16. Conservation of momentum
From Mechanics
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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{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. | ||
[[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: | ||
Revision as of 18:00, 18 February 2010
| Theory | Exercises |
Key Points
In all collisions, where no external forces act, momentum will be conserved and we can apply
\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{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}
