Example 1
A 0.20kg ball changes from +10m/sto−5.0m/s. Find impulse on it.
Δp=m(v−u)=0.20(−5−10)=−3.0kgm/s=−3.0Ns. The impulse points in the negative direction.
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Year 11 · SACE Stage 1
Momentum and impulse
Use vector momentum and force–time area for one-dimensional interactions.
Method and worked examples
Momentum is p= mv, a vector in the velocity direction with units kgm/s. Choose a positive axis before calculating. A reversal needs signed initial and final velocities: change in momentum is pfinal − pinitial, not simply a difference in speed magnitudes. For a constant mass, Newton’s second law can be written Fnet =ΔtΔp for an average force over the interval.
Impulse is the net force–time area and equals change in momentum. A rectangle represents constant force; a triangle contributes half its base times height. Distinguish average force from peak force. For the same momentum change, increasing stopping time lowers average net force. This rule explains a restraint’s force reduction without claiming that every injury outcome is determined by a single number.
Example 1
Example 2
A triangular net force–time pulse has base 0.040s and height 300N, entirely positive. Find impulse.
Impulse equals area = ½ ×0.040×300=6.0Ns. The peak 300N is not the average force.
Watch out: Momentum is a vector. Reversing velocity changes its sign and can produce a large momentum change.