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16. Conservation of momentum

From Mechanics

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Conservation of Momentum
 
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Key Results
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== '''Key Points''' ==
In all collisions, where no external forces act, momentum will be conserved and we can apply
In all collisions, where no external forces act, momentum will be conserved and we can apply
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<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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or
or
- 
<math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math>
<math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math>
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- 
- 
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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
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 <math>\text{m}{{\text{s}}^{-1}}</math>.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.
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.
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'''Solution'''
'''Solution'''
Before the collision:
Before the collision:
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<math>{{u}_{B}}=250</math>
<math>{{u}_{B}}=250</math>
and
and
<math>{{u}_{T}}=0</math>
<math>{{u}_{T}}=0</math>
- 
After the collision:
After the collision:
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<math>{{v}_{B}}={{v}_{T}}=10</math>
<math>{{v}_{B}}={{v}_{T}}=10</math>
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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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<math>{{m}_{B}}=0\textrm{.}04</math>
<math>{{m}_{B}}=0\textrm{.}04</math>
- 
Using conservation of momentum gives:
Using conservation of momentum gives:
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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}} \\
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& {{m}_{T}}=\frac{10-0\textrm{.}4}{10}=0\textrm{.}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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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.
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'''
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and
and
<math>{{u}_{C}}=0</math>
<math>{{u}_{C}}=0</math>
- 
After the collision:
After the collision:
<math>{{v}_{V}}={{v}_{C}}=v</math>
<math>{{v}_{V}}={{v}_{C}}=v</math>
- 
The masses should be converted to kilograms:
The masses should be converted to kilograms:
-
 
<math>{{m}_{V}}=2500</math>
<math>{{m}_{V}}=2500</math>
and
and
<math>{{m}_{C}}=1500</math>
<math>{{m}_{C}}=1500</math>
- 
Using conservation of momentum gives:
Using conservation of momentum gives:
-
 
<math>\begin{align}
<math>\begin{align}
& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\
& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\
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& v=\frac{30000}{4000}=7\textrm{.}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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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.
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'''
'''Solution'''
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and
and
<math>{{u}_{B}}=-u</math>
<math>{{u}_{B}}=-u</math>
- 
After the collision:
After the collision:
<math>{{v}_{A}}={{v}_{B}}=v</math>
<math>{{v}_{A}}={{v}_{B}}=v</math>
- 
Using conservation of momentum gives:
Using conservation of momentum gives:
-
 
<math>\begin{align}
<math>\begin{align}
& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\
& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\
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<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
. If the particles coalesce during the collision, find their final velocity.
. If the particles coalesce during the collision, find their final velocity.
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'''Solution'''
'''Solution'''
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and
and
<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>
<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</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:
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and
and
<math>{{m}_{B}}=3</math>
<math>{{m}_{B}}=3</math>
- 
Using conservation of momentum gives:
Using conservation of momentum gives:
- 
<math>\begin{align}
<math>\begin{align}
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& \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}
\end{align}</math>
\end{align}</math>
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[[Image:E16.4fig1.GIF]]
[[Image:E16.4fig1.GIF]]
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<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>
<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>
and
and
<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>
<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>
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This diagram shows the velocity after the collision.
This diagram shows the velocity after the collision.
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<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}</math>
- 
Using conservation of momentum gives:
Using conservation of momentum gives:
- 
<math>\begin{align}
<math>\begin{align}
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& 1200\times 15\mathbf{i}+1400\times U\mathbf{j}=2600(V\cos 20{}^\circ \mathbf{i}+V\sin 20{}^\circ \mathbf{j})
& 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}</math>
\end{align}</math>
- 
Considering the <math>\mathbf{i}</math> component gives:
Considering the <math>\mathbf{i}</math> component gives:
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<math>\begin{align}
<math>\begin{align}
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& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7\textrm{.}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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Considering the <math>\mathbf{j}</math> component gives:
Considering the <math>\mathbf{j}</math> component gives:
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<math>\begin{align}
<math>\begin{align}

Revision as of 17:21, 18 February 2010

       Theory          Exercises      


Key Points

In all collisions, where no external forces act, momentum will be conserved and we can apply

mAvA+mBvB=mAuA+mBuB

or

mAvA+mBvB=mAuA+mBuB


Example 16.1

A bullet of mass 40 grams is travelling horizontally at 250 ms−1. It hits a wooden trolley that is at rest. The bullet and trolley then move together at 10 ms−1.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.

Solution

Before the collision:

uB=250 and uT=0

After the collision:

vB=vT=10

Also the mass of the bullet should be converted to kg:

mB=0.04

Using conservation of momentum gives:

mBuB+mTuT=mBvB+mTvT0.04250+mT0=0.0410+mT1010=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 ms−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: uV=12 and uC=0

After the collision: vV=vC=v

The masses should be converted to kilograms:

mV=2500 and mC=1500

Using conservation of momentum gives:

mVuV+mCuC=mVvV+mCvC250012+15000=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: uA=4u and uB=−u

After the collision: vA=vB=v

Using conservation of momentum gives:

mAuA+mBuB=mAvA+mBvBm4u+3m(−u)=mv+3mvmu=4mvv=mu4mu=4u


Example 16.4

A particle, A, of mass 2 kg has velocity (4i+2j) ms-1 . It collides with a second particle, B, of mass 3 kg and velocity (2i−4j) ms-1 . If the particles coalesce during the collision, find their final velocity.

Solution

Before the collision: uA=4i+2j and uB=2i−4j

After the collision: vA=vB=v

The masses are defined: mA=2 and mB=3

Using conservation of momentum gives:

mAuA+mBuB=mAvA+mBvB2(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 ms−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 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.

Image:E16.4fig1.GIF

uC=15i and uV=Uj

This diagram shows the velocity after the collision.

Image:E16.4fig2.GIF

\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}

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