201. Two particles \(A\) and \(B\) have masses \(2\,\text{kg}\) and \(4\,\text{kg}\), respectively. At an instant, \(A\) exerts a force
\[
\vec F_{AB}=(8\hat i+4\hat j)\,\text{N}
\]
on \(B\). An additional external force
\[
\vec P=(2\hat i-6\hat j)\,\text{N}
\]
acts on \(A\), while no other force acts on \(B\). Which option correctly gives the accelerations?
ⓐ. \(\vec a_A=(-3\hat i-5\hat j)\,\text{m s}^{-2}\), \(\vec a_B=(2\hat i+\hat j)\,\text{m s}^{-2}\)
ⓑ. \(\vec a_A=(5\hat i-\hat j)\,\text{m s}^{-2}\), \(\vec a_B=(-2\hat i-\hat j)\,\text{m s}^{-2}\)
ⓒ. \(\vec a_A=(-4\hat i-2\hat j)\,\text{m s}^{-2}\), \(\vec a_B=(4\hat i+2\hat j)\,\text{m s}^{-2}\)
ⓓ. \(\vec a_A=(\hat i-3\hat j)\,\text{m s}^{-2}\), \(\vec a_B=(2\hat i+\hat j)\,\text{m s}^{-2}\)
Correct Answer: \(\vec a_A=(-3\hat i-5\hat j)\,\text{m s}^{-2}\), \(\vec a_B=(2\hat i+\hat j)\,\text{m s}^{-2}\)
Explanation: The force of \(A\) on \(B\) is given as
\[
\vec F_{AB}=8\hat i+4\hat j.
\]
Newton's Third Law gives the force of \(B\) on \(A\):
\[
\vec F_{BA}=-8\hat i-4\hat j.
\]
For particle \(B\), no other force is stated, so
\[
\vec a_B=\frac{\vec F_{AB}}{4}
=2\hat i+\hat j\,\text{m s}^{-2}.
\]
For particle \(A\), the total force is
\[
\vec F_A=\vec F_{BA}+\vec P.
\]
Thus,
\[
\vec F_A=(-8\hat i-4\hat j)+(2\hat i-6\hat j)
=-6\hat i-10\hat j.
\]
Therefore,
\[
\vec a_A=\frac{-6\hat i-10\hat j}{2}
=-3\hat i-5\hat j\,\text{m s}^{-2}.
\]
The mutual forces remain equal and opposite, but the accelerations are determined by the complete net force on each body and their different masses.
202. A block is held at rest between a person's hand on its left and a vertical wall on its right. The hand pushes the block rightward with \(40\,\text{N}\), while the wall pushes the block leftward with \(40\,\text{N}\). Which statement is correct?
ⓐ. The hand's force on the block and the wall's force on the block form a third-law pair because they are equal and opposite.
ⓑ. The block exerts no force on the wall because it is at rest.
ⓒ. The two forces on the block balance, but their third-law partners are respectively the block's force on the hand and the block's force on the wall.
ⓓ. The third-law partner of the hand's \(40\,\text{N}\) force is the block's weight.
Correct Answer: The two forces on the block balance, but their third-law partners are respectively the block's force on the hand and the block's force on the wall.
Explanation: The \(40\,\text{N}\) hand force and the \(40\,\text{N}\) wall force both act on the block. They therefore belong to the same free-body equation and can balance each other. However, they arise from two different contact interactions and are not a third-law pair. The partner of the hand's force on the block is the force exerted by the block on the hand. It has magnitude \(40\,\text{N}\) and points opposite to the hand's force on the block. Similarly, the partner of the wall's force on the block is the force exerted by the block on the wall. The block's state of rest does not eliminate these contact interactions. Thus equal and opposite forces on one body must be distinguished from equal and opposite mutual forces acting on different bodies.
203. Three blocks \(A\), \(B\) and \(C\) remain in contact in this order on a smooth horizontal floor. A force \(P\) pushes \(A\) towards the right. Let \(N_1\) be the magnitude of the contact force between \(A\) and \(B\), and \(N_2\) the magnitude of the contact force between \(B\) and \(C\). All blocks have common acceleration \(a\). Which set of equations is correct?
ⓐ. \(P+N_1=m_Aa,\quad N_1+N_2=m_Ba,\quad N_2=m_Ca\)
ⓑ. \(P-N_1=m_Aa,\quad N_1-N_2=m_Ba,\quad N_2=m_Ca\)
ⓒ. \(P-N_1=m_Aa,\quad N_1+N_2=m_Ba,\quad -N_2=m_Ca\)
ⓓ. \(P=m_Aa,\quad N_1=m_Ba,\quad N_2=m_Ca\)
Correct Answer: \(P-N_1=m_Aa,\quad N_1-N_2=m_Ba,\quad N_2=m_Ca\)
Explanation: On block \(A\), the applied force \(P\) acts rightward, while block \(B\) pushes \(A\) leftward with magnitude \(N_1\). Hence,
\[
P-N_1=m_Aa.
\]
On block \(B\), block \(A\) pushes rightward with magnitude \(N_1\). By the separate \(B-C\) interaction, block \(C\) pushes \(B\) leftward with magnitude \(N_2\). Therefore,
\[
N_1-N_2=m_Ba.
\]
On block \(C\), block \(B\) pushes rightward with magnitude \(N_2\), so
\[
N_2=m_Ca.
\]
The force of \(A\) on \(B\) and the force of \(B\) on \(A\) form one third-law pair. Similarly, the \(B-C\) contact forces form another pair. Each individual equation contains only the member of each pair that actually acts on the selected block. Adding the three equations cancels the internal contact terms and gives \(P=(m_A+m_B+m_C)a\).
204. A \(2\,\text{kg}\) ball is moving downward when it is in contact with a horizontal floor. At a certain instant during the contact, the floor exerts an upward force of \(50\,\text{N}\) on the ball. Take \(g=10\,\text{m s}^{-2}\). Which statement is correct at that instant?
ⓐ. The ball accelerates downward because it is moving downward, and it exerts \(20\,\text{N}\) on the floor.
ⓑ. The ball has zero acceleration because the floor force and the force of the ball on the floor are equal and opposite.
ⓒ. The ball accelerates upward at \(25\,\text{m s}^{-2}\), and the ball exerts \(20\,\text{N}\) downward on the floor.
ⓓ. The ball accelerates upward at \(15\,\text{m s}^{-2}\), and the ball exerts \(50\,\text{N}\) downward on the floor.
Correct Answer: The ball accelerates upward at \(15\,\text{m s}^{-2}\), and the ball exerts \(50\,\text{N}\) downward on the floor.
Explanation: The floor exerts \(50\,\text{N}\) upward on the ball. The ball's weight is
\[
mg=2(10)=20\,\text{N}
\]
downward. Therefore the net force on the ball is
\[
F_{\text{net}}=50-20=30\,\text{N}
\]
upward. Its acceleration is
\[
a=\frac{30}{2}=15\,\text{m s}^{-2}
\]
upward. The fact that the ball is still moving downward at that instant does not require its acceleration to be downward; the upward acceleration means its downward speed is being reduced. By Newton's Third Law, the partner of the floor's \(50\,\text{N}\) upward force is the ball's \(50\,\text{N}\) downward force on the floor. The ball's \(20\,\text{N}\) weight belongs to a different gravitational interaction. Hence option D is correct.
205. Two isolated bodies \(P\) and \(Q\), of masses \(M\) and \(4M\), attract each other through a mutual force of instantaneous magnitude \(F\). No other force acts. Which statement is correct?
ⓐ. \(Q\) exerts \(4F\) on \(P\) because \(Q\) has four times the mass.
ⓑ. The forces have equal magnitude \(F\); the acceleration magnitude of \(P\) is four times that of \(Q\).
ⓒ. The accelerations have equal magnitudes because the interaction forces are equal.
ⓓ. \(P\) exerts the larger force because it undergoes the larger acceleration.
Correct Answer: The forces have equal magnitude \(F\); the acceleration magnitude of \(P\) is four times that of \(Q\).
Explanation: Newton's Third Law applies to mutual interactions whether or not the bodies are in physical contact. If \(P\) experiences a force of magnitude \(F\) towards \(Q\), then \(Q\) simultaneously experiences a force of magnitude \(F\) towards \(P\). Thus the interaction-force magnitudes are equal despite the unequal masses. For body \(P\),
\[
a_P=\frac{F}{M}.
\]
For body \(Q\),
\[
a_Q=\frac{F}{4M}.
\]
Therefore,
\[
\frac{a_P}{a_Q}=4.
\]
The lighter body accelerates four times as strongly under the equal interaction-force magnitude. Unequal accelerations therefore do not imply unequal action-reaction forces. The mass difference affects the response to force, not the equality required by Newton's Third Law.
206. A \(60\,\text{kg}\) person stands on a \(40\,\text{kg}\) wheeled platform on a frictionless horizontal floor. A vertical wall is to the person's left. The person pushes the wall so that the wall exerts a constant \(300\,\text{N}\) force on the person towards the right. The person does not slip relative to the platform. What are the acceleration of the person-platform system and the horizontal contact force exerted by the person on the platform?
ⓐ. \(5\,\text{m s}^{-2}\) right; \(200\,\text{N}\) right
ⓑ. \(3\,\text{m s}^{-2}\) right; \(300\,\text{N}\) right
ⓒ. \(3\,\text{m s}^{-2}\) right; \(120\,\text{N}\) right
ⓓ. \(1.2\,\text{m s}^{-2}\) right; \(120\,\text{N}\) left
Correct Answer: \(3\,\text{m s}^{-2}\) right; \(120\,\text{N}\) right
Explanation: Choose the person and platform together as one system. The horizontal interaction between person and platform is internal and cancels in the system force sum. The wall exerts the only stated horizontal external force:
\[
F_{\text{ext}}=300\,\text{N}.
\]
The total mass is
\[
60+40=100\,\text{kg}.
\]
Therefore,
\[
a=\frac{300}{100}=3\,\text{m s}^{-2}
\]
towards the right. Now isolate the platform. To accelerate its \(40\,\text{kg}\) mass at \(3\,\text{m s}^{-2}\), the person must exert a horizontal contact force
\[
F_{P\to \text{platform}}=40(3)=120\,\text{N}
\]
towards the right. By Newton's Third Law, the platform exerts \(120\,\text{N}\) leftward on the person. For the person,
\[
300-120=180\,\text{N}=60(3),
\]
which confirms the result. The person's \(300\,\text{N}\) force on the wall is the third-law partner of the wall's \(300\,\text{N}\) force on the person.
207. A \(3\,\text{kg}\) block slides on a smooth plane inclined at \(37^\circ\) to the horizontal. Take \(g=10\,\text{m s}^{-2}\) and \(\cos37^\circ=0.8\). Which statement correctly identifies the normal-force third-law pair?
ⓐ. Normal pair: plane-on-block \(24\,\text{N}\) and block-on-plane \(24\,\text{N}\); \(mg\cos37^\circ\) is not the reaction partner.
ⓑ. Normal pair: plane-on-block \(24\,\text{N}\) and \(mg\cos37^\circ=24\,\text{N}\); both act on the block.
ⓒ. Normal pair: plane-on-block \(0\) and block-on-plane \(0\); sliding removes the normal interaction.
ⓓ. Normal pair: plane-on-block \(24\,\text{N}\) and the block’s weight \(30\,\text{N}\); the magnitudes need not match.
Correct Answer: Normal pair: plane-on-block \(24\,\text{N}\) and block-on-plane \(24\,\text{N}\); \(mg\cos37^\circ\) is not the reaction partner.
Explanation: The component of the block's weight perpendicular to the plane is
\[
mg\cos37^\circ=3(10)(0.8)=24\,\text{N}.
\]
Because there is no acceleration perpendicular to the plane, the normal reaction also has magnitude
\[
N=24\,\text{N}.
\]
These two equal and opposite contributions both belong to the block's force balance in the perpendicular direction. They are therefore not a Newton's Third Law pair. The partner of the plane's \(24\,\text{N}\) force on the block is the \(24\,\text{N}\) contact force exerted by the block on the plane in the opposite direction. The perpendicular component \(mg\cos37^\circ\) is not a separate interaction; it is merely a component of Earth's gravitational force on the block. Its third-law partner belongs to the Earth-block gravitational interaction. Hence option A correctly distinguishes balance from action-reaction pairing.
208. Masses \(3\,\text{kg}\) and \(2\,\text{kg}\) are connected by a light string passing over a smooth ideal pulley. Take \(g=10\,\text{m s}^{-2}\). Which statement about the motion and the string forces is correct?
ⓐ. \(a=5\,\text{m s}^{-2}\), \(T=15\,\text{N}\); the two upward tensions on the masses are a third-law pair.
ⓑ. \(a=2\,\text{m s}^{-2}\), \(T=30\,\text{N}\); the two upward tensions on the masses are a third-law pair.
ⓒ. \(a=0\), \(T=30\,\text{N}\); equal string pulls on the two masses prevent acceleration.
ⓓ. \(a=2\,\text{m s}^{-2}\), \(T=24\,\text{N}\); the two upward tensions are not a third-law pair.
Correct Answer: \(a=2\,\text{m s}^{-2}\), \(T=24\,\text{N}\); the two upward tensions are not a third-law pair.
Explanation: The difference between the two weights provides the driving force:
\[
F_{\text{drive}}=(3-2)g=10\,\text{N}.
\]
The total mass is
\[
3+2=5\,\text{kg},
\]
so
\[
a=\frac{10}{5}=2\,\text{m s}^{-2}.
\]
For the \(3\,\text{kg}\) mass descending,
\[
30-T=3(2).
\]
Therefore,
\[
T=24\,\text{N}.
\]
An ideal light string transmits the same tension magnitude in this arrangement, so the string exerts \(24\,\text{N}\) upward on each mass. However, those two upward forces act on two different masses and are both exerted by the string; they are not forces that the same two bodies exert on each other. The partner of the string's force on the \(3\,\text{kg}\) mass is that mass's force on the string, and similarly for the \(2\,\text{kg}\) mass. Equal tension values therefore must not be mistaken for one third-law pair.
209. Two pucks \(A\) and \(B\), of masses \(2\,\text{kg}\) and \(4\,\text{kg}\), move on a frictionless horizontal plane. During a \(0.10\,\text{s}\) mutual interaction, the average force exerted by \(A\) on \(B\) is
\[
\vec F_{AB}=(60\hat i-80\hat j)\,\text{N}.
\]
External horizontal impulse during the interaction is negligible. Just before the interaction,
\[
\vec p_{A,i}=10\hat i\,\text{kg m s}^{-1},
\qquad
\vec p_{B,i}=12\hat j\,\text{kg m s}^{-1}.
\]
What are the velocities immediately afterward?
ⓐ. \(\vec v_A=(8\hat i-4\hat j)\,\text{m s}^{-1}\), \(\vec v_B=(-1.5\hat i+5\hat j)\,\text{m s}^{-1}\)
ⓑ. \(\vec v_A=(2\hat i+4\hat j)\,\text{m s}^{-1}\), \(\vec v_B=(1.5\hat i+\hat j)\,\text{m s}^{-1}\)
ⓒ. \(\vec v_A=(5\hat i+4\hat j)\,\text{m s}^{-1}\), \(\vec v_B=(1.5\hat i-\hat j)\,\text{m s}^{-1}\)
ⓓ. \(\vec v_A=(-2\hat i+4\hat j)\,\text{m s}^{-1}\), \(\vec v_B=(4\hat i+\hat j)\,\text{m s}^{-1}\)
Correct Answer: \(\vec v_A=(2\hat i+4\hat j)\,\text{m s}^{-1}\), \(\vec v_B=(1.5\hat i+\hat j)\,\text{m s}^{-1}\)
Explanation: The impulse delivered by \(A\) to \(B\) is
\[
\vec J_{AB}=\vec F_{AB}\Delta t.
\]
Therefore,
\[
\vec J_{AB}=(60\hat i-80\hat j)(0.10)
=(6\hat i-8\hat j)\,\text{N s}.
\]
By Newton's Third Law, the average force on \(A\) due to \(B\) is opposite, so
\[
\vec J_{BA}=(-6\hat i+8\hat j)\,\text{N s}.
\]
For puck \(A\),
\[
\vec p_{A,f}=10\hat i-6\hat i+8\hat j
=4\hat i+8\hat j.
\]
Thus,
\[
\vec v_{A,f}=\frac{\vec p_{A,f}}{2}
=2\hat i+4\hat j\,\text{m s}^{-1}.
\]
For puck \(B\),
\[
\vec p_{B,f}=12\hat j+6\hat i-8\hat j
=6\hat i+4\hat j.
\]
Hence,
\[
\vec v_{B,f}=\frac{\vec p_{B,f}}{4}
=1.5\hat i+\hat j\,\text{m s}^{-1}.
\]
The mutual impulses are equal and opposite, but the velocity changes are unequal because the puck masses differ.
210. During a \(0.50\,\text{s}\) interaction between bodies \(A\) and \(B\), the force exerted by \(A\) on \(B\) is described as follows.
The force is directed eastward. Its magnitude starts from zero at \(t=0\), rises linearly to \(120\,\text{N}\) at \(t=0.20\,\text{s}\), and then falls linearly to zero at \(t=0.50\,\text{s}\). The masses of \(A\) and \(B\) are \(3\,\text{kg}\) and \(5\,\text{kg}\), respectively.
Which statement is correct if no other impulse acts during this interval?
ⓐ. Partner force peaks later than \(0.20\,\text{s}\); each impulse has magnitude \(30\,\text{N s}\); \(|\Delta v_A|:|\Delta v_B|=3:5\).
ⓑ. Partner force is westward with equal instantaneous magnitude; each impulse has magnitude \(30\,\text{N s}\); \(|\Delta v_A|:|\Delta v_B|=5:3\).
ⓒ. Partner force is westward with a smaller peak; each impulse has magnitude \(18\,\text{N s}\); \(|\Delta v_A|:|\Delta v_B|=5:3\).
ⓓ. Both impulses are eastward with magnitude \(30\,\text{N s}\); the velocity changes are opposite in direction.
Correct Answer: Partner force is westward with equal instantaneous magnitude; each impulse has magnitude \(30\,\text{N s}\); \(|\Delta v_A|:|\Delta v_B|=5:3\).
Explanation: Newton's Third Law applies at every instant of the interaction:
\[
\vec F_{BA}(t)=-\vec F_{AB}(t).
\]
Therefore the force on \(A\) is westward whenever the force on \(B\) is eastward, and both reach their peak magnitude at the same instant \(t=0.20\,\text{s}\). There is no delay between action and reaction. The impulse magnitude is the area of the triangular force-time graph:
\[
J=\frac12(0.50)(120)=30\,\text{N s}.
\]
Thus the two bodies receive equal-magnitude opposite impulses. For body \(A\),
\[
|\Delta v_A|=\frac{J}{m_A}=\frac{30}{3}=10\,\text{m s}^{-1}.
\]
For body \(B\),
\[
|\Delta v_B|=\frac{30}{5}=6\,\text{m s}^{-1}.
\]
Hence,
\[
|\Delta v_A|:|\Delta v_B|=10:6=5:3.
\]
Equal interaction forces and equal impulses therefore do not require equal velocity changes when the masses differ.
211. A stone of mass \(m\) is attracted gravitationally by Earth of mass \(M\). At some instant, the magnitude of the gravitational force exerted by Earth on the stone is \(F\). Neglect all other forces. Which statement is correct?
ⓐ. Stone-on-Earth force \(=F\); \(a_{\text{stone}}:a_{\text{Earth}}=1:1\).
ⓑ. Stone-on-Earth force \(=(m/M)F\); \(a_{\text{stone}}:a_{\text{Earth}}=1:1\).
ⓒ. Stone-on-Earth force \(=F\); \(a_{\text{stone}}:a_{\text{Earth}}=M:m\).
ⓓ. Stone-on-Earth force \(\approx0\); \(a_{\text{stone}}:a_{\text{Earth}}=M:m\).
Correct Answer: Stone-on-Earth force \(=F\); \(a_{\text{stone}}:a_{\text{Earth}}=M:m\).
Explanation: The gravitational interaction between Earth and the stone is mutual. Newton's Third Law requires
\[
|\vec F_{\text{Earth on stone}}|
=
|\vec F_{\text{stone on Earth}}|
=F.
\]
The forces have opposite directions and act on different bodies. For the stone,
\[
a_{\text{stone}}=\frac{F}{m}.
\]
For Earth,
\[
a_{\text{Earth}}=\frac{F}{M}.
\]
Therefore,
\[
\frac{a_{\text{stone}}}{a_{\text{Earth}}}
=
\frac{F/m}{F/M}
=
\frac{M}{m}.
\]
Since \(M\gg m\), Earth's acceleration is extremely small, but its interaction force is not smaller. The difference in observed accelerations arises from the mass difference, not from unequal third-law forces. Hence the acceleration ratio is \(M:m\).
212. Two skaters \(P\) and \(Q\) push each other apart on frictionless ice. In Trial I, their masses are \(40\,\text{kg}\) and \(60\,\text{kg}\). In Trial II, \(P\) carries a \(40\,\text{kg}\) backpack rigidly with him, while \(Q\)'s mass remains unchanged. Force sensors show that the interaction-force versus time graph is exactly the same in the two trials. Both trials start with the skaters at rest. Which statement is correct?
ⓐ. Trial II: \(P\)’s speed change is unchanged; \(Q\)’s speed change is unchanged.
ⓑ. Trial II: \(P\)’s speed change is halved; \(Q\)’s speed change is unchanged.
ⓒ. Trial II: both speed changes are halved because the total mass is larger.
ⓓ. Trial II: \(P\)’s speed change is halved; \(Q\)’s speed change doubles because the force on \(Q\) doubles.
Correct Answer: Trial II: \(P\)’s speed change is halved; \(Q\)’s speed change is unchanged.
Explanation: In each trial, Newton's Third Law gives equal and opposite interaction forces at every instant. The measured force-time graph is also stated to be identical in the two trials. Therefore the impulse magnitude delivered to each skater is the same in both trials; call it \(J\). In Trial I,
\[
\Delta v_P=\frac{J}{40},
\qquad
\Delta v_Q=\frac{J}{60}.
\]
In Trial II, the effective mass moving with \(P\) is
\[
40+40=80\,\text{kg}.
\]
Hence,
\[
\Delta v'_P=\frac{J}{80}
=\frac12\Delta v_P.
\]
Skater \(Q\) still has mass \(60\,\text{kg}\) and receives the same impulse \(J\), so
\[
\Delta v'_Q=\frac{J}{60}
=\Delta v_Q.
\]
Increasing one participant's mass changes that participant's acceleration and velocity response without changing the equality of the mutual interaction forces.
213. A \(50\,\text{kg}\) person and a \(150\,\text{kg}\) boat are initially moving together eastward at \(2\,\text{m s}^{-1}\) on still water. During a short push-off, the boat exerts a westward impulse of \(100\,\text{N s}\) on the person. Neglect external horizontal impulse during the push. What are their velocities immediately afterward?
ⓐ. Person at rest; boat at \(\dfrac{8}{3}\,\text{m s}^{-1}\) eastward
ⓑ. Person \(2\,\text{m s}^{-1}\) westward; boat \(2\,\text{m s}^{-1}\) eastward
ⓒ. Person at rest; boat \(4\,\text{m s}^{-1}\) eastward
ⓓ. Person \(4\,\text{m s}^{-1}\) westward; boat \(8/3\,\text{m s}^{-1}\) eastward
Correct Answer: Person at rest; boat at \(\dfrac{8}{3}\,\text{m s}^{-1}\) eastward
Explanation: Take east as positive. The person's initial momentum is
\[
p_{P,i}=50(2)=100\,\text{kg m s}^{-1}.
\]
The boat gives the person a westward impulse
\[
J_P=-100\,\text{N s}.
\]
Therefore,
\[
p_{P,f}=100-100=0,
\]
so the person is momentarily at rest relative to the shore. By Newton's Third Law, the person exerts an equal eastward impulse on the boat:
\[
J_B=+100\,\text{N s}.
\]
The boat initially has momentum
\[
p_{B,i}=150(2)=300\,\text{kg m s}^{-1}.
\]
Hence,
\[
p_{B,f}=300+100=400\,\text{kg m s}^{-1}.
\]
Its final speed is
\[
v_{B,f}=\frac{400}{150}=\frac83\,\text{m s}^{-1}.
\]
The push-off impulses are equal and opposite even though the two bodies already had non-zero momentum before they interacted.
214. Block \(A\) rests on block \(B\), and \(B\) rests on a horizontal floor. Consider the chosen system consisting only of \(A+B\). Which statement is correct?
ⓐ. \(B\)-on-\(A\) is external; floor-on-\(B\) and Earth-on-block forces are external.
ⓑ. The \(A\)-\(B\) contact pair is internal; floor-on-\(B\) and Earth-on-block forces are external.
ⓒ. \(B\)-on-floor is external; the \(A\)-\(B\) contact pair is internal.
ⓓ. All normal forces are internal; only the gravitational forces are external.
Correct Answer: The \(A\)-\(B\) contact pair is internal; floor-on-\(B\) and Earth-on-block forces are external.
Explanation: System classification depends on the chosen boundary. Since both \(A\) and \(B\) are included, the contact interaction between them lies entirely inside the system. The force of \(B\) on \(A\) and the force of \(A\) on \(B\) form a third-law pair:
\[
\vec F_{BA}=-\vec F_{AB}.
\]
Their vector sum is therefore zero in the whole-system force equation. They have not disappeared physically; they remain essential if either block is analysed separately. The floor, however, lies outside the selected \(A+B\) system, so the floor's force on \(B\) is external. Earth also lies outside the selected system, so the gravitational forces exerted by Earth on \(A\) and \(B\) are external. The force of \(B\) on the floor acts on the floor, not on the selected system, so it does not belong on the system's free-body diagram. Hence option B is correct.
215. A fixed smooth pulley is supported from the ceiling. A light string passes over it so that two vertical string segments pull downward on the pulley, each with tension \(20\,\text{N}\). Neglect the pulley's weight. Which statement is correct?
ⓐ. Ceiling-on-pulley \(=20\,\text{N}\) upward; the two downward tensions cancel as a third-law pair.
ⓑ. Ceiling-on-pulley \(=40\,\text{N}\) upward; pulley-on-support \(=40\,\text{N}\) downward is its third-law partner.
ⓒ. Ceiling-on-pulley \(=40\,\text{N}\) upward; the two \(20\,\text{N}\) downward tensions are a third-law pair.
ⓓ. Ceiling-on-pulley \(=0\); a fixed pulley exerts no force on its support.
Correct Answer: Ceiling-on-pulley \(=40\,\text{N}\) upward; pulley-on-support \(=40\,\text{N}\) downward is its third-law partner.
Explanation: Each vertical string segment exerts a \(20\,\text{N}\) downward force on the pulley. Both forces act on the same body and in the same direction, so their resultant is
\[
20+20=40\,\text{N}
\]
downward. Since the pulley is fixed and its weight is neglected, the ceiling support must exert
\[
40\,\text{N}
\]
upward on the pulley. The third-law partner of this support force is the force exerted by the pulley on the support. It has equal magnitude \(40\,\text{N}\) and points downward. The two downward string forces are not a third-law pair because they are both exerted on the pulley. Each has its own partner force exerted by the pulley on the corresponding string segment. Thus option B correctly separates equilibrium forces from third-law partners.
216. Person \(A\) of mass \(50\,\text{kg}\) stands on a frictionless platform, while person \(B\) is braced against a rigid wall. They push directly on each other. A force sensor between their hands reads \(210\,\text{N}\). Person \(B\) remains at rest. Which statement is correct?
ⓐ. \(A\) accelerates at \(4.2\,\text{m s}^{-2}\); wall-on-\(B\) is the third-law partner of \(B\)-on-\(A\).
ⓑ. \(A\) has zero acceleration; because \(B\) remains at rest, \(B\)-on-\(A\) is zero.
ⓒ. \(A\) accelerates at \(4.2\,\text{m s}^{-2}\); \(A\)-on-\(B\) is larger than \(B\)-on-\(A\).
ⓓ. \(A\) accelerates at \(4.2\,\text{m s}^{-2}\); the wall supplies \(210\,\text{N}\) on \(B\) opposite to \(A\)-on-\(B\).
Correct Answer: \(A\) accelerates at \(4.2\,\text{m s}^{-2}\); the wall supplies \(210\,\text{N}\) on \(B\) opposite to \(A\)-on-\(B\).
Explanation: The hand-to-hand interaction obeys Newton's Third Law regardless of whether one person is moving. If \(A\) pushes \(B\) with \(210\,\text{N}\), then \(B\) simultaneously pushes \(A\) with \(210\,\text{N}\) in the opposite direction. Person \(A\) is on a frictionless support, so this interaction force produces
\[
a_A=\frac{210}{50}=4.2\,\text{m s}^{-2}.
\]
Person \(B\) remains at rest, so the net horizontal force on \(B\) must be zero. Therefore the wall must exert a force of \(210\,\text{N}\) opposite to the hand force exerted by \(A\) on \(B\). The wall force and the force from \(A\) both act on \(B\), so they are balancing forces rather than a third-law pair. The reaction partner of \(A\)'s force on \(B\) is \(B\)'s force on \(A\). Thus equal interaction forces can coexist with completely different accelerations when other external forces differ.
217. A runner of total mass \(60\,\text{kg}\) accelerates on a horizontal road while the ground exerts a forward frictional force of \(240\,\text{N}\) on the runner. In a second trial, the runner carries a \(20\,\text{kg}\) load, and the measured forward frictional force from the ground is again \(240\,\text{N}\). Neglect other horizontal forces. Which statement is correct?
ⓐ. Ground force \(240\,\text{N}\) forward in both trials; accelerations are \(4\) and \(3\,\text{m s}^{-2}\).
ⓑ. Ground force \(240\,\text{N}\) first and \(320\,\text{N}\) second; both accelerations are \(4\,\text{m s}^{-2}\).
ⓒ. Ground force is less than \(240\,\text{N}\) first and \(240\,\text{N}\) second; the runner’s backward force is larger.
ⓓ. Ground force \(240\,\text{N}\) backward in both trials; friction must oppose the runner’s forward motion.
Correct Answer: Ground force \(240\,\text{N}\) forward in both trials; accelerations are \(4\) and \(3\,\text{m s}^{-2}\).
Explanation: In the first trial, the horizontal resultant on the runner is
\[
F=240\,\text{N}.
\]
Hence,
\[
a_1=\frac{240}{60}=4\,\text{m s}^{-2}.
\]
In the second trial, the total moving mass is
\[
60+20=80\,\text{kg}.
\]
The measured horizontal force remains \(240\,\text{N}\), so
\[
a_2=\frac{240}{80}=3\,\text{m s}^{-2}.
\]
The forward force of the ground on the runner belongs to the foot-ground interaction. By Newton's Third Law, the runner exerts an equal-magnitude \(240\,\text{N}\) force backward on the ground in each trial. The increased mass changes the acceleration response, not the equality of the action-reaction forces. Friction on a walking or running person can act forward because its direction is determined by the tendency of relative slipping at the contact, not simply by the person's overall direction of motion.
218. A gun of mass \(4\,\text{kg}\) and a bullet of mass \(0.020\,\text{kg}\) are initially moving together eastward at \(3\,\text{m s}^{-1}\). During firing, the gun exerts an average eastward force of \(800\,\text{N}\) on the bullet for \(0.010\,\text{s}\). Neglect external horizontal impulse during this interval. What are the velocities immediately afterward?
ⓐ. Bullet \(403\,\text{m s}^{-1}\) east; gun \(1\,\text{m s}^{-1}\) east
ⓑ. Bullet \(400\,\text{m s}^{-1}\) east; gun \(2\,\text{m s}^{-1}\) west
ⓒ. Bullet \(403\,\text{m s}^{-1}\) east; gun \(2\,\text{m s}^{-1}\) west
ⓓ. Bullet \(397\,\text{m s}^{-1}\) east; gun \(1\,\text{m s}^{-1}\) west
Correct Answer: Bullet \(403\,\text{m s}^{-1}\) east; gun \(1\,\text{m s}^{-1}\) east
Explanation: Take east as positive. The impulse delivered by the gun to the bullet is
\[
J=800(0.010)=8\,\text{N s}.
\]
The bullet's initial momentum is
\[
p_{b,i}=0.020(3)=0.060\,\text{kg m s}^{-1}.
\]
Therefore,
\[
p_{b,f}=0.060+8=8.060\,\text{kg m s}^{-1}.
\]
Thus,
\[
v_{b,f}=\frac{8.060}{0.020}=403\,\text{m s}^{-1}.
\]
By Newton's Third Law, the bullet exerts an equal westward impulse of \(8\,\text{N s}\) on the gun. The gun's initial momentum is
\[
p_{g,i}=4(3)=12\,\text{kg m s}^{-1}.
\]
Hence,
\[
p_{g,f}=12-8=4\,\text{kg m s}^{-1},
\]
and
\[
v_{g,f}=\frac44=1\,\text{m s}^{-1}
\]
eastward. Recoil means that the gun's velocity changes opposite to the bullet's impulse; it need not mean that the gun reverses its absolute direction of motion.
219. Block \(A\) rests on block \(B\), which rests on floor \(C\). Consider only the contact forces among \(A\), \(B\) and \(C\). Which statement is correct when the selected system is \(A+B\)?
ⓐ. For \(A+B\), \(B\)-on-\(A\) is external while \(A\)-on-\(B\) is internal; \(C\)-on-\(B\) is external.
ⓑ. For \(A+B\), the \(A\)-\(B\) pair is internal; \(C\)-on-\(B\) is external, while \(B\)-on-\(C\) does not act on the system.
ⓒ. For \(A+B\), the \(A\)-\(B\) pair is internal; both \(C\)-on-\(B\) and \(B\)-on-\(C\) act on the system.
ⓓ. For \(A+B\), every third-law pair involving either \(A\) or \(B\) is internal and cancels from the system force sum.
Correct Answer: For \(A+B\), the \(A\)-\(B\) pair is internal; \(C\)-on-\(B\) is external, while \(B\)-on-\(C\) does not act on the system.
Explanation: The selected system contains both \(A\) and \(B\). Therefore the two forces belonging to their mutual contact interaction are internal:
\[
\vec F_{A\to B}=-\vec F_{B\to A}.
\]
Both act between parts of the chosen system, so they cancel when forces on the complete \(A+B\) system are summed. Floor \(C\), however, lies outside the system boundary. Thus the force exerted by \(C\) on \(B\) crosses the boundary and is an external force on the system. Its third-law partner is the force exerted by \(B\) on \(C\). That partner acts on the floor, not on the selected \(A+B\) system, so it is not placed on the system's free-body diagram. Third-law forces cancel in a system equation only when both members act on bodies lying inside that selected system. Hence option B is correct.
220. Two blocks of masses \(2\,\text{kg}\) and \(3\,\text{kg}\) are initially at rest on a frictionless horizontal surface with a compressed spring between them. At a certain instant during release, the spring exerts a force of magnitude \(12\,\text{N}\) on each block. At that same instant, an additional external force of \(10\,\text{N}\) acts to the right on the \(3\,\text{kg}\) block. Which statement is correct?
ⓐ. The spring forces cancel on each block separately, so only the \(3\,\text{kg}\) block accelerates.
ⓑ. The spring forces are internal to the two-block system, so \(d\vec p_{\rm sys}/dt=10\,\text{N}\) rightward; each block still responds to its spring force.
ⓒ. The spring forces are equal and opposite, so the two blocks must have equal-magnitude opposite accelerations despite their unequal masses.
ⓓ. The external \(10\,\text{N}\) force changes only one member of the spring-force pair, so the two spring forces need not remain equal.
Correct Answer: The spring forces are internal to the two-block system, so \(d\vec p_{\rm sys}/dt=10\,\text{N}\) rightward; each block still responds to its spring force.
Explanation: The spring exerts equal and opposite forces of \(12\,\text{N}\) on the two blocks. These forces form an internal third-law pair when both blocks are selected as one system. Their vector sum is therefore zero in the whole-system force balance. The only stated horizontal external force on the two-block system is \(10\,\text{N}\) to the right. Thus the rate of change of the total momentum of the system is governed by
\[
F_{\text{ext}}=10\,\text{N}.
\]
However, the spring forces still act physically on the individual blocks. The \(2\,\text{kg}\) block accelerates due to its \(12\,\text{N}\) spring force, while the \(3\,\text{kg}\) block responds to the vector sum of its spring force and the external \(10\,\text{N}\) force. Their individual accelerations therefore need not be equal in magnitude or opposite in direction. Internal cancellation is a system-level simplification, not a claim that internal forces have no effect on subsystem motion.
