Explosions and recoil — Physics — Mathemagicians Library

Before this

This leads to

Circular motion

A collision run backwards

Remember the two skaters from the momentum lectures? One of mass 50 kg and one of 70 kg stood still on the ice, then pushed each other apart. The lighter skater moved off at 1.4 m/s, the heavier one at 1.0 m/s the other way. Each had 70 kg m/s of momentum, in opposite directions. Nothing exploded, but physically that push was an explosion: things that start together and at rest, flying apart.

The impulse lecture ended with a promise: the next lecture runs a collision backwards. In a sticky collision, separate objects come together into one. In an explosion, one object (or a group at rest) comes apart into pieces. A gun firing, a firework bursting, a rocket throwing out gas, a person jumping off a boat: the same law covers them all.

Starting from zero

The key fact about an explosion that starts from rest is its starting momentum: zero. Nothing is moving, so the total momentum is zero. The forces of the explosion are internal: each piece pushes on the others, in third-law pairs. So conservation of momentum says the total must still be zero afterwards. For two pieces:

m_{1} v_{1} + m_{2} v_{2} m_{1} v_{1} = 0 = - m_{2} v_{2}

Explosions from rest

The pieces fly apart with momenta that add up to zero. For two pieces, the momenta are equal in size and opposite in direction: $m_{1} v_{1} = - m_{2} v_{2}$ .

So their speeds are in the inverse ratio of their masses: $\frac{speed _{1}}{speed _{2}} = \frac{m _{2}}{m _{1}}$ . The lighter piece moves faster.

Check it on the skaters: $1.4 \div 1.0 = 1.4$ , and $70 \div 50 = 1.4$ . The lighter skater is 1.4 times faster, because the heavier one is 1.4 times heavier.

Recoil

When a gun fires a bullet, the gun moves backwards. This backward motion is called recoil. It is not a side effect that better design could remove. It is required by conservation of momentum: if the bullet gains momentum forwards, the gun must gain the same momentum backwards.

Worked Example 1 — A rifle's recoil

A 3 kg rifle fires a 0.02 kg bullet at 400 m/s. Find the rifle's recoil velocity.

Take the bullet's direction as positive. Before: total momentum 0. After: the bullet has $0.02 \times 400 = + 8$ kg m/s. So the rifle must have $- 8$ kg m/s: $3 v = - 8$ , giving $v = - \frac{8}{3} \approx - 2.67$ m/s. The rifle moves backwards at about 2.67 m/s.

The speed ratio is $400 \div \frac{8}{3} = 150$ , and the mass ratio is $3 \div 0.02 = 150$ . Equal, as the rule says.

Here is something the momentum equation hides. The two momenta are equal in size, but the kinetic energies are not. The bullet has $\frac{1}{2} \times 0.02 \times 40 0^{2} = 1600$ J. The rifle has only $\frac{1}{2} \times 3 \times (\frac{8}{3})^{2} \approx 10.7$ J. The light piece takes almost all the kinetic energy. That is why the bullet is dangerous and the recoil only pushes the shooter's shoulder. Holding the rifle firmly against the shoulder also helps in another way: the shooter's body joins the "gun" side of the explosion, adding mass and making the recoil slower, and a padded stock makes the stop take longer, as in the impulse lecture.

Where did all that kinetic energy come from, if the total momentum stayed at zero? From the chemical energy stored in the explosive. Momentum must be conserved, but kinetic energy can be created from stored energy in an explosion, just as it was lost to heat in a sticky collision. The energy split follows a simple rule too: each piece's share of the kinetic energy is in the ratio of the other piece's mass. Explore it with two carts pushed apart by the same spring.

Explore — two carts and a spring

A compressed spring between two carts at rest releases 24 J of energy and pushes them apart. Choose the masses. Compare the momenta (always equal and opposite) with the speeds and the kinetic energies.

Mass of A: 2 kg

Mass of B: 2 kg

The same 24 J spring with different pairs of carts (right is positive):

Masses	Velocities	Momenta	Kinetic energy
2 kg, 2 kg	≈ −3.46, +3.46 m/s	≈ ∓6.93 kg m/s	12 J, 12 J
1 kg, 3 kg	−6, +2 m/s	∓6 kg m/s	18 J, 6 J
1 kg, 9 kg	≈ −6.57, +0.73 m/s	≈ ∓6.57 kg m/s	21.6 J, 2.4 J

Recoil in everyday life

You meet recoil more often than you might think. Step off a small rowing boat onto a dock, and the boat slides away behind you. You pushed the boat backwards to push yourself forwards. Suppose you have a mass of 60 kg and the boat 120 kg, and you leave it at 2 m/s. The total momentum was zero, so the boat moves back with $60 \times 2 = 120$ kg m/s, which is $120 \div 120 = 1$ m/s. Some of the push you meant for your jump went into the boat instead. That is why people sometimes fall short of the dock and into the water. Jumping from a large ship feels completely different: the ship is so heavy that it barely moves, and almost all the push goes into you.

Stop and think

When you jump off the ground, you push the Earth down. Why do you never notice the Earth recoil?

Show the answer

The Earth does recoil, with exactly the same momentum as you. If a 60 kg person leaves the ground at 3 m/s, they have 180 kg m/s upwards, so the Earth gains 180 kg m/s downwards. But the Earth's mass is about $6 \times 1 0^{24}$ kg, so its speed is about $180 \div (6 \times 1 0^{24}) = 3 \times 1 0^{- 23}$ m/s. That is far too small ever to measure. And when you land, the Earth's momentum is handed back. The law still holds; the Earth's huge mass hides it.

Pieces of a firework

Worked Example 2 — A firework splits in two

A 0.4 kg firework is momentarily at rest at the top of its flight when it explodes into two pieces. Piece A, 0.15 kg, flies left at 30 m/s. Find the velocity of piece B, 0.25 kg.

Before: the firework is at rest, so the total momentum is 0. After: piece A has −4.5 kg m/s and piece B has +4.5 kg m/s. The total is still 0. The heavier piece, B, moves more slowly.

Take right as positive. A has $0.15 \times (- 30) = - 4.5$ kg m/s. So B must have $+ 4.5$ kg m/s: $0.25 v = 4.5$ , giving $v = 18$ m/s to the right.

Gravity acts on the firework the whole time, so it is not a perfectly closed system. But the explosion lasts a tiny fraction of a second, and in that time gravity's impulse is tiny compared with the explosion's. So momentum is conserved across the explosion itself, just as for collisions on a road.

Worked Example 3 — An explosion that does not start at rest

A 2 kg object moving right at 5 m/s splits into two pieces. A 0.5 kg piece moves right at 20 m/s. Find the velocity of the 1.5 kg piece.

Now the starting momentum is not zero: $2 \times 5 = 10$ kg m/s. After: $0.5 \times 20 + 1.5 v = 10 + 1.5 v$ . Setting this equal to 10 gives $1.5 v = 0$ , so $v = 0$ . The heavy piece stops dead, because the light piece carried off all the momentum. The rule is the same as always: total before = total after. "Zero before" is just the most common special case.

Rockets: pushing on what you throw away

The impulse lecture left rockets for this lecture. A rocket in space has no ground and no air to push against. So how does it speed up? It carries its own "something to push against": its propellant. The engine throws exhaust gas backwards at very high speed. The gas gains momentum backwards, so the rocket gains the same momentum forwards. By the third law, the rocket pushes the gas back and the gas pushes the rocket forward. It is a continuous recoil.

Stop and think

A student says: "In space there is nothing for the exhaust gas to push against, so a rocket engine cannot work there." What is wrong with this?

Show the answer

The rocket does not need the gas to push on anything outside. The rocket and the gas push on each other. Before the engine fires, rocket plus propellant has some total momentum. Throwing gas backwards gives the gas backward momentum, so the rocket must gain forward momentum to keep the total the same. A rocket actually works slightly better in space than in air, where the surrounding air gets in the way of the exhaust.

Worked Example 4 — A spacecraft's thruster

A 500 kg spacecraft fires its thruster for 2 s, throwing out 10 kg of gas each second at 200 m/s backwards. Estimate its change in velocity.

Gas thrown out: $10 \times 2 = 20$ kg. Its momentum: $20 \times 200 = 4000$ kg m/s backwards. So the spacecraft gains about 4000 kg m/s forwards, and its velocity changes by about $4000 \div 500 = 8$ m/s.

This is an estimate, because the spacecraft gets lighter as it throws out gas: it ends at 480 kg. Using 480 kg instead gives $4000 \div 480 \approx 8.33$ m/s. The true answer lies between the two. The exact calculation, called the rocket equation, comes in more advanced courses. It explains why large rockets are mostly propellant: to go faster, a rocket must throw away a great deal of its own mass.

Now make a prediction about recoil in space.

Predict, then reveal — an astronaut throws a tool

An 80 kg astronaut floats at rest outside a space station, not touching anything. She throws a 2 kg tool away from her at 10 m/s. What happens to her?

She drifts backwards at 0.25 m/s. Before the throw, the total momentum is 0. The tool gains 2 × 10 = 20 kg m/s, so she must gain 20 kg m/s the other way: 80v = 20, so v = 0.25 m/s.

Three classic mistakes

Trap 1 — Equal speeds instead of equal momenta

In an explosion from rest, the pieces have equal and opposite momenta, not speeds. The 3 kg rifle does not recoil at 400 m/s; it recoils at about 2.67 m/s.

Trap 2 — Forgetting the minus sign

$m_{1} v_{1} + m_{2} v_{2} = 0$ means the velocities have opposite signs. If you get both velocities positive, you have lost a sign: the pieces would be flying the same way, and the total could not be zero.

Trap 3 — "Kinetic energy is shared equally too"

Momentum is shared equally; kinetic energy is not. The lighter piece gets most of it: the bullet gets 1600 J, the rifle about 10.7 J. And the total kinetic energy is not conserved in an explosion: it comes from stored energy.

Try it yourself

Choose a positive direction for every question.

A 500 kg cannon on smooth ground fires a 5 kg ball at 100 m/s. Find the cannon's recoil velocity.

Hint 1
The total momentum is 0 before and after.

Hint 2
$5 \times 100 + 500 v = 0$ .

Full answer
$500 v = - 500$ , so $v = - 1$ m/s. The cannon recoils at 1 m/s, opposite to the ball.
A 60 kg skater and a 40 kg skater stand still together and push apart. The 40 kg skater moves off at 1.5 m/s. Find the other skater's speed, and the kinetic energy each gains.

Hint 1
$60 v = 40 \times 1.5$ .

Hint 2
Kinetic energy $= \frac{1}{2} m v^{2}$ for each skater.

Full answer
$60 v = 60$ , so $v = 1$ m/s, the other way. Kinetic energy: $\frac{1}{2} \times 40 \times 1. 5^{2} = 45$ J and $\frac{1}{2} \times 60 \times 1^{2} = 30$ J, a total of 75 J, supplied by their muscles. The lighter skater gets the larger share, in the ratio 60 : 40.
A 3 kg object at rest explodes into three pieces moving along one line. A 1 kg piece moves right at 12 m/s. A 0.5 kg piece moves right at 6 m/s. Find the velocity of the third piece.

Hint 1
The third piece has mass $3 - 1 - 0.5 = 1.5$ kg.

Hint 2
$12 + 3 + 1.5 v = 0$ .

Full answer
$1.5 v = - 15$ , so $v = - 10$ m/s: 10 m/s to the left. With more than two pieces, the rule is unchanged: all the momenta must add to zero.
In Worked Example 4, the spacecraft's speed change was estimated as 8 m/s using 500 kg and 8.33 m/s using 480 kg. Explain why the true value lies between these, and why a real rocket's engineers cannot simply use the starting mass.

Hint 1
At the start of the burn the spacecraft has 500 kg; at the end, 480 kg.

Hint 2
Each kilogram of gas gives the same push, but it changes the velocity of a lighter craft as the burn goes on.

Full answer
Every kilogram of gas thrown out carries 200 kg m/s of momentum, so each gives the craft the same push. Early in the burn, that push acts on nearly 500 kg; late in the burn, on nearly 480 kg. So the first kilograms change the velocity a little less than the last ones. Using 500 kg for the whole burn underestimates the change; using 480 kg overestimates it. The truth is between, about 8.16 m/s. For a short burn the difference is small, but a real rocket burns most of its mass, so its mass changes enormously. That is why engineers need the full rocket equation.

The bigger idea

Collisions and explosions are the same law in opposite directions. In both, the total momentum never changes. Kinetic energy is what differs: collisions usually turn some of it into heat and sound, and explosions create it from stored energy. Momentum is the bookkeeping that never fails; energy tells you how violent the event was.

This completes the topic on momentum and impulse. Rockets, recoil, crash design and collisions all follow from one idea: forces come in equal and opposite pairs. The next topic, circular motion, asks a new question: how can an object moving at a constant speed still be accelerating?

Key concepts in this lesson

An explosion from rest starts with zero total momentum, so the momenta of the pieces add to zero afterwards.
For two pieces, m₁v₁ = −m₂v₂: equal and opposite momenta, with the lighter piece moving faster.
Recoil is required by conservation of momentum; the heavier the gun, the slower it recoils.
Kinetic energy is not conserved in an explosion; it comes from stored energy, and the lighter piece takes the larger share.
A rocket moves by throwing propellant backwards: rocket and exhaust push on each other, so no outside surface is needed.