15. Momentum and impulse
From Mechanics
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| - | 15. Momentum and Impulse | ||
| - | Key | + | |
| + | == '''Key Points''' == | ||
Momentum is defined as the product of the mass and velocity of a body. | Momentum is defined as the product of the mass and velocity of a body. | ||
Momentum = | Momentum = | ||
| - | <math>mv</math> | + | <math>mv</math>, |
| - | or Momentum = | + | or in vector form, Momentum = |
<math>m\mathbf{v}</math> | <math>m\mathbf{v}</math> | ||
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If the velocity of a body, of mass | If the velocity of a body, of mass | ||
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<math>mv</math> | <math>mv</math> | ||
- | - | ||
| - | <math>mu</math> | + | <math>mu</math>, |
| + | or in vector form, | ||
| + | <math>m\mathbf{v}-m\mathbf{u}</math> | ||
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<math>I=mv-mu</math> | <math>I=mv-mu</math> | ||
| - | or | + | , |
| + | or in vector form, | ||
<math>\mathbf{I}=m\mathbf{v}-m\mathbf{u}</math> | <math>\mathbf{I}=m\mathbf{v}-m\mathbf{u}</math> | ||
The relationships | The relationships | ||
| - | <math>I=Ft</math> | + | <math>I=Ft</math>, |
| - | or | + | or in vector form, |
<math>\mathbf{I}=\mathbf{F}t</math> | <math>\mathbf{I}=\mathbf{F}t</math> | ||
can be used where | can be used where | ||
| - | <math>F</math> | + | <math>F</math>, |
| - | + | or in vector form, <math>\mathbf{F}</math> | |
| - | <math>F</math> | + | is constant or may |
| - | is assumed to be constant and represents the average force. | + | be assumed to be constant and represents the average force. |
| - | + | ||
'''[[Example 15.1]]''' | '''[[Example 15.1]]''' | ||
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Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line | Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line | ||
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Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives: | Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives: | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \text{Momentum}=1200\times 30 \\ | & \text{Momentum}=1200\times 30 \\ | ||
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& =-6600\text{ Ns} | & =-6600\text{ Ns} | ||
\end{align}</math> | \end{align}</math> | ||
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Use the formula | Use the formula | ||
<math>\text{Impulse }=Ft</math> | <math>\text{Impulse }=Ft</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& -6600=F\times 2\textrm{.}2 \\ | & -6600=F\times 2\textrm{.}2 \\ | ||
& F=-3000\text{ N} | & F=-3000\text{ N} | ||
\end{align}</math> | \end{align}</math> | ||
| - | |||
The average force has magnitude 3000 N | The average force has magnitude 3000 N | ||
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'''[[Example 15.3]]''' | '''[[Example 15.3]]''' | ||
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The speed of a car, travelling in a straight line, is reduced from 20 <math>\text{m}{{\text{s}}^{-1}}</math> to | The speed of a car, travelling in a straight line, is reduced from 20 <math>\text{m}{{\text{s}}^{-1}}</math> to | ||
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First find the impulse (change in momentum). | First find the impulse (change in momentum). | ||
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<math>\begin{align} | <math>\begin{align} | ||
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& =-6000\text{ Ns} | & =-6000\text{ Ns} | ||
\end{align}</math> | \end{align}</math> | ||
| - | |||
Use the formula | Use the formula | ||
<math>\text{Impulse }=Ft</math> | <math>\text{Impulse }=Ft</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
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[[Image:E15.4.GIF]] | [[Image:E15.4.GIF]] | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \text{Initial Momentum }=\text{ m}\mathbf{u} \\ | & \text{Initial Momentum }=\text{ m}\mathbf{u} \\ | ||
& =0\textrm{.}2\times 10\mathbf{i} \\ | & =0\textrm{.}2\times 10\mathbf{i} \\ | ||
| - | & =2\mathbf{i} | + | & =2\mathbf{i} \text{ Ns} |
\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \text{Final Momentum }=m\mathbf{v} \\ | & \text{Final Momentum }=m\mathbf{v} \\ | ||
& =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\ | & =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\ | ||
| - | & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j} | + | & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j} \text{ Ns} |
\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\ | & \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\ | ||
& =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\ | & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\ | ||
| - | & =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j} | + | & =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j} \text{ Ns} |
\end{align}</math> | \end{align}</math> | ||
| + | |||
'''[[Example 15.5]]''' | '''[[Example 15.5]]''' | ||
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& =2\textrm{.}4\mathbf{i} | & =2\textrm{.}4\mathbf{i} | ||
\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
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\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\ | & \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\ | ||
& =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i} | & =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i} | ||
\end{align}</math> | \end{align}</math> | ||
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The diagram shows the impulse. | The diagram shows the impulse. | ||
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using Pythagoras: | using Pythagoras: | ||
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<math>I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}</math> | <math>I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}</math> | ||
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The angle, <math>\alpha </math>, can be found using trigonometry: | The angle, <math>\alpha </math>, can be found using trigonometry: | ||
Current revision
| Theory | Exercises |
Key Points
Momentum is defined as the product of the mass and velocity of a body.
Momentum = \displaystyle mv, or in vector form, Momentum = \displaystyle m\mathbf{v}
If the velocity of a body, of mass \displaystyle m , changes from \displaystyle u to \displaystyle v , then
Change in momentum or impulse = \displaystyle mv - \displaystyle mu, or in vector form, \displaystyle m\mathbf{v}-m\mathbf{u}
We often use
\displaystyle I=mv-mu , or in vector form, \displaystyle \mathbf{I}=m\mathbf{v}-m\mathbf{u}
The relationships
\displaystyle I=Ft,
or in vector form,
\displaystyle \mathbf{I}=\mathbf{F}t
can be used where
\displaystyle F,
or in vector form, \displaystyle \mathbf{F}
is constant or may
be assumed to be constant and represents the average force.
Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line at 30 \displaystyle \text{m}{{\text{s}}^{-1}}.
Solution
Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:
\displaystyle \begin{align} & \text{Momentum}=1200\times 30 \\ & =36000\text{ Ns} \end{align}
A car, of mass 1.1 tonnes, which was initially travelling at 6 \displaystyle \text{m}{{\text{s}}^{-1}} is brought to rest in 2.2 seconds by a wall. Find the average force exerted by the wall on the car.
Solution
First find the impulse (change in momentum).
\displaystyle \begin{align} & \text{Impulse }=mv-mu \\ & =1100\times 0-1100\times 6 \\ & =-6600\text{ Ns} \end{align}
Use the formula \displaystyle \text{Impulse }=Ft
\displaystyle \begin{align} & -6600=F\times 2\textrm{.}2 \\ & F=-3000\text{ N} \end{align}
The average force has magnitude 3000 N
The speed of a car, travelling in a straight line, is reduced from 20 \displaystyle \text{m}{{\text{s}}^{-1}} to 15 \displaystyle \text{m}{{\text{s}}^{-1}} in 20 seconds. The mass of the car is 1200 kg. Find the average force acting on the car.
Solution
First find the impulse (change in momentum).
\displaystyle \begin{align} & \text{Impulse }=mv-mu \\ & 1200\times 15-1200\times 20 \\ & =-6000\text{ Ns} \end{align}
Use the formula \displaystyle \text{Impulse }=Ft
\displaystyle \begin{align} & -6000=F\times 20 \\ & F=-300\text{ N} \\ \end{align}
A ball has a mass of 200 grams. It initially travels horizontally at 10 \displaystyle \text{m}{{\text{s}}^{-1}}. After being hit it travels at 16 \displaystyle \text{m}{{\text{s}}^{-1}} at an angle of 60\displaystyle {}^\circ above the horizontal. Find the initial and final momentum of the ball and the change in momentum.
Solution
The diagrams show the initial and final velocities of the ball and the unit vectors \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.
\displaystyle \begin{align} & \text{Initial Momentum }=\text{ m}\mathbf{u} \\ & =0\textrm{.}2\times 10\mathbf{i} \\ & =2\mathbf{i} \text{ Ns} \end{align}
\displaystyle \begin{align} & \text{Final Momentum }=m\mathbf{v} \\ & =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\ & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j} \text{ Ns} \end{align}
\displaystyle \begin{align} & \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\ & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\ & =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j} \text{ Ns} \end{align}
A ball is travelling horizontally at 8 \displaystyle \text{m}{{\text{s}}^{-1}}, when it is hit. After being hit it initially travels upwards at 6 \displaystyle \text{m}{{\text{s}}^{-1}}. The mass of the ball is 300 grams. Find the magnitude and direction of the impulse on the ball.
Solution
The diagrams show the initial and final velocities of the ball and the unit vectors \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.
\displaystyle \begin{align} & \text{Initial Momentum }=\text{ }m\mathbf{u} \\ & =0\textrm{.}3\times 8\mathbf{i} \\ & =2\textrm{.}4\mathbf{i} \end{align}
\displaystyle \begin{align} & \text{Final Momentum }=m\mathbf{v} \\ & =0\textrm{.}3\times 6\mathbf{j} \\ & =1\textrm{.}8\mathbf{j} \end{align}
\displaystyle \begin{align} & \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\ & =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i} \end{align}
The diagram shows the impulse.
The magnitude of the impulse, \displaystyle I, is found using Pythagoras:
\displaystyle I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}
The angle, \displaystyle \alpha , can be found using trigonometry:
