ImpulseAQA A-Level Further Maths: Revision notes

The impulse of a constant force $F$ acting for a time $t$ is $I=Ft$. It is a vector in the direction of the force, measured in N s, which is the same as kg m s⁻¹. Impulses are most useful for short interactions such as a bat striking a ball, when the force is large and acts for a very short time. The force in such cases is often called an impulsive force, and the weight of the object is negligible in comparison. Example: a force of $72$ N acting for $0.05$ s has impulse $72\times0.05=3.6$ N s.

From Newton's second law, $F=ma=m\frac{v-u}{t}$, so $Ft=mv-mu.$ The impulse of the resultant force equals the change in momentum: $I=mv-mu$. This is a vector equation. In one dimension choose a positive direction: if a ball reverses direction, $u$ and $v$ have opposite signs. Example: a ball of mass $0.06$ kg travelling at $20$ m s⁻¹ and hit back at $30$ m s⁻¹ has $I=0.06(30-(-20))=3$ N s. Then the average force over $0.004$ s is $\frac{3}{0.004}=750$ N.

When the force varies with time in one dimension, the impulse is the integral of the force over time: $I=\int_{t_1}^{t_2}F\,dt,$ which is the area under the force–time graph. Then use $I=mv-mu$. Example: a $2$ kg particle starts at $3$ m s⁻¹ and a force $F=6t^2$ acts for $4$ s. $I=\int_0^4 6t^2\,dt=\left[2t^3\right]_0^4=128$ N s, so $128=2(v-3)$ and $v=67$ m s⁻¹. Work in one dimension only for variable forces.

When two bodies collide, each exerts an impulse on the other. These impulses are equal in magnitude and opposite in direction, because the forces are equal and opposite (Newton's third law) and act for the same time. This links impulse to conservation of momentum: the gain in momentum of one body equals the loss in momentum of the other. Example: if a bat gives a ball an impulse of $3$ N s forwards, the ball gives the bat an impulse of $3$ N s backwards.

When velocities are given as vectors $a\mathbf{i}+b\mathbf{j}$, apply $\mathbf{I}=m\mathbf{v}-m\mathbf{u}$ as a vector equation, subtracting the $\mathbf{i}$ and $\mathbf{j}$ components separately. At AS level you are not required to resolve forces or velocities at angles. Example: $m=0.5$, $\mathbf{u}=6\mathbf{i}-2\mathbf{j}$, $\mathbf{v}=-2\mathbf{i}+4\mathbf{j}$. Then $\mathbf{I}=0.5(-8\mathbf{i}+6\mathbf{j})=-4\mathbf{i}+3\mathbf{j}$ N s, with magnitude $\sqrt{16+9}=5$ N s. To find the force from a constant impulse, divide by the time: $\mathbf{F}=\frac{\mathbf{I}}{t}$.