Easy -1.2 This is a straightforward application of the impulse-momentum theorem requiring only a single integration of a linear function. The formula (impulse = ∫F dt) is directly applicable with no problem-solving or conceptual insight needed, and the calculation ∫₀³ (t/5) dt = [t²/10]₀³ = 0.9 is routine. Being multiple choice with 1 mark confirms this is below average difficulty.
As a particle moves along a straight horizontal line, it is subjected to a force \(F\) newtons that acts in the direction of motion of the particle.
At time \(t\) seconds, \(F = \frac{t}{5}\)
Calculate the magnitude of the impulse on the particle between \(t = 0\) and \(t = 3\)
Circle your answer.
[1 mark]
0.3 N s \quad 0.6 N s \quad 0.9 N s \quad 1.8 N s
As a particle moves along a straight horizontal line, it is subjected to a force $F$ newtons that acts in the direction of motion of the particle.
At time $t$ seconds, $F = \frac{t}{5}$
Calculate the magnitude of the impulse on the particle between $t = 0$ and $t = 3$
Circle your answer.
[1 mark]
0.3 N s \quad 0.6 N s \quad 0.9 N s \quad 1.8 N s
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q2 [1]}}