At a railway yard, a shunting engine gives a goods wagon a push and lets it roll. The wagon runs into a standing wagon with a loud clank. The round buffers on the ends of the two wagons squeeze and spring out again. Then the standing wagon rolls away and the first one slows almost to a stop.
A large force for a short time
Two bodies come close, touch, push on each other and separate. This brief event is a collision. The push between them is a contact force. It grows from zero to a very large value and falls to zero again, all within a few thousandths of a second.
A force of this kind is an impulsive force. In the chapter on Accelerated Objects you met the impulse of a force, \(J = F\Delta t\), the area under its force-time graph. You also met the impulse-momentum theorem: the impulse on a body equals the change in its momentum, \(J = \Delta p\), where the momentum is \(p = mv\).
Look at the two curves of Fig 1.1. At every instant the force of A on B and the force of B on A are equal in size and opposite in direction. This is Newton's third law. So the two impulses are equal and opposite too: B gains \(3\,\text{N s}\) of momentum and A loses exactly \(3\,\text{N s}\).
Why momentum survives every collision
Add the two changes. The momentum of A changes by \(-J\) and the momentum of B changes by \(+J\). The total momentum \(p_A + p_B\) of the pair does not change at all. However hard the bodies hit each other, the internal pair of forces only moves momentum from one body to the other.
What about outside forces, such as gravity or friction with the floor? They do act during the collision, but they are small and last only as long as the contact. Their impulse is tiny next to the impulse of the contact force. So we write the total momentum just before the collision equal to the total momentum just after it.
In every collision, the total momentum of the colliding bodies just after the collision equals the total momentum just before it. The contact forces are an internal third-law pair, and any outside force is too small to act in the brief time of contact.
The chapter on Momentum and Its Conservation develops this law for any system. Here it is the one tool that works for every collision. Kinetic energy, as the next lesson shows, is not so reliable.
A \(0.5\,\text{kg}\) ball strikes a wall. The contact lasts \(0.002\,\text{s}\), and the average contact force is \(1500\,\text{N}\). The weight of the ball is \(5\,\text{N}\). How many times larger is the impulse of the contact force than the impulse of the weight during the contact?
Two trolleys on a smooth track collide and stick together. Why is their total momentum the same just before and just after?
A collision along one line
Take two balls moving along the line that joins their centres. After they collide, they still move along that line. Such a collision is a head-on collision, or a one-dimensional collision. The line joining the centres at the instant of contact is the line of centres. We label the velocities before the collision \(u_1\) and \(u_2\), and the velocities after it \(v_1\) and \(v_2\).
These are signed numbers, as in the chapter on Motion in One Dimension. Choose one direction along the line as positive. A velocity towards the other side then carries a minus sign. With this rule the whole collision is written in one dimension.
Three stages
Freeze a collision into stages. First, the bodies approach, each with its own velocity. Then the surfaces touch and press into each other. The bodies are squashed a little at the contact; this is deformation. As long as the rear body is faster than the front one, the gap keeps closing and the deformation keeps growing.
Block A of \(1\,\text{kg}\) slides at \(4\,\text{m/s}\) on a smooth floor into block B of \(3\,\text{kg}\) at rest. B carries a spring bumper. At the instant the spring is squeezed the most, how do the velocities of A and B compare?
Play the figure above, or choose a chip. The spring is squeezed while A is faster than B. The squeeze stops growing at the instant the two velocities are equal. At that instant the relative motion is zero, and both blocks move with a common velocity. Then the spring pushes on. A slows further and B gains speed, so the gap opens again.
The second stage, from first contact to the greatest deformation, is the stage of compression. The third stage is restitution: the bodies regain some or all of their shape and push each other apart. For the spring bumper, restitution is complete. A leaves backwards at \(2\,\text{m/s}\) and B forwards at \(2\,\text{m/s}\).
The common velocity
At the greatest deformation the two bodies move as one body of mass \(m_1 + m_2\). Momentum is conserved, so the common velocity \(v\) follows at once.
\[ m_1u_1 + m_2u_2 = (m_1 + m_2)\,v \]
This is also the velocity of the centre of mass of the pair, which you meet in the chapter on Centre of Mass. The centre of mass moves steadily through the whole collision. At the greatest deformation both bodies simply move with it.
Watch the \(K\) bar in the figure above. The kinetic energy starts at \(8\,\text{J}\). At the greatest compression it falls to \(\tfrac12 \times 4 \times 1^2 = 2\,\text{J}\), its least value. The missing \(6\,\text{J}\) sits in the squeezed spring as \(U\). During restitution the spring returns it, and the kinetic energy climbs back to \(8\,\text{J}\).
A \(2\,\text{kg}\) ball moving at \(6\,\text{m/s}\) strikes a \(1\,\text{kg}\) ball at rest, head-on. At the instant of greatest deformation, what is the velocity of the \(1\,\text{kg}\) ball?
How far the bumper squeezes
A spring bumper makes the stages visible. The spring is light, and the floor is smooth, so no energy is lost. The kinetic energy that disappears during compression is stored in the spring as \(\tfrac12 kx^2\), where \(k\) is the spring constant and \(x\) the compression. From the chapter on Potential Energy, this is the energy of a squeezed spring.
So two laws give the greatest compression. Momentum gives the common velocity. Energy then gives the compression: the kinetic energy at the start equals the kinetic energy at the common velocity plus the energy in the spring.
\[ \tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 = \tfrac12 (m_1 + m_2)v^2 + \tfrac12 kx_{\max}^2 \]
Block A of \(1\,\text{kg}\) moves at \(4\,\text{m/s}\) on a smooth floor towards block B of \(3\,\text{kg}\) at rest. B carries a light spring of constant \(300\,\text{N/m}\). Find the greatest compression of the spring.
Take A's direction as positive. At the greatest compression both blocks move with the common velocity \(v\).
Momentum: \(1 \times 4 + 3 \times 0 = 4v\). Energy: \(\tfrac12 \times 1 \times 4^2 = \tfrac12 \times 4 \times v^2 + \tfrac12 \times 300 \times x^2\).
\(v = 1\,\text{m/s}\). Then \(8 = 2 + 150x^2\), so \(x^2 = 0.04\) and \(x = 0.2\,\text{m}\).
The spring is compressed by at most \(20\,\text{cm}\). If B were fixed to a wall, all \(8\,\text{J}\) would go into the spring, and \(x\) would be about \(23\,\text{cm}\).
Do not put all the kinetic energy into the spring. The pair still moves at the common velocity when the spring is shortest, and that motion keeps \(\tfrac12 (m_1 + m_2)v^2\) of the energy. Only the rest is stored in the spring.
A \(2\,\text{kg}\) block moving at \(3\,\text{m/s}\) on a smooth floor strikes a \(2\,\text{kg}\) block at rest. The front of the second block carries a light spring of constant \(900\,\text{N/m}\). Find the greatest compression of the spring, in centimetres.
A block strikes a second block, at rest, whose front carries a light spring, on a smooth floor. Which statements are true?
Now return to the railway yard. Take two wagons of \(20\) tonnes each, and let one roll at \(1\,\text{m/s}\) into the other. At the greatest squeeze both move at \(0.5\,\text{m/s}\). The kinetic energy falls from \(10\,\text{kJ}\) to \(5\,\text{kJ}\), so \(5\,\text{kJ}\) sits in the buffers. With buffers of combined stiffness \(10^6\,\text{N/m}\), \(\tfrac12 \times 10^6 \times x^2 = 5000\) gives a squeeze of \(10\,\text{cm}\).
In a crash test, a new car is driven into a barrier at a set speed while dummies inside record the forces. The Bharat NCAP programme, started in India in 2023, tests cars at \(64\,\text{km/h}\) into a barrier that crumples. The front of a good car crumples in a planned way. A longer crumple spreads the change in momentum over a longer time, so the force on the passengers is smaller.
- Setting both velocities to zero at the greatest deformation.The pair shares the common velocity \((m_1u_1 + m_2u_2)/(m_1 + m_2)\). It is zero only when the total momentum is zero.
- Putting all the starting kinetic energy into the spring.At the greatest compression the pair still moves; subtract \(\tfrac12(m_1 + m_2)v^2\) first.
- Including the weight or the floor's friction in the momentum of a collision.Over a few milliseconds their impulse is negligible. Momentum just before equals momentum just after.