16. Conservation of momentum
From Mechanics
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== '''Key Points''' == | == '''Key Points''' == | ||
| - | In all collisions, where no external forces act, momentum will be conserved and | + | In all collisions, where no external forces act, momentum will be conserved and thus the total momentum just after the collision will be the same as the total momentum just before the collision, which gives, |
<math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> | <math>{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}</math> | ||
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Before the collision: | Before the collision: | ||
| - | <math>{{u}_{B}}=250</math> | + | <math>{{u}_{B}}=250\text{ m}{{\text{s}}^{\text{-1}}}</math> |
and | and | ||
<math>{{u}_{T}}=0</math> | <math>{{u}_{T}}=0</math> | ||
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After the collision: | After the collision: | ||
| - | <math>{{v}_{B}}={{v}_{T}}=10</math> | + | <math>{{v}_{B}}={{v}_{T}}=10\text{ m}{{\text{s}}^{\text{-1}}}</math> |
Also the mass of the bullet should be converted to kg: | Also the mass of the bullet should be converted to kg: | ||
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Before the collision: | Before the collision: | ||
| - | <math>{{u}_{V}}=12</math> | + | <math>{{u}_{V}}=12\text{ m}{{\text{s}}^{\text{-1}}}</math> |
and | and | ||
<math>{{u}_{C}}=0</math> | <math>{{u}_{C}}=0</math> | ||
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Before the collision: | Before the collision: | ||
| - | <math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math> | + | <math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> |
and | and | ||
| - | <math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math> | + | <math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> |
After the collision: | After the collision: | ||
<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math> | <math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math> | ||
| - | The masses are defined: | + | The masses are defined (in kg): |
<math>{{m}_{A}}=2</math> | <math>{{m}_{A}}=2</math> | ||
and | and | ||
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& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\ | & 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\ | ||
& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\ | & 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\ | ||
| - | & \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} | + | & \mathbf{v}=\frac{\smash{14\mathbf{i}-8\mathbf{j}}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}} |
\end{align}</math> | \end{align}</math> | ||
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[[Image:E16.4fig1.GIF]] | [[Image:E16.4fig1.GIF]] | ||
| - | <math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math> | + | <math>{{\mathbf{u}}_{C}}=15\mathbf{i}\text{ m}{{\text{s}}^{\text{-1}}}</math> |
and | and | ||
| - | <math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math> | + | <math>{{\mathbf{u}}_{V}}=U\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> |
This diagram shows the velocity after the collision. | This diagram shows the velocity after the collision. | ||
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[[Image:E16.4fig2.GIF]] | [[Image:E16.4fig2.GIF]] | ||
| - | <math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}</math> | + | <math>{{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> |
Using conservation of momentum gives: | Using conservation of momentum gives: | ||
Current revision
| Theory | Exercises |
Key Points
In all collisions, where no external forces act, momentum will be conserved and thus the total momentum just after the collision will be the same as the total momentum just before the collision, which gives,
\displaystyle {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}
or
\displaystyle {{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}
A bullet of mass 40 grams is travelling horizontally at 250 \displaystyle \text{m}{{\text{s}}^{-1}}. It hits a wooden trolley that is at rest. The bullet and trolley then move together at 10 \displaystyle \text{m}{{\text{s}}^{-1}}.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.
Solution
Before the collision:
\displaystyle {{u}_{B}}=250\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{u}_{T}}=0
After the collision:
\displaystyle {{v}_{B}}={{v}_{T}}=10\text{ m}{{\text{s}}^{\text{-1}}}
Also the mass of the bullet should be converted to kg:
\displaystyle {{m}_{B}}=0\textrm{.}04
Using conservation of momentum gives:
\displaystyle \begin{align} & {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\ & 0\textrm{.}04\times 250+{{m}_{T}}\times 0=0\textrm{.}04\times 10+{{m}_{T}}\times 10 \\ & 10=0\textrm{.}4+10{{m}_{T}} \\ & {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}96\text{ kg} \end{align}
A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 \displaystyle \text{m}{{\text{s}}^{-1}} and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.
Solution
Before the collision: \displaystyle {{u}_{V}}=12\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{u}_{C}}=0
After the collision: \displaystyle {{v}_{V}}={{v}_{C}}=v
The masses should be converted to kilograms:
\displaystyle {{m}_{V}}=2500 and \displaystyle {{m}_{C}}=1500
Using conservation of momentum gives:
\displaystyle \begin{align} & {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\ & 2500\times 12+1500\times 0=2500v+1500v \\ & 30000=4000v \\ & v=\frac{30000}{4000}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}} \end{align}
Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.
Solution
Before the collision: \displaystyle {{u}_{A}}=4u and \displaystyle {{u}_{B}}=-u
After the collision: \displaystyle {{v}_{A}}={{v}_{B}}=v
Using conservation of momentum gives:
\displaystyle \begin{align} & {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\ & m\times 4u+3m\times (-u)=mv+3mv \\ & mu=4mv \\ & v=\frac{mu}{4mu}=\frac{u}{4} \end{align}
A particle, A, of mass 2 kg has velocity \displaystyle (4\mathbf{i}+2\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} . It collides with a second particle, B, of mass 3 kg and velocity \displaystyle (2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} . If the particles coalesce during the collision, find their final velocity.
Solution
Before the collision: \displaystyle {{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}
After the collision: \displaystyle {{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}
The masses are defined (in kg): \displaystyle {{m}_{A}}=2 and \displaystyle {{m}_{B}}=3
Using conservation of momentum gives:
\displaystyle \begin{align} & {{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}={{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}} \\ & 2\times (4\mathbf{i}+2\mathbf{j})+3\times (2\mathbf{i}-4\mathbf{j})=2\mathbf{v}+3\mathbf{v} \\ & 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\ & 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\ & \mathbf{v}=\frac{\smash{14\mathbf{i}-8\mathbf{j}}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} \text{ m}{{\text{s}}^{\text{-1}}} \end{align}
A car, of mass 1.2 tonnes, is travelling at 15 \displaystyle \text{m}{{\text{s}}^{-1}}, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20\displaystyle {}^\circ to the original motion of the car. Find the speed of the heavier van just before the collision.
Solution
This diagram shows the velocities before the collision.
\displaystyle {{\mathbf{u}}_{C}}=15\mathbf{i}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\mathbf{u}}_{V}}=U\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}
This diagram shows the velocity after the collision.
\displaystyle {{\mathbf{v}}_{C}}={{\mathbf{v}}_{V}}=V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}
Using conservation of momentum gives:
\displaystyle \begin{align} & {{m}_{C}}{{\mathbf{u}}_{C}}+{{m}_{V}}{{\mathbf{u}}_{V}}={{m}_{C}}{{\mathbf{v}}_{C}}+{{m}_{V}}{{\mathbf{v}}_{V}} \\ & 1200\times 15\mathbf{i}+1400\times U\mathbf{j}=2600(V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j}) \end{align}
Considering the \displaystyle \mathbf{i} component gives:
\displaystyle \begin{align} & 1200\times 15=2600V\cos 20{}^\circ \\ & V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7\textrm{.}36\text{ m}{{\text{s}}^{\text{-1}}} \end{align}
Considering the \displaystyle \mathbf{j} component gives:
\displaystyle \begin{align} & 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\ & U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4\textrm{.}68\text{ m}{{\text{s}}^{\text{-1}}} \end{align}
