac{38.71}{14.52}
\]
\[
\theta = \tan^{-1}\left(\frac{38.71}{14.52}\right) \approx 69.6^\circ
\]
This angle is measured north of west.

### Final Answer:
The velocity of the combined masses just after impact is approximately \( 41.35 \, \text{m/s} \) at an angle of \( 69.6^\circ \) north of west.

Do you have any questions, or would you like further details?

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Here are five related questions:
1. How does the mass of each object affect the final velocity?
2. What happens if the two objects have identical masses but different velocities?
3. How would the result change if one of the objects were initially at rest?
4. How do inelastic collisions differ from elastic collisions?
5. What if the collision happened at an angle other than 90 degrees?

**Tip:** When dealing with collisions, always check whether momentum is conserved in all directions to avoid errors.

A 60 kg object moves north at 1200 m/s and collides with an 1800 kg object moving west at 15 m/s. What is the velocity of the combined masses just after impact?

Explore the detailed step-by-step solution to a momentum conservation problem, where a 60 kg object moving north at 1200 m/s collides with an 1800 kg object moving west at 15 m/s. Learn how to calculate the velocity of the combined masses just after impact using key physics concepts and formulas. Suitable for high school students in Grades 11-12.

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