# Chapter 7: Linear momentum

## 7.4 Two-dimensional collisions

When you are dealing with objects that are not moving along a single axis, you must remember that velocity—and therefore momentum—is a vector. This means that both the $$x$$ and $$y$$ components of momentum are conserved in a collision:

\begin{align*} p_{ix} &= p_{fx} \tag{7.5} \\ p_{iy} &= p_{fy} \tag{7.6} \end{align*}

#### Example

A 1500 kg car is moving west at 25 m/s, and collides with a 4000 kg truck moving 10 m/s in a direction 50° North from West. The two vehicles stick together after the collision. What is the velocity of the wreckage?

It is convention to define East as the positive $$x$$ direction, and North as the positive $$y$$ direction. We’ll need to separate the initial velocities into $$x$$ and $$y$$ components. Since the car is initially traveling due West, it’s velocity is only in the $$-x$$ direction. The truck has velocity components in both the $$-x$$ and $$+y$$ directions; we can use trigonometry to determine these components in terms of the truck’s speed and direction of travel. Expressed as vectors, the initial velocities are

$\vec{v}_c = \underbrace{-v_c}_{v_{cx}}\hat{x} \quad\text{and}\quad \vec{v}_t = \underbrace{-v_t\cos\theta}_{v_{tx}}\hat{x} + \underbrace{v_t\sin\theta}_{v_{ty}}\hat{y}$

Momentum is conserved. Whenever we work with vectors, we work with the $$x$$ and $$y$$ components separately.

\begin{align*} p_{i,x} &= p_{f,x} & p_{i,y} &= p_{f,y} \\ m_cv_{cx} + m_tv_{tx} &= (m_c + m_t)v_{f,x} & m_c(0) + m_tv_{ty} &= (m_c + m_t)v_{f,y} \\ -m_cv_c – m_tv_t\cos\theta &= (m_c + m_t)v_{f,x} & m_tv_t\sin\theta &= (m_c + m_t)v_{f,y} \\ \hookrightarrow v_{f,x} &= -\frac{m_cv_c + m_tv_t\cos\theta}{m_c + m_t} & \hookrightarrow v_{f,y} &= \frac{m_tv_t\sin\theta}{m_c + m_t} \\ &= -11.5\ \textrm{m/s} & &= 5.57\ \textrm{m/s} \end{align*}

So, the final velocity is

$\vec{v}_f = \left(-11.5\ \textrm{m/s}\right)\hat{x} + \left(5.57\ \textrm{m/s}\right)\hat{y}$