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AP Physics B : Understanding Conservation of Momentum

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Example Question #1 : Understanding Conservation Of Momentum

A baseball pitcher throws a baseball at $35\frac{m}{s}$ $\displaystyle 35\frac{m}{s}$ horizontally in the positive direction to a batter, who hits the ball in the opposite direction at $40\frac{m}{s}$ $\displaystyle 40\frac{m}{s}$. What is the change in momentum of the baseball if the baseball has a mass of 0.45kg $\displaystyle 0.45kg$?

Possible Answers:

$-18\frac{kg*m}{s}$ $\displaystyle -18\frac{kg*m}{s}$

$33\frac{kg*m}{s}$ $\displaystyle 33\frac{kg*m}{s}$

$-33\frac{kg*m}{s}$ $\displaystyle -33\frac{kg*m}{s}$

$-15\frac{kg*m}{s}$ $\displaystyle -15\frac{kg*m}{s}$

$18\frac{kg*m}{s}$ $\displaystyle 18\frac{kg*m}{s}$

Correct answer:

$-33\frac{kg*m}{s}$ $\displaystyle -33\frac{kg*m}{s}$

Explanation:

In order to calculate the change in momentum, we must find the initial and final momentum of the baseball, and then find the difference.

p=mv $\displaystyle p=mv$

Use the given velocities and mass to calculate the initial and final values.

$p_1=mv_{1}=(0.45kg)(35\frac{m}{s})=15\frac{kg*m}{s}$ $\displaystyle p_1=mv_{1}=(0.45kg)(35\frac{m}{s})=15\frac{kg*m}{s}$

$p_2=mv_{2}=(0.45kg)(-40\frac{m}{s})=-18\frac{kg*m}{s}$ $\displaystyle p_2=mv_{2}=(0.45kg)(-40\frac{m}{s})=-18\frac{kg*m}{s}$

The initial momentum is positive because the problem states that the ball was originally thrown in the positive direction. The final momentum is negative due to the change in direction.

Now we find the change in momentum:

$\Delta p=p_2-p_1$ $\displaystyle \Delta p=p_2-p_1$

$\Delta p=(-18\frac{kg*m}{s})-(15\frac{kg*m}{s})=-33\frac{kg*m}{s}$ $\displaystyle \Delta p=(-18\frac{kg*m}{s})-(15\frac{kg*m}{s})=-33\frac{kg*m}{s}$

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AP Physics B : Understanding Conservation of Momentum

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Example Question #1 : Understanding Conservation Of Momentum

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