16. Conservation of momentum
From Mechanics
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- | <math>{{m}_{B}}=0.04</math> | + | <math>{{m}_{B}}=0\textrm{.}04</math> |
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<math>\begin{align} | <math>\begin{align} | ||
& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\ | & {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\ | ||
- | & 0.04\times 250+{{m}_{T}}\times 0=0.04\times 10+{{m}_{T}}\times 10 \\ | + | & 0\textrm{.}04\times 250+{{m}_{T}}\times 0=0\textrm{.}04\times 10+{{m}_{T}}\times 10 \\ |
- | & 10=0.4+10{{m}_{T}} \\ | + | & 10=0\textrm{.}4+10{{m}_{T}} \\ |
- | & {{m}_{T}}=\frac{10-0.4}{10}=0.96\text{ kg} | + | & {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}96\text{ kg} |
\end{align}</math> | \end{align}</math> | ||
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& 2500\times 12+1500\times 0=2500v+1500v \\ | & 2500\times 12+1500\times 0=2500v+1500v \\ | ||
& 30000=4000v \\ | & 30000=4000v \\ | ||
- | & v=\frac{30000}{4000}=7.5\text{ m}{{\text{s}}^{\text{-1}}} | + | & v=\frac{30000}{4000}=7\textrm{.}5\text{ m}{{\text{s}}^{\text{-1}}} |
\end{align}</math> | \end{align}</math> | ||
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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}=2.8\mathbf{i}-1.6\mathbf{j} | + | & \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2\textrm{.}8\mathbf{i}-1\textrm{.}6\mathbf{j} |
\end{align}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& 1200\times 15=2600V\cos 20{}^\circ \\ | & 1200\times 15=2600V\cos 20{}^\circ \\ | ||
- | & V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7.36\text{ m}{{\text{s}}^{\text{-1}}} | + | & 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}</math> | \end{align}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
& 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\ | & 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\ | ||
- | & U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4.68\text{ m}{{\text{s}}^{\text{-1}}} | + | & U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4\textrm{.}68\text{ m}{{\text{s}}^{\text{-1}}} |
\end{align}</math> | \end{align}</math> |
Revision as of 15:46, 29 September 2009
Theory | Exercises |
Conservation of Momentum
Key Results
In all collisions, where no external forces act, momentum will be conserved and we can apply
or
Example 16.1
A bullet of mass 40 grams is travelling horizontally at 250
Solution Before the collision:
After the collision:
Also the mass of the bullet should be converted to kg:
Using conservation of momentum gives:
250+mT
0=0.04
10+mT
1010=0.4+10mTmT=1010−0.4=0.96 kg
Example 16.2
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
Solution
Before the collision:
After the collision:
The masses should be converted to kilograms:
Using conservation of momentum gives:
12+1500
0=2500v+1500v30000=4000vv=400030000=7.5 ms-1
Example 16.3
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:
After the collision:
Using conservation of momentum gives:
4u+3m
(−u)=mv+3mvmu=4mvv=mu4mu=4u
Example 16.4
A particle, A, of mass 2 kg has velocity
Solution
Before the collision:
After the collision:
The masses are defined:
Using conservation of momentum gives:
(4i+2j)+3
(2i−4j)=2v+3v8i+4j+6i−12j=5v14i−8j=5vv=514i−8j=2.8i−1.6j
Example 16.5
A car, of mass 1.2 tonnes, is travelling at 15
Solution This diagram shows the velocities before the collision.
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}
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}