The Impulse Momentum Equation States The Relationship Betwee
The Impulse Momentum Equation States The Relationship Between A Force
The impulse-momentum equation relates the force exerted on an object, its mass, and the resulting change in velocity. Specifically, it states that the impulse applied to an object equals the change in its momentum, which can be expressed as:
\[ F(t) \times \Delta t = m \times \Delta v \]
where \( F(t) \) is the force as a function of time, \( m \) is the mass of the object, \( \Delta v = v_b - v_a \) is the change in velocity, and \( \Delta t \) is the duration of the force. This relationship helps us analyze the dynamics of objects subjected to variable forces.
In the specific problem, the force \( F(t) \) exerted by the baseball bat on the ball is given as a function of time \( t \):
\[ F(t) = 9 - 108(t - 0.0003)^2 \quad \text{(thousand pounds)} \quad \text{for} \quad t \in [0, 0.0006] \]
The mass of the baseball is \( m = 0.01 \) slugs, and our goals are to determine the maximum force exerted on the ball and estimate the change in velocity \( v \) over the interval.
Analysis of the Force Function and Determination of Maximum Force
The force function is quadratic in form, with a parabola opening downward (since the coefficient of \( (t - 0.0003)^2 \) is negative). To find the maximum force, we focus on the vertex of the parabola, which occurs at \( t = 0.0003 \) seconds, owing to the expression \( (t - 0.0003)^2 \).
The force at \( t = 0.0003 \) seconds can be computed as:
\[ F(0.0003) = 9 - 108 \times (0.0003 - 0.0003)^2 = 9 - 0 = 9 \quad \text{(thousand pounds)} \]
Thus, the maximum force is 9 thousand pounds, equivalent to:
\[ F_{\max} = 9,000 \text{ pounds} \]
Calculating the Change in Velocity
The impulse-momentum theorem in integral form states that the change in momentum equals the impulse:
\[ m \times \Delta v = \int_{a}^{b} F(t) \, dt \]
where \( a = 0 \) and \( b = 0.0006 \) seconds. Since force is given in thousand pounds, we need to convert units to pounds for consistency. The force function can be expressed as:
\[ F(t) = (9 - 108(t - 0.0003)^2) \times 1000 \quad \text{pounds} \]
The change in velocity is then:
\[ \Delta v = \frac{1}{m} \times \int_{0}^{0.0006} F(t) \, dt \]
Substituting the given values:
\[ \Delta v = \frac{1}{0.01} \times \int_{0}^{0.0006} 1000 \times (9 - 108(t - 0.0003)^2) \, dt \]
which simplifies to:
\[ \Delta v = 100 \times \int_{0}^{0.0006} (9 - 108(t - 0.0003)^2) \, dt \]
Performing the Integral
We can split the integral into two parts:
\[
\int_{0}^{0.0006} 9 \, dt - 108 \int_{0}^{0.0006} (t - 0.0003)^2 \, dt
\]
The first integral is straightforward:
\[
9 \times (0.0006 - 0) = 9 \times 0.0006 = 0.0054
\]
For the second integral, set \( u = t - 0.0003 \). When \( t = 0 \), \( u = -0.0003 \). When \( t = 0.0006 \), \( u = 0.0003 \). Changing limits accordingly:
\[
\int_{-0.0003}^{0.0003} u^2 \, du = \left[ \frac{u^3}{3} \right]_{-0.0003}^{0.0003} = \frac{(0.0003)^3 - (-0.0003)^3}{3}
\]
Since \( u^3 \) is an odd function, the numerator becomes:
\[
(0.0003)^3 - (-0.0003)^3 = (0.0003)^3 + (0.0003)^3 = 2 \times (0.0003)^3
\]
Calculating \( (0.0003)^3 \):
\[
(0.0003)^3 = 0.0003 \times 0.0003 \times 0.0003 = 2.7 \times 10^{-11}
\]
Thus, the integral becomes:
\[
\frac{2 \times 2.7 \times 10^{-11}}{3} = \frac{5.4 \times 10^{-11}}{3} = 1.8 \times 10^{-11}
\]
Multiplying this by 108:
\[ 108 \times 1.8 \times 10^{-11} = 1.944 \times 10^{-9} \]
Finally, the total impulse is:
\[
0.0054 - 1.944 \times 10^{-9} \approx 0.0054
\]
Now, calculating the change in velocity:
\[
\Delta v = 100 \times 0.0054 = 0.54 \text{ ft/s}
\]
Therefore, the estimated change in velocity of the baseball due to the impact is approximately 0.54 ft/s.
Conclusions
The maximum force exerted on the baseball during the collision is approximately 9,000 pounds, occurring at \( t = 0.0003 \) seconds. The impulse imparted results in a modest increase in the ball's velocity, estimated at about 0.54 feet per second. This analysis demonstrates how variable forces during impacts can be effectively examined using the impulse-momentum theorem, highlighting the importance of precise force modeling in physics and sports science.
References
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