Questions: A catcher's mitt recoils a distance of 12.9 cm in bringing a 142-gram baseball to a stop. If the applied force is 588 N, then what was the speed of the baseball at the moment of contact with the catcher's mitt?

Solution Steps

We need to find the initial speed of the baseball when it makes contact with the catcher's mitt. The given data includes:

• Recoil distance of the mitt: \( d = 12.9 \, \text{cm} = 0.129 \, \text{m} \)
• Mass of the baseball: \( m = 142 \, \text{g} = 0.142 \, \text{kg} \)
• Applied force: \( F = 588 \, \text{N} \)

The work-energy principle states that the work done by external forces is equal to the change in kinetic energy plus the change in potential energy. Since the baseball comes to a stop, the final kinetic energy is zero, and we assume no change in potential energy. Thus, the equation simplifies to:

\[ \mathrm{KE}_{\mathbf{i}} + \mathbf{W}_{\mathbf{ext}} = \mathrm{KE}_{\mathbf{f}} \]

Since \(\mathrm{KE}_{\mathbf{f}} = 0\), we have:

\[ \mathrm{KE}_{\mathbf{i}} = -\mathbf{W}_{\mathbf{ext}} \]

The work done by the force is given by:

\[ \mathbf{W}_{\mathbf{ext}} = F \cdot d \]

Substituting the given values:

\[ \mathbf{W}_{\mathbf{ext}} = 588 \, \text{N} \times 0.129 \, \text{m} = 75.852 \, \text{J} \]

The initial kinetic energy \(\mathrm{KE}_{\mathbf{i}}\) is equal to the negative of the work done:

\[ \mathrm{KE}_{\mathbf{i}} = 75.852 \, \text{J} \]

The initial kinetic energy is also given by:

\[ \mathrm{KE}_{\mathbf{i}} = \frac{1}{2} m v^2 \]

Solving for \(v\):

\[ v^2 = \frac{2 \cdot \mathrm{KE}_{\mathbf{i}}}{m} = \frac{2 \cdot 75.852}{0.142} \]

\[ v^2 = 1068.338 \]

\[ v = \sqrt{1068.338} \approx 32.678 \, \text{m/s} \]

Final Answer