World Science Scholars
42.9 Force and Energy

In this lecture, we considered a collision between two particles (identical except for their color—one red, the other blue), from the perspective of two reference frames: in one frame the red particle had no horizontal motion and in the other the blue particle had no horizontal motion. Comparing the two yielded the formula for relativistic momentum. Here you undertake a similar calculation (phrased in terms of the jousters), but comparing the stadium frame, in which the jousters have equal but opposite velocities, and the frame of one of the jousters.

Module 42 – Force and Energy: Problem 3

Imagine that George and Evil George are in combat as described in this module’s lecture. Specifically, in the ground frame, let George have velocity $v_x$ and Evil George have velocity $-v_x$. When they jab at each other, have them do so in the $y$-direction with velocities $v_y$ and $-v_y$ respectively. Assume that both George and Evil George time their jousts perfectly, so each weapon hits the other squarely. For ease, assume $v_y \ll v_x$.

From Newtonian physics, we expect the impact of a jab to involve a product of the mass of the weapon times its speed, $m v_y$. Relativity, and the parable of the jousters, inspires us to update this formula in a specific way: we allow the mass of an object to depend on its velocity. Let’s denote this dependence by writing $m_0[v_x]$, where we are using our assumption of $v_y \ll v_x$ to only include $v_x$ dependence, and $m_0[v_x]$ is the function of velocity that we want to determine. Using this notation, the momentum of George’s jab in the $y$-direction is expressed as $m_0[v_x] v_y$. Let’s figure out the explicit form of $m_0[v_x]$ in a manner similar to what we did in lecture, but now comparing observations in the stadium frame and in George’s frame.

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