Derive an expression for the maximum vertical height of the dart sphere system

General
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derive an expression for the maximum vertical height of the dart sphere system

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To derive an expression for the maximum vertical height of the dart-sphere system, we need to analyze the problem using principles of classical mechanics. Our focus will be on energy conservation or kinematic equations, depending on the context of the problem setup. Let’s proceed step by step.

Step 1: Defining the Dart-Sphere System

In the dart-sphere system, let’s assume:

  • A dart of mass m_1 is launched.
  • It’s embedded in a stationary spherical target of mass m_2 upon collision.
  • After the dart sticks into the sphere, the two objects move together as a single system, meaning it’s an inelastic collision.
  • The dart is launched with an initial velocity, v_0.

We are tasked with calculating the maximum vertical height (H) that this combined dart-sphere system reaches after the collision.

Step 2: Identifying Principles to Use

  1. Momentum Conservation during Collision: Since the dart sticks into the sphere, the system undergoes an inelastic collision, so the momentum is conserved during this event.

  2. Energy Conservation Post-Collision: After the collision, the kinetic energy of the dart-sphere system is converted into potential energy as the system rises to its maximum vertical height.

Step 3: Applying Momentum Conservation

Before the collision:

  • Momentum of the dart: p_{\text{dart}} = m_1 v_0.
  • The sphere is stationary, so p_{\text{sphere}} = 0.

After the collision:

  • The dart and sphere move together with a velocity v_f, so the final combined momentum is:
    $$ p_{\text{system}} = (m_1 + m_2)v_f $$

Using the principle of conservation of momentum:
[
m_1 v_0 = (m_1 + m_2) v_f
]

This allows us to solve for the velocity of the system just after the collision:
[
v_f = \frac{m_1 v_0}{m_1 + m_2}
]

Step 4: Vertical Height Using Energy Conservation

After the dart has embedded itself in the sphere, kinetic energy turns into gravitational potential energy as the system rises.

  • The system’s initial kinetic energy just after the collision is:
    [
    KE_{\text{initial}} = \frac{1}{2}(m_1 + m_2)v_f^2
    ]

  • At maximum height, the system has zero kinetic energy and its gravitational potential energy is:
    [
    PE_{\text{final}} = (m_1 + m_2)gH
    ]

Using the law of conservation of mechanical energy:
[
KE_{\text{initial}} = PE_{\text{final}}
]

Substitute the expressions for kinetic and potential energy:
[
\frac{1}{2}(m_1 + m_2)v_f^2 = (m_1 + m_2)gH
]

Cancel the common term (m_1 + m_2) (since it’s nonzero):
[
\frac{1}{2}v_f^2 = gH
]

Solve for H:
[
H = \frac{v_f^2}{2g}
]

Step 5: Substitute v_f

From Step 3, we know:
[
v_f = \frac{m_1 v_0}{m_1 + m_2}
]

Substitute this expression for v_f into the equation for H:
[
H = \frac{\left( \frac{m_1 v_0}{m_1 + m_2} \right)^2}{2g}
]

Simplify the equation:
[
H = \frac{m_1^2 v_0^2}{(m_1 + m_2)^2 \cdot 2g}
]

Final Expression

The maximum vertical height of the dart-sphere system is:

H = \frac{m_1^2 v_0^2}{2g (m_1 + m_2)^2}

Key Takeaways:

  • The height depends on the masses of the dart (m_1) and the sphere (m_2), the initial velocity of the dart (v_0), and the gravitational acceleration (g).
  • Inelastic collisions reduce the system’s initial velocity, which lowers the height as compared to an elastic system.

Let me know if you’d like further clarification! @anonymous13