Newton's Second Law for a System of Particles — Question 1

Question 1

Internal tension disappears from the system equation

Two carts of $2.0$ and $3.0\,\mathrm{kg}$ are connected by a light taut cord on a frictionless horizontal track. A $15\,\mathrm N$ pull acts rightward on the $3.0\,\mathrm{kg}$ cart.

1. Find the center-of-mass acceleration and cord tension.

2. Explain why including both the applied pull and the cord tension in the total external force gives a wrong result.

Question 1 - Solution

Internal tension disappears from the system equation

(a) Derivation and calculation

For the two-cart system, the cord forces cancel: $$\sum F_{\mathrm{ext}}=15=(2+3)a_C\Rightarrow\boxed{a_C=3.0\ \mathrm{m/s^2}}.$$ The taut cord imposes $a_1=a_2=a_C$. For the lighter cart, $$\boxed{T=m_1a_1=2(3)=6.0\ \mathrm N}.$$

(b) Result and interpretation

The separate equations add to the system equation: $$T=2a,\qquad15-T=3a\quad\Rightarrow\quad15=5a.$$ Counting one internal tension in the total external force omits its opposite partner. The heavier cart independently gives $15-6=3(3)$.

Physical check

The cord redistributes the acceleration-producing force without adding to the external force sum.