16. Conservation of momentum

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Conservation of Momentum
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Key Results
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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>
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or
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<math>{{m}_{A}}{{\mathbf{v}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{v}}_{\mathbf{B}}}={{m}_{A}}{{\mathbf{u}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{u}}_{\mathbf{B}}}</math>
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Example 16.1
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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
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10 ms-1.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.
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Solution
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Before the collision:
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<math>{{u}_{B}}=250</math>
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and
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<math>{{u}_{T}}=0</math>
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After the collision:
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<math>{{v}_{B}}={{v}_{T}}=10</math>
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Also the mass of the bullet should be converted to kg:
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<math>{{m}_{B}}=0.04</math>
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Using conservation of momentum gives:
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<math>\begin{align}
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& {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\
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& 0.04\times 250+{{m}_{T}}\times 0=0.04\times 10+{{m}_{T}}\times 10 \\
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& 10=0.4+10{{m}_{T}} \\
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& {{m}_{T}}=\frac{10-0.4}{10}=0.96\text{ kg}
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\end{align}</math>
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Example 16.2
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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 ms-1 and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.
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Solution
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Before the collision:
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<math>{{u}_{V}}=12</math>
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and
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<math>{{u}_{C}}=0</math>
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After the collision:
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<math>{{v}_{V}}={{v}_{C}}=v</math>
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The masses should be converted to kilograms:
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<math>{{m}_{V}}=2500</math>
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and
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<math>{{m}_{C}}=1500</math>
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Using conservation of momentum gives:
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<math>\begin{align}
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& {{m}_{V}}{{u}_{V}}+{{m}_{C}}{{u}_{C}}={{m}_{V}}{{v}_{V}}+{{m}_{C}}{{v}_{C}} \\
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& 2500\times 12+1500\times 0=2500v+1500v \\
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& 30000=4000v \\
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& v=\frac{30000}{4000}=7.5\text{ m}{{\text{s}}^{\text{-1}}}
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\end{align}</math>
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Example 16.3
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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.
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Solution
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Before the collision:
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<math>{{u}_{A}}=4u</math>
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and
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<math>{{u}_{B}}=-u</math>
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After the collision:
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<math>{{v}_{A}}={{v}_{B}}=v</math>
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Using conservation of momentum gives:
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<math>\begin{align}
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& {{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}={{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}} \\
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& m\times 4u+3m\times (-u)=mv+3mv \\
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& mu=4mv \\
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& v=\frac{mu}{4mu}=\frac{u}{4}
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\end{align}</math>
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Example 16.4
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A particle, A, of mass 2 kg has velocity
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<math>(4\mathbf{i}+2\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
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. It collides with a second particle, B, of mass 3 kg and velocity
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<math>(2\mathbf{i}-4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}}</math>
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. If the particles coalesce during the collision, find their final velocity.
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Solution
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Before the collision:
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<math>{{\mathbf{u}}_{A}}=4\mathbf{i}+2\mathbf{j}</math>
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and
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<math>{{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}</math>
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After the collision:
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<math>{{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}</math>
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The masses are defined:
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<math>{{m}_{A}}=2</math>
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and
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<math>{{m}_{B}}=3</math>
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Using conservation of momentum gives:
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<math>\begin{align}
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& {{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}={{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}} \\
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& 2\times (4\mathbf{i}+2\mathbf{j})+3\times (2\mathbf{i}-4\mathbf{j})=2\mathbf{v}+3\mathbf{v} \\
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& 8\mathbf{i}+4\mathbf{j}+6\mathbf{i}-12\mathbf{j}=5\mathbf{v} \\
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& 14\mathbf{i}-8\mathbf{j}=5\mathbf{v} \\
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& \mathbf{v}=\frac{14\mathbf{i}-8\mathbf{j}}{5}=2.8\mathbf{i}-1.6\mathbf{j}
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\end{align}</math>
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Example 16.5
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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.
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Solution
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This diagram shows the velocities before the collision.
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<math>{{\mathbf{u}}_{C}}=15\mathbf{i}</math>
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and
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<math>{{\mathbf{u}}_{V}}=U\mathbf{j}</math>
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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>
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Using conservation of momentum gives:
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<math>\begin{align}
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& {{m}_{C}}{{\mathbf{u}}_{C}}+{{m}_{V}}{{\mathbf{u}}_{V}}={{m}_{C}}{{\mathbf{v}}_{C}}+{{m}_{V}}{{\mathbf{v}}_{V}} \\
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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})
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\end{align}</math>
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Considering the i component gives:
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<math>\begin{align}
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& 1200\times 15=2600V\cos 20{}^\circ \\
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& V=\frac{1200\times 15}{2600\cos 20{}^\circ }=\frac{180}{26\cos 20{}^\circ }=7.36\text{ m}{{\text{s}}^{\text{-1}}}
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\end{align}</math>
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Considering the j component gives:
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<math>\begin{align}
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& 1400U=2600\times \frac{180}{26\cos 20{}^\circ }\times \sin 20{}^\circ \\
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& U=\frac{2600\times 180}{1400\times 26}\tan 20{}^\circ =4.68\text{ m}{{\text{s}}^{\text{-1}}}
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\end{align}</math>

Revision as of 15:18, 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


\displaystyle {{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}

or

\displaystyle {{m}_{A}}{{\mathbf{v}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{v}}_{\mathbf{B}}}={{m}_{A}}{{\mathbf{u}}_{\mathbf{A}}}+{{m}_{B}}{{\mathbf{u}}_{\mathbf{B}}}


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:


\displaystyle {{u}_{B}}=250 and \displaystyle {{u}_{T}}=0


After the collision:


\displaystyle {{v}_{B}}={{v}_{T}}=10


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


\displaystyle {{m}_{B}}=0.04


Using conservation of momentum gives:


\displaystyle \begin{align} & {{m}_{B}}{{u}_{B}}+{{m}_{T}}{{u}_{T}}={{m}_{B}}{{v}_{B}}+{{m}_{T}}{{v}_{T}} \\ & 0.04\times 250+{{m}_{T}}\times 0=0.04\times 10+{{m}_{T}}\times 10 \\ & 10=0.4+10{{m}_{T}} \\ & {{m}_{T}}=\frac{10-0.4}{10}=0.96\text{ kg} \end{align}


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: \displaystyle {{u}_{V}}=12 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}=7.5\text{ m}{{\text{s}}^{\text{-1}}} \end{align}


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: \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}


Example 16.4

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} and \displaystyle {{\mathbf{u}}_{B}}=2\mathbf{i}-4\mathbf{j}


After the collision: \displaystyle {{\mathbf{v}}_{A}}={{\mathbf{v}}_{B}}=\mathbf{v}


The masses are defined: \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}=2.8\mathbf{i}-1.6\mathbf{j} \end{align}


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.





\displaystyle {{\mathbf{u}}_{C}}=15\mathbf{i} and \displaystyle {{\mathbf{u}}_{V}}=U\mathbf{j}


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 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 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}

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