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

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

Practice

Practice

Practice

Practice