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Impulse

Rockets and landing · level 5, lesson 4 of 6

A question to start

The airbags of the Spirit rover hit Mars at about 13 m/s, the speed of a fast bicycle, and the rover inside was fine. The same rover dropped onto bare rock at that speed would have smashed. Both had to lose the same momentum. What was different?

The idea

From \(F = \Delta p / \Delta t\):

\[F\,\Delta t = \Delta p.\]

Force times time is the impulse, and it equals the change of momentum. To stop a body you must give it an impulse equal to its momentum. You can do it with a big force for a short time or a small force for a long time.

  • On bare rock, the stop takes perhaps a thousandth of a second: a huge force.
  • In airbags, or on squashing legs, or with bent knees, the stop takes much longer: a much smaller force.
hard: short and bigsoft: long and smallforceforcesame area

On a force-time graph, the impulse is the area under the curve. A soft landing spreads the same area over a longer time.

Over a distance. If a lander's legs squash by a distance \(d\) while it stops from speed \(v\) at a steady deceleration, then \(v^2 = 2ad\), so \(a = v^2/2d\), and the force (besides holding up its weight) is \(ma\). Longer legs, softer landing.

Worked examples

1. Perseverance touching down. 1,025 kg at 0.75 m/s: momentum 769 kg·m/s. If its wheels and suspension stop it in 0.2 s, the average extra force is \(769/0.2 = 3{,}800\) N, about its own weight on Mars. A firm but gentle bump.

2. Airbags. A 540 kg airbag ball at 13 m/s: momentum 7,020 kg·m/s. If the airbags squash and stop it in 0.1 s, the force is 70,000 N, about 13 g. On bare rock, stopped in 0.002 s, it would be 3.5 MN, over 600 g.

3. Lander legs. A 1,000 kg lander touches at 2 m/s and its legs squash 0.3 m. \(a = 4/0.6 = 6.7\) m/s²; force from the legs \(= m(a + g) = 1000 \times (6.7 + 3.71) = 10{,}400\) N.

Try it

  1. A 0.06 kg egg falls at 5 m/s and stops in 0.002 s on a hard floor. Find the force.
  2. On a cushion it stops in 0.05 s. Find the force now.
  3. What impulse stops a 2,000 kg lander moving at 3 m/s?
  4. A lander's legs squash 0.5 m to stop it from 3 m/s on Mars. Find the deceleration.
  5. Why do crumple zones in cars save lives?
Answers
  1. \(\Delta p = 0.3\) kg·m/s; \(F = 0.3/0.002 = 150\) N.
  2. \(0.3/0.05 = 6\) N.
  3. 6,000 N·s.
  4. \(a = 9/1.0 = 9\) m/s².
  5. They make the car stop over a longer time and distance, so the force on the people is smaller.

Common mistake

Thinking a soft landing removes less momentum. The momentum lost is the same; only the time it takes changes. Softness is about spreading out the force.

Where it leads