Use vector momentum changes and the area under a force–time graph.
Method and worked examples
Momentum p= mv points along velocity. Impulse equals the vector change in momentum: J=pf− pi. Retain signs when subtracting; a rebound can require more impulse than stopping from the same initial velocity. For constant resultant force, J=FΔt. For a varying force, use the signed area under its force–time graph.
Average resultant force is ΔtΔp. For an unchanged initial and final momentum, increasing the collision duration reduces average force but leaves impulse unchanged. A triangular force pulse has area half base times height; a rectangular pulse has area base times height. Average force does not determine the peak force or exact shape of the pulse.
Example 1
A 0.200kg ball changes velocity from +10.0to−5.00ms−1. Find impulse.
Δp=0.200(−5.00−10.0)=−3.00kgms−1=−3.00Ns. The impulse is opposite the positive direction.
Example 2
A triangular force pulse rises to 400N and returns to zero over 0.0200s. Find impulse.
Area = ½ × base × height = ½ ×0.0200×400=4.00Ns.
Watch out:For the same initial and final momenta, impulse is unchanged; doubling duration halves average resultant force.
4. Impulse and changing momentum | Year 12 Physics | Ascent Arcade