MB1 - Conservation of momentum

MB1 - Conservation of momentum

Momentum combines mass with velocity and retains its direction. You will conserve signed or vector momentum during an interaction, choose a suitable system and interpret the direction of the resulting motion.

Signed and vector momentum

A 2 kg2\,\mathrm{kg} trolley moving right at 4 m s−14\,\mathrm{m\,s}^{-1} sticks to a 1 kg1\,\mathrm{kg} trolley moving left at 2 m s−12\,\mathrm{m\,s}^{-1}. Which way will they move? Adding their speeds would hide the competition between their directions.

Take right as positive. Their initial momenta are 2(4)=82(4)=8 and 1(−2)=−21(-2)=-2, in kg m s−1\mathrm{kg\,m\,s}^{-1}. The total is +6+6, directed right. The same idea works in a plane: multiply each component of velocity by mass, then add the momentum vectors.

Momentum

For a particle of mass mm and velocity v\mathbf v, its momentum is p=mv\mathbf p=m\mathbf v. Momentum is a vector with SI unit kg m s−1\mathrm{kg\,m\,s}^{-1}. In one dimension, a signed value represents the component along a declared positive direction.

For a system of bodies, total momentum is the vector sum of their momenta. When the resultant external impulse over an interaction is zero, total momentum before equals total momentum after. During a short collision, we often model any external impulse in the directions considered as negligible. Part 2 explains that condition.

Conservation of momentum

For two constant-mass particles with negligible resultant external impulse during the interaction,

mAuA+mBuB=mAvA+mBvBm_A\mathbf u_A+m_B\mathbf u_B =m_A\mathbf v_A+m_B\mathbf v_B

Here u\mathbf u denotes velocity immediately before and v\mathbf v immediately after. Use a common coordinate system for every vector.

Not in the formula booklet: learn it.

One vector equation gives a separate scalar equation in each coordinate direction. If one final velocity is known, solve for the other; no restitution rule is needed for that task.

Worked example 1: recovering a signed velocity

Question. On a smooth horizontal track, AA has mass 2 kg2\,\mathrm{kg} and velocity +3 m s−1+3\,\mathrm{m\,s}^{-1}; BB has mass 1 kg1\,\mathrm{kg} and velocity −2 m s−1-2\,\mathrm{m\,s}^{-1}. After their collision, AA's velocity is +1 m s−1+1\,\mathrm{m\,s}^{-1}. Find BB's velocity, neglecting external impulse along the track.

Plan. Conserve the signed momentum of A+BA+B, using right as positive.

2(3)+1(−2)=2(1)+1vB4=2+vBvB=2\begin{aligned} 2(3)+1(-2)&=2(1)+1v_B\\ 4&=2+v_B\\ v_B&=2 \end{aligned}

Answer. BB moves right at 2 m s−12\,\mathrm{m\,s}^{-1}.

Check. Final momentum is 2+2=42+2=4, equal to the initial 6−2=46-2=4. With AA initially left of BB, vB>vAv_B>v_A, so they separate after impact.

Keep the masses and horizontal components unchanged, but add vertical components on a horizontal table. The same calculation now has a second coordinate equation.

Worked example 2: recovering a two-dimensional velocity

Question. On a smooth horizontal table, particles of masses 2 kg2\,\mathrm{kg} and 1 kg1\,\mathrm{kg} have velocities uA=3i+2j\mathbf u_A=3\mathbf i+2\mathbf j and uB=−2i−j\mathbf u_B=-2\mathbf i-\mathbf j, in m s−1\mathrm{m\,s}^{-1}. After collision, vA=i+j\mathbf v_A=\mathbf i+\mathbf j. Find vB\mathbf v_B, neglecting external impulse in the table's plane.

Plan. Use the same conservation law, retaining both components.

2(3i+2j)+(−2i−j)=2(i+j)+vB4i+3j=2i+2j+vBvB=2i+j\begin{aligned} 2(3\mathbf i+2\mathbf j)+(-2\mathbf i-\mathbf j) &=2(\mathbf i+\mathbf j)+\mathbf v_B\\ 4\mathbf i+3\mathbf j&=2\mathbf i+2\mathbf j+\mathbf v_B\\ \mathbf v_B&=2\mathbf i+\mathbf j \end{aligned}

Answer. vB=(2i+j) m s−1\mathbf v_B=(2\mathbf i+\mathbf j)\,\mathrm{m\,s}^{-1}.

Check. The final i\mathbf i-momentum is 2+2=42+2=4; the final j\mathbf j-momentum is 2+1=32+1=3. Each matches its initial total. Taking magnitudes before adding would fail to preserve those two totals.

A total momentum vector may be zero even when both particles move. For example, equal masses with opposite velocities have cancelling momenta; zero total does not mean zero individual motion.

Watch out

Wrong: total momentum is mA∣uA∣+mB∣uB∣m_A|\mathbf u_A|+m_B|\mathbf u_B|. That adds magnitudes and loses directions. Add the signed components or vectors first; take a magnitude only if the question asks for it.

Use a new pair of vectors for your turn.

Continue this lesson.

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