Newton's Third Law — Question 3

Question 3

Internal force cancellation depends on the chosen system

A 2 kg cart A pushes a 3 kg cart B on a frictionless horizontal track. A hand applies 15 N rightward to A; the carts remain in contact.

1. Find the common acceleration of A+B.

2. Find the contact force on B and its partner on A.

3. Why may the contact pair be canceled for A+B but not on A alone?

Question 3 - Solution

Internal force cancellation depends on the chosen system

(a) Sum external forces on the pair.

$$15=(m_A+m_B)a\ \Rightarrow\ \boxed{a=15/5=3\,\mathrm{m/s^2}}$$ The total mass is 5 kg and external horizontal force is 15 N, so $a=3$ m/s$^2$.

(b) Isolate B.

$$C=m_Ba=3(3)=9\,\mathrm N,\quad15-C=2(3)=6\,\mathrm N$$ B needs $F_{A\to B}=3(3)=9$ N rightward. B exerts 9 N leftward on A. A’s resultant is $15-9=6$ N, matching $2(3)$.

(c) Respect recipients and boundary.

$$(15-C)+C=m_Aa+m_Ba\ \Rightarrow\ 15=(m_A+m_B)a$$ The two contact forces appear on different cart equations. They cancel when those equations are added for the combined system. On A alone only B’s force acts; canceling it against the force on B would omit a real contribution to A’s motion.