8867 / 2027
Collisions overview

Lesson 1 of 3

Impulse and a change of momentum

  • I can use the area under a body's force-time graph to calculate impulse and solve problems.Syllabus

    Syllabus 8867, 6(a). Use the area under a body's force-time graph to calculate impulse and solve problems.

Make a guess. You are not marked.

Two identical eggs are dropped from the same height. One lands on a cushion and survives; the other lands on concrete and breaks. Why?

  1. The concrete gives the egg a larger impulse because it is harder.
  2. The cushion makes the stop last longer, so the force is smaller.
  3. The cushion reduces the egg's change in momentum when it stops.
Show the answer

The cushion makes the stop last longer, so the force is smaller.

Both eggs lose the same momentum, so the impulse, force x time, is the same. The cushion spreads the stop over a longer time, so the average force is smaller.

Impulse measures the effect of a force over an interval. Its direction matters: choose the body and positive direction before finding the area or the momentum change.

Momentum is the vector p = mv. In one dimension, use signed velocities along one axis. Newton's second law relates the resultant force component to the rate of change of that momentum component.

Net impulse = signed area under the resultant force-time graph
= Δp = pfinal - pinitial

For constant mass, Δp = m(v - u). Impulse has units N s = kg m/s. It is force multiplied by time, not force divided by time.

Check what force the graph represents. A contact-force pulse alone gives the net impulse in that direction only when other impulses in that direction are negligible. For a graph that crosses zero, add its signed positive and negative areas; do not add all area magnitudes.

Estimate an impact force

Assume a roughly 0.1 kg ball is stopped from about 10 m/s. Its momentum change has magnitude of order 1 N s. If contact lasts about 0.01 s, the average force has magnitude of order 100 N, opposite the incoming direction.

These rough mass, speed and duration assumptions give a size check, not a universal impact force. A different contact duration changes the average force even if the momentum change is the same.

Worked rebound and force pulse

A ball rebounds from a wall

A 0.150 kg ball approaches a fixed wall at +20.0 m/s and rebounds at -12.0 m/s. Take motion towards the wall as positive. Other horizontal impulses during contact are negligible.

pinitial = 0.150 × 20.0 = +3.00 kg m/s
pfinal = 0.150 × (-12.0) = -1.80 kg m/s
Δp = -1.80 - 3.00 = -4.80 N s

The impulse is negative because the ball's momentum changes opposite the positive incoming direction. Subtracting speeds, 12 - 20, would miss the reversal.

Keep the same body and positive direction throughout

A 0.150 kg ball rebounds from a wall

The ball changes from positive twenty to negative twelve metres per secondTwo separately pictured moments show the same 0.150-kilogram ball and fixed wall on its right. Right, towards the wall, is positive. Before impact the blue velocity arrow points right and is one hundred drawing units long, representing positive 20.0 metres per second. After impact a sixty-unit blue arrow points left, representing negative 12.0 metres per second. Both arrows use five drawing units per metre per second. The ball positions and elapsed time between snapshots are schematic; no position or contact-time scale is assigned here. Blue arrows represent velocities, not forces. The ball's momentum changes from positive 3.00 to negative 1.80 kilogram metres per second, so its horizontal impulse is negative 4.80 newton seconds.Positive directionWallBefore+20.0 m/sp = +3.00 kg m/sAfter-12.0 m/sp = -1.80 kg m/s

Within these two ball views, the blue arrows share one velocity scale. The positions are schematic. The change is final minus initial: -1.80 - (+3.00) = -4.80 kg m/s.

The signed force-time area gives the same impulse

An asymmetric negative force pulse has impulse negative 4.80 newton secondsThe vertical axis is horizontal force on the ball in newtons. The arrowed time axis is at zero force. Time is labelled in milliseconds along the grid, from zero to sixty. The supplied triangular pulse joins time-force pairs zero milliseconds and zero newtons, twenty milliseconds and negative 160 newtons, and sixty milliseconds and zero newtons. The peak is one third of the way through the contact interval, not at its midpoint. The shaded triangle is below the zero-force axis. Converting sixty milliseconds to 0.060 seconds, its signed area is negative one half times 160 times 0.060, or negative 4.80 newton seconds. It describes force on the ball, with other horizontal impulses negligible, not force on the wall.02040600-40-80-120-160Horizontal force on ball / NContact time / msSigned area-4.80 N s

60 ms = 0.060 s. The area is -½ × 160 N × 0.060 s = -4.80 N s. Moving the peak away from the midpoint does not change this triangle's area when its base and height are unchanged.

This is the horizontal force on the ball, with other horizontal impulses negligible. The ball's opposite force on the wall acts on a different body and does not cancel from the ball's own momentum change.

The motion arrows show the ball's velocities before and after contact. The graph separately shows horizontal force on the ball during contact; its negative area is an impulse opposite the incoming direction.

The supplied force model joins (0 ms, 0 N), (20 ms, -160 N) and (60 ms, 0 N) with straight lines. First check its size: a constant -160 N for the whole 0.060 s would give -9.6 N s. The actual triangular pulse has a smaller area magnitude, so |impulse| < 9.6 N s.

Convert milliseconds to seconds before calculating:

60 ms = 0.060 s
Impulse = -½ × 160 × 0.060 = -4.80 N s

This agrees with the ball's momentum change. The peak occurs at 20 ms, not halfway through the 60 ms interval. The triangle's area still uses its full base and perpendicular height.

Try it step by step

Work it out in steps: Change in momentum of the ball in N s, then Average force on the ball F = change in p / t in N.

  1. Problem 1. A 0.16 kg ball hits a wall at 25 m/s and rebounds along the same line at 20 m/s. It is in contact with the wall for 0.050 s. Take the direction towards the wall as positive. Find the average force on the ball.
  2. Problem 2. A 0.45 kg ball hits a wall at 12 m/s and rebounds along the same line at 8.0 m/s, in contact for 0.020 s. Take towards the wall as positive. Find the average force on the ball.
  3. Problem 3. A 0.060 kg tennis ball hits a racket at 30 m/s and leaves along the same line at 24 m/s, in contact for 5.0 ms. Take towards the racket as positive. Find the average force on the ball.
Show the answer

Problem 1.

Change in momentum of the ball in N s: change in p = m(v - u) = 0.16 x (-20 - 25) = -7.2 N s

Average force on the ball F = change in p / t in N: F = -7.2 / 0.050 = -144 N, that is 144 N away from the wall (3 sf)

Problem 2.

Change in momentum of the ball in N s: change in p = 0.45 x (-8.0 - 12) = -9.0 N s

Average force on the ball F = change in p / t in N: F = -9.0 / 0.020 = -450 N, that is 450 N away from the wall (2 sf)

Problem 3.

Change in momentum of the ball in N s: change in p = 0.060 x (-24 - 30) = -3.24 N s

Average force on the ball F = change in p / t in N: F = -3.24 / 0.0050 = -648 N, that is 648 N away from the racket (3 sf)

Average force is not peak force

Faverage = net impulse / contact time
= -4.80/0.060 = -80.0 N

The average force gives the same impulse when multiplied by the full interval. Its magnitude, 80.0 N, differs from the pulse's 160 N peak.

If the same -4.80 N s momentum change occurred in 0.030 s, the average force would be -160 N. Increasing contact time can reduce the magnitude of average force for a fixed momentum change. It does not, by itself, specify the peak or the exact pulse shape.

If the original ball merely stopped, its momentum change would be 0 - 3.00 = -3.00 N s. Rebounding adds momentum in the opposite direction, increasing the change's magnitude to 4.80 N s.

The ball also exerts an opposite force on the wall. These Newton's-third-law forces act on different bodies. They do not cancel in the ball's individual momentum equation.

Check your understanding

A 0.150 kg ball travels towards a wall at +20.0 m/s and rebounds at -12.0 m/s. What is the impulse on the ball in this chosen direction?

  1. +4.80 N s
  2. -1.20 N s
  3. -4.80 N s
Show the answer

-4.80 N s

Impulse is delta p = m(v - u) = 0.150(-12.0 - 20.0) = -4.80 N s. Include the sign of the rebound velocity before subtracting.

Compare a sensor trace with momentum change

To compare a recorded contact pulse with m(v - u), identify the force recipient and use the same positive direction for both measurements. A sensor at a fixed wall may report the ball's force on the wall. With motion towards the wall positive, that force is positive. The force on the ball is opposite, so reverse the reported sign before comparing it with the ball's negative momentum change. Confirm the instrument's calibrated sign convention rather than assuming its displayed sign already refers to the ball.

  1. Calibrate and zero the force channel. Check its response to a known force in a known direction, and use a range that captures the pulse without clipping.
  2. Record the whole contact interval. Include the start, changing force, peak and return to the baseline. Use actual sample times and a sampling interval short enough to resolve the pulse.
  3. Measure the same body's mass and velocities. Use signed velocities just before and just after the contact, so the momentum change and force area refer to the same event.
  4. Compare quantities with units and uncertainty. Sum signed force-time areas in N s and compare with m(v - u) in kg m/s. Include or justify neglecting any other impulse in that direction.

Sparse samples can miss a negative peak and underestimate the impulse magnitude. A force-zero shift changes the area throughout the interval. For example, an uncorrected +10 N offset lasting 0.060 s adds +0.60 N s, making this -4.80 N s pulse appear as -4.20 N s. Repeating the same wrongly zeroed measurement does not remove that bias.

A supplied straight-segment model gives an exact geometric area. A sum formed from samples of a real curved pulse is an estimate whose quality depends on timing, response and sampling. The three points below describe the model, not an actual sensor recording.

Vertices of the supplied force model
Time / msForce on ball / N
00
20-160
600

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Explain how an airbag reduces the force on a driver's head in a crash. [3]

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Exam-style question

A ball of mass 0.20 kg is at rest. It is kicked, and the force-time graph of the kick is a triangle that rises from zero to a peak of 40 N and returns to zero after 0.050 s. Calculate the speed of the ball as it leaves the foot. [3]

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Exam-style question

Data-based question: Testing a crash barrier [15]

A test car of mass 1200 kg is driven into a roadside barrier at 15 m/s and comes to rest. A sensor records the horizontal force of the barrier on the car every 0.020 s.

Table 1. Force on the car during the impact
t / s00.0200.0400.0600.0800.100
F / kN01503003001500
  1. (a) Calculate the momentum of the car just before the impact. [1]
  2. (b) Use Table 1 to estimate the impulse on the car, and show that it is consistent with (a). [3]
  3. (c) Calculate the average force on the car during the impact. [1]
  4. (d) Calculate the average deceleration of the car, and express it as a multiple of g (g = 9.81 m s-2). [2]
  5. (e) A softer barrier stops the same car from the same speed in 0.25 s, with a force-time graph of the same shape. Calculate the peak force. [2]
  6. (f) Calculate the kinetic energy of the car before the impact and state where this energy goes. [2]
  7. (g) A designer says: "A barrier that crumples twice as far halves the average force on the car." Assuming the force is constant, use energy ideas to evaluate this claim. [3]
  8. (h) State one assumption you made in (b). [1]

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