Linear Momentum — Question 10

Question 10

Spatial momentum data determine the force indirectly

A particle of constant mass $2.0\,\mathrm{kg}$ moves rightward where $p_x(x)=\sqrt{16+8x}\,\mathrm{kg\,m/s}$, using SI numerical $x\ge0$.

1. Find the net force as a function of position.

2. Check the result using kinetic energy and explain why $dp/dx$ alone is not force.

Question 10 - Solution

Spatial momentum data determine the force indirectly

(a) Derivation and calculation

Convert the spatial momentum derivative into a time derivative: $$F_x=\frac{dp_x}{dt}=\frac{dp_x}{dx}\frac{dx}{dt},\qquad \frac{dp_x}{dx}=\frac4{\sqrt{16+8x}},\quad v_x=\frac{p_x}{2}.$$ $$\boxed{F_x=\frac4{p_x}\frac{p_x}{2}=2.0\ \mathrm N}.$$

(b) Result and interpretation

The energy gives an independent spatial check: $$K(x)=\frac{p_x^2}{2m}=\frac{16+8x}{4}=4+2x,\qquad \boxed{\frac{dK}{dx}=2.0\ \mathrm N}.$$ The derivative $dp/dx$ alone lacks the velocity factor needed for $dp/dt$.

Physical check

The increasing momentum and energy are consistent with a positive constant force along the motion.