91Ó°ÊÓ

Collision problems are a fundamental aspect of physics. They help us understand how objects interact with each other. In this scenario, we explore an ice-skating collision between a man and a child. By analyzing these interactions, we can determine critical details like the velocity of the child just before the collision.
The key principle used in collision problems is the conservation of momentum. Collisions can be elastic or inelastic. In this problem, since the man and child stick together post-collision, it is an inelastic collision. The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision.

• Momentum is a vector quantity. It has both magnitude and direction.
• Factors to consider include masses of the colliding bodies and their velocities.
• In inelastic collisions, kinetic energy is not conserved, but momentum is.

To solve collision problems, we must consider the momentum in different directions separately. First, let's focus on the x direction.
For this exercise, the man is initially moving north, so his x component of velocity is zero. The x component of momentum for any object is given by its mass times the x component of velocity.
In the x direction, using the conservation of momentum, the equation is: \[ m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2) v_{fx} \]
Here:

• \( m_1 \) and \( m_2 \) are the masses of the man and child respectively.
• \( v_{1x} \) and \( v_{2x} \) are the initial velocities of the man and child in the x direction.
• \( v_{fx} \) is the final combined velocity in the x direction.

Given the man's initial velocity in the x direction \( (v_{1x}) \) is zero, the equation simplifies. The exercise specifies the final velocity angle, so we use cosine to find the x component of the final velocity. \[ v_{fx} = v_f \times \text{cos}(35^{\text{°}}) = 2.46 \text{ m/s} \]
Plug these into the equation to find the child's initial x component velocity (\( v_{2x} \)): \[ v_{2x} = \frac{(m_1 + m_2) v_{fx}}{m_2} = 6.4 \text{ m/s} \]

Next, let's analyze the momentum in the y direction. Initially, the man is moving north, meaning all his velocity is in the y direction. The conservation of momentum in the y direction is given by: \[ m_1 v_{1y} + m_2 v_{2y} = (m_1 + m_2) v_{fy} \]
In this scenario:

• \(v_{1y} = 6 \text{ m/s} \)
• \(v_{2y} \) is the initial y component of velocity of the child.
• \(v_{fy} \) is the combined y component of velocity post-collision.

Using the given data and solving for \(v_{fy} \): \[ v_{fy} = v_f \times \text{sin}(35^{\text{°}}) = 1.72 \text{ m/s} \]
We then plug these values into the conservation equation to find \(v_{2y} \): \[ 60 \times 6 + 38 \times v_{2y} = (60+38) \times 1.72 \]
After solving for \( v_{2y} \), we get: \[ v_{2y} = -1.24 \text{ m/s} \] This negative sign indicates that before the collision, the child was moving in the opposite y direction compared to the man.

Velocity components help break down complex problems involving directions. By analyzing the x and y components separately, it becomes easier to apply physics principles accurately.
To find the magnitude of the child's initial velocity, we use the Pythagorean theorem. The two components we've calculated are \(v_{2x} = 6.4 \text{ m/s} \) and \( v_{2y} = -1.24 \text{ m/s} \).
Plug these into the equation: \[ v_2 = \text{sqrt}(v_{2x}^2 + v_{2y}^2) \]
Substituting the values, we get: \[ v_2 = \text{sqrt}(6.4^2 + (-1.24)^2) = 6.52 \text{ m/s} \]
To determine the direction, we use trigonometry. The direction angle \(\theta \) is found using the tangent function: \[ \theta = \text{tan}^{-1}\frac{v_{2y}}{v_{2x}} = \text{tan}^{-1}\frac{-1.24}{6.4} = -11^{\text{°}} \]
The negative sign indicates the direction of the child's initial velocity is south of east.