16. Conservation of momentum
From Mechanics
| Line 15: | Line 15: | ||
<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> | ||
| + | |||
or | or | ||
| - | <math>{{m}_{A}}{{\mathbf{v}}_ | + | |
| + | <math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math> | ||
| + | |||
| + | |||
Example 16.1 | Example 16.1 | ||
| - | A bullet of mass 40 grams is travelling horizontally at 250 | + | A bullet of mass 40 grams is travelling horizontally at 250 <math>\text{m}{{\text{s}}^{-1}}</math>. It hits a wooden trolley that is at rest. The bullet and trolley then move together at |
| - | 10 | + | 10 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley. |
Solution | Solution | ||
| Line 59: | Line 63: | ||
Example 16.2 | 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 | + | 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 <math>\text{m}{{\text{s}}^{-1}}</math> and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision. |
Solution | Solution | ||
| Line 154: | Line 158: | ||
Example 16.5 | Example 16.5 | ||
| - | A car, of mass 1.2 tonnes, is travelling at 15 | + | A car, of mass 1.2 tonnes, is travelling at 15 <math>\text{m}{{\text{s}}^{-1}}</math>, 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<math>{}^\circ </math> to the original motion of the car. Find the speed of the heavier van just before the collision. |
Solution | Solution | ||
| Line 189: | Line 193: | ||
| - | Considering the i component gives: | + | Considering the <math>\mathbf{i}</math> component gives: |
| Line 198: | Line 202: | ||
| - | Considering the j component gives: | + | Considering the <math>\mathbf{j}</math> component gives: |
Revision as of 15:39, 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:
04
Using conservation of momentum gives:
04
250+mT0=00410+mT1010=04+10mTmT=1010−04=096 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+15000=2500v+1500v30000=4000vv=400030000=75 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=28i−16j
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 }=7.36\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 =4.68\text{ m}{{\text{s}}^{\text{-1}}}
\end{align}
