On an air track in a school laboratory in Kota, a \(2\,\text{kg}\) trolley moving at \(4\,\text{m/s}\) strikes a \(1\,\text{kg}\) trolley at rest. Momentum is conserved, so \(2 \times 4 = 2v_1 + v_2\). This one equation has two unknowns. The pair \(v_1 = 2\), \(v_2 = 4\) fits the equation. So do \(v_1 = 1\), \(v_2 = 6\) and endless other pairs. Which pair really happens?
Two unknowns need two equations
Simultaneous equations must all hold at once. One linear equation in two unknowns fixes only a relation between them. For every value of \(v_1\) there is a \(v_2\) that fits. To fix both, a second fact is needed, and it must be a new fact, not the first one written again. For the trolleys the second fact is the coefficient of restitution, from the chapter Collisions: \(v_2 - v_1 = e(u_1 - u_2)\). The trolleys have \(e = 0.5\), so \(v_2 - v_1 = 2\).
Elimination
Now there are two equations: \(2v_1 + v_2 = 8\) and \(v_2 - v_1 = 2\). Both contain \(v_2\) once, with the same sign. Subtract the second equation from the first, side by side. The \(v_2\) terms cancel and leave \(3v_1 = 6\), so \(v_1 = 2\,\text{m/s}\). Put this back into either equation: \(v_2 = 2 + 2 = 4\,\text{m/s}\). This is elimination: add or subtract whole equations, multiplied by suitable numbers, until one unknown cancels.
Substitution
A second route, substitution, gives the same answer. Make one unknown the subject of one equation, and substitute it into the other. From \(v_2 - v_1 = 2\), \(v_2 = v_1 + 2\). Then \(2v_1 + (v_1 + 2) = 8\), so \(3v_1 = 6\) and \(v_1 = 2\,\text{m/s}\) again. Use whichever route is shorter for the equations in front of you.
Two independent linear equations in two unknowns have exactly one solution. Eliminate one unknown, by adding or subtracting multiples of the equations or by substitution, and solve for the other. Substitute that value back to find the first unknown. Check the pair in both of the original equations.
\[ a_1b_2\,x + b_1b_2\,y = c_1b_2 \]
Why. multiply the first equation by \(b_2\), so that \(y\) carries the factor \(b_1b_2\).
\[ a_2b_1\,x + b_1b_2\,y = c_2b_1 \]
Why. multiply the second equation by \(b_1\). Now \(y\) carries the same factor in both.
\[ (a_1b_2 - a_2b_1)\,x = c_1b_2 - c_2b_1 \]
Why. subtract the second line from the first. The terms in \(y\) cancel.
\[ x = \dfrac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1} \]
Read it. divide by \(a_1b_2 - a_2b_1\), which is allowed only when it is not zero. Eliminating \(x\) in the same way gives \(y\).
For the trolleys, \(a_1 = 2\), \(b_1 = 1\), \(c_1 = 8\) and \(a_2 = -1\), \(b_2 = 1\), \(c_2 = 2\). The denominator is \(2 \times 1 - (-1) \times 1 = 3\). Then \(v_1 = (8 - 2)/3 = 2\) and \(v_2 = (4 + 8)/3 = 4\), as before. The formula is useful to check a hand solution, but writing the equations and eliminating by hand keeps the physics in view.
The Atwood machine of the chapter Accelerated Objects is solved by elimination too. Its two equations are \(m_1g - T = m_1a\) and \(T - m_2g = m_2a\). The tension appears with opposite signs, so adding the equations removes it: \((m_1 - m_2)g = (m_1 + m_2)a\). The acceleration follows, and then the tension from either equation.
For the Atwood machine, \(m_1g - T = m_1a\) and \(T - m_2g = m_2a\). Which single step removes the tension \(T\)?
Blocks of \(5\,\text{kg}\) and \(3\,\text{kg}\) hang over a smooth light pulley. Take \(g = 10\,\text{m/s}^2\). Solve the two equations of the blocks together and find the tension, in newtons.
Each equation is a straight line
Plot \(v_2\) against \(v_1\). Every pair that obeys \(2v_1 + v_2 = 8\) lies on one straight line. Every pair that obeys \(v_2 - v_1 = 4e\) lies on another. The only pair that obeys both is the point where the two lines cross. In the figure below, a slider changes \(e\). The restitution line moves, and the crossing slides along the momentum line.
The picture also shows when two equations fail. Two parallel lines never cross, so the equations contradict each other and have no solution. Two lines that lie on top of each other cross everywhere. Then the second equation is the first one written again, such as \(4v_1 + 2v_2 = 16\), and it adds nothing. In both cases \(a_1b_2 - a_2b_1 = 0\), the denominator of the formula.
A \(2\,\text{kg}\) ball moving at \(4\,\text{m/s}\) strikes a \(1\,\text{kg}\) ball at rest, head-on, with \(e = 0.5\). Find both velocities after the collision and the kinetic energy lost.
Take the direction of the moving ball as positive: \(u_1 = 4\,\text{m/s}\), \(u_2 = 0\). The unknowns are \(v_1\) and \(v_2\).
Momentum: \(2v_1 + v_2 = 8\). Restitution: \(v_2 - v_1 = 0.5 \times 4 = 2\).
Subtracting: \(3v_1 = 6\), so \(v_1 = 2\,\text{m/s}\), and \(v_2 = 2 + 2 = 4\,\text{m/s}\). Kinetic energy: \(\tfrac12 \times 2 \times 16 = 16\,\text{J}\) before, \(4 + 8 = 12\,\text{J}\) after.
The \(2\,\text{kg}\) ball goes on at \(2\,\text{m/s}\), the \(1\,\text{kg}\) ball leaves at \(4\,\text{m/s}\), and \(4\,\text{J}\) of kinetic energy is lost.
What can you say about the pair of equations \(2x + y = 8\) and \(4x + 2y = 10\)?
A \(1\,\text{kg}\) ball moving at \(6\,\text{m/s}\) strikes a \(2\,\text{kg}\) ball at rest, head-on, with \(e = 0.5\). Find the velocity of the \(2\,\text{kg}\) ball after the collision, in m/s.
Three unknowns
Three unknowns need three independent equations. The method is the same, applied twice. Use one equation to remove one unknown from the other two. That leaves two equations in two unknowns, which you already know how to solve. Often a lucky sum removes two unknowns at once.
A train of three blocks shows the lucky sum. Blocks of \(1\), \(2\) and \(3\,\text{kg}\) lie on a smooth table, joined by light strings, and \(30\,\text{N}\) pulls the front block. The unknowns are the tensions \(T_1\) and \(T_2\) and the common acceleration \(a\). Fig 2.1 draws each block alone with its own forces and its own equation.
Add the three equations. Each tension appears once with a plus sign and once with a minus sign, so both cancel: \(30 = 6a\), and \(a = 5\,\text{m/s}^2\). The rear block's equation then gives \(T_1 = 1 \times 5 = 5\,\text{N}\). The middle block's gives \(T_2 = T_1 + 2a = 15\,\text{N}\). The front block checks the answer: \(30 - 15 = 15 = 3 \times 5\).
A \(10\,\text{V}\) cell with an internal resistance of \(1\,\Omega\) and a \(4\,\text{V}\) cell with \(2\,\Omega\) are joined in parallel across a \(2\,\Omega\) load. Their positive terminals are joined together. Find the three branch currents.
Let \(I_1\) flow up through the \(10\,\text{V}\) cell, \(I_2\) up through the \(4\,\text{V}\) cell and \(I_3\) down through the load. Three unknowns need three equations.
Junction: \(I_1 + I_2 = I_3\). Loop through the \(10\,\text{V}\) cell and the load: \(10 - I_1 - 2I_3 = 0\). Loop through the \(4\,\text{V}\) cell and the load: \(4 - 2I_2 - 2I_3 = 0\).
The loops give \(I_1 = 10 - 2I_3\) and \(I_2 = 2 - I_3\). Substitute into the junction equation: \(12 - 3I_3 = I_3\), so \(I_3 = 3\,\text{A}\), \(I_1 = 4\,\text{A}\) and \(I_2 = -1\,\text{A}\).
The load carries \(3\,\text{A}\). The \(10\,\text{V}\) cell gives \(4\,\text{A}\). The minus sign of \(I_2\) means \(1\,\text{A}\) flows down, into the positive terminal of the \(4\,\text{V}\) cell, which is being charged.
Blocks of \(2\), \(3\) and \(5\,\text{kg}\) lie in a line on a smooth table, joined by light strings. A force of \(40\,\text{N}\) pulls the \(5\,\text{kg}\) block at the front. Find the tension in the string between the \(3\,\text{kg}\) and \(5\,\text{kg}\) blocks, in newtons.
For an Atwood machine, a student writes an equation for each block. The student then adds a third equation for the two blocks together. How many unknowns can these three equations fix?
Count only independent equations. An equation made by adding or scaling others, such as the equation of a whole system, tells nothing new. Also keep one sign convention for every unknown. A negative answer, like \(I_2 = -1\,\text{A}\) above, is not an error: it means the unknown points the other way.
Linear equations solved together run through mechanics. The Atwood machine, a block on a table pulled by a hanging mass and trains of blocks all give one equation per body. The same holds for a heavy pulley, a block hanging from a wheel, a yo-yo and a cylinder on a moving plank. There the equation of rotation joins the equations of motion. Every head-on collision is two equations, momentum and restitution. In circuits, Kirchhoff's rules and the nodal method give one equation per loop or junction, and a cell's internal resistance comes from two readings. Two resonance lengths give the speed of sound, and an orbit's radius and speed come from two conditions of the Bohr atom.
- Subtracting when the unknown has opposite signs.Look at the signs first. Add the equations when the unknown appears as \(+T\) in one and \(-T\) in the other; subtract when the signs are the same.
- Stopping after finding one unknown.Substitute it back into one equation for the other unknown, then check the pair in the other equation.
- Counting an equation that is the sum of others.Only independent equations count. The equation of a whole system is the sum of the equations of its parts.