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AP Physics C: Mechanics : Momentum and Impulse

Study concepts, example questions & explanations for AP Physics C: Mechanics

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Example Questions

Example Question #143 : Mechanics Exam

A baseball player hits a $\small 150g$ $\displaystyle \small 150g$ baseball initially moving at $\small 35\frac{m}{s}$ $\displaystyle \small 35\frac{m}{s}$, returning it at a speed of $\small 45\frac{m}{s}$ $\displaystyle \small 45\frac{m}{s}$ along the same path. If the ball was in contact with the bat for $\small 1ms$ $\displaystyle \small 1ms$, what magnitude of force did the ball experience during the moment of contact?

Possible Answers:

42kN $\displaystyle 42kN$

12kN $\displaystyle 12kN$

530N $\displaystyle 530N$

1.5kN $\displaystyle 1.5kN$

80kN $\displaystyle 80kN$

Correct answer:

12kN $\displaystyle 12kN$

Explanation:

Relevant equations:

$J = m\Delta v$ $\displaystyle J = m\Delta v$

$J = F_{avg}\Delta t$ $\displaystyle J = F_{avg}\Delta t$

Evaluate the impulse based on the mass and change of velocity.

$J = m\Delta v = (0.150kg)(80\frac{m}{s})=12 \frac{kg*m}{s}$ $\displaystyle J = m\Delta v = (0.150kg)(80\frac{m}{s})=12 \frac{kg*m}{s}$

Use the total impulse and time in the second equation.

$m\Delta v = F_{avg}\Delta t$ $\displaystyle m\Delta v = F_{avg}\Delta t$

$12\frac{kg*m}{s}=F_{avg}(0.001s)$ $\displaystyle 12\frac{kg*m}{s}=F_{avg}(0.001s)$

Solve for the average force.

$F_{avg}=\frac{12\frac{kg*m}{s}}{0.001s}=12,000N = 12kN$ $\displaystyle F_{avg}=\frac{12\frac{kg*m}{s}}{0.001s}=12,000N = 12kN$

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Example Question #11 : Momentum

Which of the following could be used as units of impulse?

Possible Answers:

$\frac{kg*m^2}{s}$ $\displaystyle \frac{kg*m^2}{s}$

$\frac{kg*m}{s^3}$ $\displaystyle \frac{kg*m}{s^3}$

$\frac{kg*m^2}{s^2}$ $\displaystyle \frac{kg*m^2}{s^2}$

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

$\frac{kg*m}{s^2}$ $\displaystyle \frac{kg*m}{s^2}$

Correct answer:

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

Explanation:

Relevant equations:

$J = \Delta p = m\Delta v$ $\displaystyle J = \Delta p = m\Delta v$

$J = F \Delta t$ $\displaystyle J = F \Delta t$

Impulse is defined as change in momentum, so has the same units as momentum. These units can easily be found using the given equations.

$J=m\Delta v=(kg)(\frac{m}{s})=\frac{kg*m}{s}$ $\displaystyle J=m\Delta v=(kg)(\frac{m}{s})=\frac{kg*m}{s}$

$J=F\Delta t=(N)(s)=N*s$ $\displaystyle J=F\Delta t=(N)(s)=N*s$

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Example Question #142 : Ap Physics C

A pitcher throws a 0.15kg baseball at $40\frac{m}{s}$ $\displaystyle 40\frac{m}{s}$ towards the batter and the batter hits the ball with his bat. The ball leaves the bat in the opposite direction at a speed of $45\frac{m}{s}$ $\displaystyle 45\frac{m}{s}$. Calculate the impulse experienced by the baseball.

Possible Answers:

$-17.1N\cdot s$ $\displaystyle -17.1N\cdot s$

$-6.25N\cdot s$ $\displaystyle -6.25N\cdot s$

$-3.5N\cdot s$ $\displaystyle -3.5N\cdot s$

$0N\cdot s$ $\displaystyle 0N\cdot s$

$-12.5N\cdot s$ $\displaystyle -12.5N\cdot s$

Correct answer:

$-12.5N\cdot s$ $\displaystyle -12.5N\cdot s$

Explanation:

To calculate the impulse, know that it is equal to the change in momentum.

$\textbf{J}=\Delta\textbf{p}=\textbf{p}_f-\textbf{p}_i$ $\displaystyle \textbf{J}=\Delta\textbf{p}=\textbf{p}_f-\textbf{p}_i$

Write the impulse equation in terms of mass and velocity.

$\textbf{J}=m\textbf{v}_f-m\textbf{v}_i$ $\displaystyle \textbf{J}=m\textbf{v}_f-m\textbf{v}_i$

In our case, the initial velocity of the baseball is $\textbf{v}_i=40\ \text{m/s}$ $\displaystyle \textbf{v}_i=40\ \text{m/s}$ and its final velocity is $\textbf{v}_f=-45\ \text{m/s}$ $\displaystyle \textbf{v}_f=-45\ \text{m/s}$, where the negative sign indicates that the ball is traveling in the opposite direction. Also, m = 0.15kg. The impulse on the ball can be calculated below.

$\textbf{J}=(0.15\ \text{kg})(-45\ \text{m/s})-(0.15\ \text{kg})(40\ \text{m/s})=-12.75{N\cdot s}$ $\displaystyle \textbf{J}=(0.15\ \text{kg})(-45\ \text{m/s})-(0.15\ \text{kg})(40\ \text{m/s})=-12.75{N\cdot s}$

The negative sign tells that the force on the baseball is opposed to the original momentum.

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Momentum and Impulse

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #143 : Mechanics Exam

Example Question #11 : Momentum

Example Question #142 : Ap Physics C

All AP Physics C: Mechanics Resources

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