16. Conservation of momentum

From Mechanics

Current revision

Key Points

In all collisions, where no external forces act, momentum will be conserved and thus the total momentum just after the collision will be the same as the total momentum just before the collision, which gives,

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 ms-1 and uT=0

After the collision:

vB=vT=10 ms-1

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 ms-1 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: \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}\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{\smash{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}

Example 16.5

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}