Exercises — Newton's third law — action-reaction, common misconceptions

Two forces act on the ball:

1. Gravity, Earth pulling the ball down: ${\vec{F}}_{E\to b}$.
2. Normal force, floor pushing the ball up: ${\vec{N}}_{f\to b}$.

The reaction partner always swaps the two bodies and is the same type of force:

• Partner of "Earth pulls ball down" (${\vec{F}}_{E\to b}$) is "ball pulls Earth up" ${\vec{F}}_{b\to E}$ — both gravity.
• Partner of "floor pushes ball up" (${\vec{N}}_{f\to b}$) is "ball pushes floor down" ${\vec{N}}_{b\to f}$ — both normal/contact.

Note the partners act on Earth and on the floor, not on the ball. That is rule 5.

False. Weight and normal force here both act on the same object (the ball) and are different types of force (gravity vs. contact). A real 3rd-law pair must be the same type and act on two different bodies. They only happen to be equal because the ball is in equilibrium — that is Newton's Second Law ($\sum F=0$), not the third law.

Nothing external acts, so total momentum is conserved (this is Conservation of Momentum). Start at rest: total $p=0$. $0=m_w{v}_w+m_a{v}_a$ Take "wrench direction" as positive. $m_w=2$, $v_w=+6$, $m_a=70$: $0=(2)(6)+(70)v_a\Rightarrow v_a=-\frac{12}{70}=-0.171\text{m/s}$ The minus means backward (opposite the wrench). Speed $\approx 0.171\text{m/s}$. Third-law view: the astronaut pushes the wrench for a short time $t$ with force $F$; the wrench pushes back equally. Same impulse $Ft$ on each, so $m_a|v_a|=m_w|v_w|$ — the heavier astronaut moves slower.

By the third law the wrench pushes the hand with exactly $30\text{N}$, opposite direction. Astronaut's acceleration from Newton's Second Law: $a=F/m=30/70=0.4286{\text{m/s}}^2$, backward. (The wrench's acceleration is $30/2=15{\text{m/s}}^2$ — same force, much larger acceleration because its mass is tiny.)

(a) Treat both blocks as one system of mass $m_1+m_2=5\text{kg}$. Only $F$ is external horizontally: $a=\frac{F}{m_1+m_2}=\frac{20}{5}=4{\text{m/s}}^2.$ (b) Now isolate block 2 (mass 2 kg). The only horizontal force on it is the contact push from block 1, call it $C$ (block 1 → block 2): $C=m_{2}a=(2)(4)=8\text{N}\text{(forward)}.$ By the third law, block 2 pushes back on block 1 with 8 N backward. Check block 1 directly: $m_{1}a=F-C=20-8=12=(3)(4)=12✓$ The two contact forces ($1\to 2$ forward, $2\to 1$ backward) have equal size 8 N, opposite directions, same type (contact/normal), swapped labels, on different blocks — a genuine 3rd-law pair. See the red arrows in the figure: they live on different free-body diagrams (this is why they never cancel — Free Body Diagrams).

The pair still accelerates at $a=20/5=4{\text{m/s}}^2$ (leftward now). Isolate block 1 (3 kg), whose only horizontal force is the contact push from block 2: $C^{\prime }=m_{1}a=(3)(4)=12\text{N}.$ The contact force changed (12 N vs. the earlier 8 N) even though the total force and total mass are identical! Why: the contact force must accelerate whichever block is behind the push. Pushing from the left, it drives the 2 kg block (→ 8 N); pushing from the right, it drives the 3 kg block (→ 12 N).

(a) Momentum before = 0 (everything at rest). Positive = ball's direction. $0=m_{\text{ball}}{v}_{\text{ball}}+m_{\text{cannon}}{v}_{\text{cannon}}=(4)(80)+50v_c$ $v_c=-\frac{320}{50}=-6.4\text{m/s}\text{(backward)}.$ (b) Impulse on the ball: $J=\Delta p=(4)(80)=320\text{kg⋅m/s}$ over $t=0.02\text{s}$: $F=\frac{J}{t}=\frac{320}{0.02}=16000\text{N}.$ By the third law the ball pushes the cannon back with the same 16000 N. Check the cannon's impulse: $Ft=16000\times 0.02=320=50\times 6.4✓$. (c) Total momentum after $=(4)(80)+(50)(-6.4)=320-320=0$, same as before — conserved (Conservation of Momentum). Centre of mass: total momentum is 0, so the centre of mass stays put — it never moves, before or after firing (Center of Mass Motion). The internal explosion (a 3rd-law pair) cannot shift it.

Thrust = rate of momentum carried off by the exhaust (this is the Rocket Propulsion & Variable Mass result, itself a 3rd-law statement — gas pushed back, rocket pushed forward): $F_{\text{thrust}}=\left(\frac{\Delta m}{\Delta t}\right)v_{\text{exhaust}}=\left(\frac{5\text{kg}}{1\text{s}}\right)(2000\text{m/s})=10000\text{N}.$ Instantaneous acceleration: $a=\frac{F_{\text{thrust}}}{m_{\text{rocket}}}=\frac{10000}{1000}=10{\text{m/s}}^2.$ The reaction partner is the exhaust gas, not any surrounding medium — which is exactly why rockets work in vacuum.

Positive = rightward. The force on $B$ from $A$ is forward, giving impulse $+J=+4$ to $B$; by the third law the force on $A$ from $B$ is backward, impulse $-J=-4$ on $A$. (Impulse = force × time = $\Delta p$; equal-and-opposite forces at every instant integrate to equal-and-opposite impulses.)

Block $B$ ($m_B=1$, started at rest): $v_B^{\prime }=0+\frac{J}{m_B}=0+\frac{4}{1}=4\text{m/s}.$ Block $A$ ($m_A=2$, started at $3$): $v_A^{\prime }=3+\frac{-J}{m_A}=3-\frac{4}{2}=1\text{m/s}.$ Impulse check: on $A$ it is $-4$, on $B$ it is $+4$ — exact negatives, as the third law demands. Momentum check: before $=(2)(3)+(1)(0)=6$. After $=(2)(1)+(1)(4)=2+4=6✓$. Conserved.

The child pushes the raft backward (left) with their feet; the raft pushes the child forward (right) — a 3rd-law pair, internal to the child+raft system. With no external horizontal force, the centre of mass stays fixed (Center of Mass Motion).

Let the raft slide left by distance $d$. Then the child (who walks 6 m along the raft) moves right by $6-d$ relative to the water. For the centre of mass to stay put, the two mass-weighted displacements must cancel: $m_{\text{child}}(6-d)=m_{\text{raft}}d$ $40(6-d)=120d\Rightarrow 240-40d=120d\Rightarrow 240=160d\Rightarrow d=1.5\text{m}.$ The raft slides 1.5 m to the left, and the child ends up $6-1.5=4.5\text{m}$ right of their start (relative to water). Check: $40(4.5)=180=120(1.5)=180✓$.

L1: a 3rd-law pair swaps bodies, same type — not just "equal and opposite." L2–L4: same force ⇒ unequal accelerations, and the reaction partner is the ejected mass, not any medium. L5: internal 3rd-law pairs keep the centre of mass fixed, so lighter parts move farther.