Sharp swerve (impulse)

Problem

A $1\,200\;\text{kg}$ car swerves sharply: its velocity goes from $(20;\, 0)$ to $(15;\, 5)\;\text{m/s}$. Calculate the net impulse applied to the car (vector, magnitude and direction).

Solution

The impulse-momentum theorem states that the net impulse is the change in momentum: $\vv{J} = \Delta\vv{p} = m(\vv{v}_f - \vv{v}_i)$. It is a vector, so work component by component. Data: $m = 1\,200\;\text{kg}$, $\vv{v}_i = (20;\, 0)\;\text{m/s}$, $\vv{v}_f = (15;\, 5)\;\text{m/s}$. $x$ component: $J_x = m\,(v_{fx} - v_{ix}) = 1\,200\,(15 - 20) = -6000\ \mathrm{N\,s}$ $y$ component: $J_y = m\,(v_{fy} - v_{iy}) = 1\,200\,(5 - 0) = 6000\ \mathrm{N\,s}$ Impulse vector: $\vv{J} = (-6000;\, 6000)\ \mathrm{N\,s}$ Magnitude: $\norm{\vv{J}} = \sqrt{J_x^2 + J_y^2} = \sqrt{6000^2 + 6000^2} = 6000\sqrt{2} \approx 8485\ \mathrm{N\,s}$ Direction: negative $x$ component and positive $y$, so the vector points into the second quadrant; since $\abs{J_x} = \abs{J_y}$ the angle relative to the $x$ axis is: $\theta = 180^\circ - 45^\circ = 135^\circ$ The impulse slows the car along $x$ and pushes it along $y$: it is the lateral force from the tyres that makes the car turn. $\ev{\vv{J} = (-6000;\, 6000)\ \mathrm{N\,s},\quad \norm{\vv{J}} \approx 8485\ \mathrm{N\,s},\quad \theta = 135^\circ}$

Links

Topics: Momentum and collisions Concepts: Impulse · Momentum Skills: Symbolic setup