Linear Momentum — Question 4

Question 4

One velocity shift changes both momentum components

A $3.0\,\mathrm{kg}$ particle has laboratory velocity $(4,-2)\,\mathrm{m/s}$. Another inertial frame moves at $(1,3)\,\mathrm{m/s}$ relative to the lab with parallel axes.

1. Find the momentum in both frames.

2. If the particle feels $(6,0)\,\mathrm N$, compare $d\vec p/dt$ in the two frames.

Question 4 - Solution

One velocity shift changes both momentum components

(a) Derivation and calculation

Compute momentum in each inertial frame: $$\boxed{\vec p=3(4,-2)=(12,-6)\ \mathrm{kg\,m/s}}.$$ $$\vec v^{\prime}=\vec v-\vec U=(3,-5),\qquad \boxed{\vec p^{\prime}=3(3,-5)=(9,-15)\ \mathrm{kg\,m/s}}.$$

(b) Result and interpretation

Because both $m$ and the frame velocity are constant, $$\frac{d\vec p^{\prime}}{dt}=\frac{d}{dt}(\vec p-m\vec U) =\frac{d\vec p}{dt}=\boxed{(6,0)\ \mathrm N}.$$ Momenta differ, but their time derivatives agree in these Newtonian inertial frames.

Physical check

The difference $\vec p-\vec p^{\prime}=m\vec U=(3,9)$ kg m/s is constant.