201. The total linear momentum of a system of particles is found by
ⓐ. adding only the speeds of all particles
ⓑ. vector addition of all particle momenta
ⓒ. multiplying total mass by total time
ⓓ. taking the largest single momentum only
Correct Answer: vector addition of all particle momenta
Explanation: Linear momentum is a vector quantity, so total momentum must be found by vector addition. For a system, the total momentum can be written as \(\vec{P}=\vec{p}_1+\vec{p}_2+\vec{p}_3+\cdots\). If all particles move along one straight line, signs can be used to represent direction. Adding only speeds would ignore mass and direction. Taking only the largest momentum would ignore the contribution of other particles. The word “total” in momentum conservation always refers to the vector sum for the chosen system.
202. Momentum of a system is conserved when the net external force on the system is
ⓐ. equal to the largest internal force
ⓑ. zero or negligible
ⓒ. always equal to its total mass
ⓓ. opposite to every velocity in the system
Correct Answer: zero or negligible
Explanation: Conservation of linear momentum applies to a chosen system when the net external force on that system is zero or negligible. Internal forces may act between parts of the system, but they occur in equal and opposite pairs. These internal forces cannot change the total momentum of the whole system. External forces are different because they come from bodies outside the chosen system. If the net external force is non-zero, the total momentum of the system changes with time. The condition for conservation is about net external force, not about the absence of all forces.
203. Two blocks collide on a smooth horizontal surface and are treated together as one system. During the collision, the contact force exerted by one block on the other is
ⓐ. an internal force of the two-block system
ⓑ. an external force of the two-block system
ⓒ. the weight of the two-block system
ⓓ. a force that acts on neither block
Correct Answer: an internal force of the two-block system
Explanation: The classification of a force depends on the system chosen. If both colliding blocks are included in one system, the force between them is an interaction between parts of the same system. Such a force is internal to the two-block system. The force may change the momentum of each block separately, but it does not change the total momentum of the two-block system by itself. Weight would be a gravitational force due to Earth and is not the same as the contact force between the blocks. This system choice is the reason momentum conservation is useful in collisions.
204. Study the table and identify the row that correctly states the conservation condition.
| Row | System condition | Momentum conclusion |
|---|
| P | \(\sum\vec{F}_{\text{ext}}=0\) | Total momentum remains constant |
| Q | Internal forces are large | Total momentum must become zero |
| R | One object has large mass | Total momentum is always conserved |
| S | Objects are moving fast | External force is impossible |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row S
ⓓ. Row P
Correct Answer: Row P
Explanation: Row P gives the correct condition for conservation of total linear momentum. If \(\sum\vec{F}_{\text{ext}}=0\), then the total momentum of the chosen system remains constant. Large internal forces can change individual momenta strongly, but they do not by themselves change the system’s total momentum. A large mass does not automatically guarantee momentum conservation if an external force acts. High speed also does not remove the possibility of external forces. The conservation condition is decided by the net external force on the chosen system.
205. A two-particle system has momenta \(+6\,\text{kg m s}^{-1}\) and \(-6\,\text{kg m s}^{-1}\) along the same straight line. The total momentum of the system is
ⓐ. \(+12\,\text{kg m s}^{-1}\)
ⓑ. \(-12\,\text{kg m s}^{-1}\)
ⓒ. \(36\,\text{kg m s}^{-1}\)
ⓓ. \(0\)
Correct Answer: \(0\)
Explanation: \( \textbf{Given momenta:} \) \(p_1=+6\,\text{kg m s}^{-1}\) and \(p_2=-6\,\text{kg m s}^{-1}\).
The signs show opposite directions along the same line.
\( \textbf{Total momentum relation:} \)
\[P=p_1+p_2\]
\( \textbf{Substitution:} \)
\[P=(+6)+(-6)\]
\( \textbf{Calculation:} \)
\[P=0\]
The two momenta cancel as vectors because they have equal magnitudes and opposite directions.
This does not mean the particles are individually at rest.
\( \textbf{Final answer:} \) The total momentum of the system is zero.
206. A system is called isolated for momentum-conservation purposes when
ⓐ. no internal interaction occurs inside it
ⓑ. all bodies in it are at rest
ⓒ. every body in it has zero mass
ⓓ. no net external force acts on it
Correct Answer: no net external force acts on it
Explanation: An isolated system, in the momentum sense, is one for which the net external force is zero or negligible. The bodies inside the system may still interact strongly with each other. Those internal forces can redistribute momentum among the bodies but cannot change the total momentum of the whole isolated system. The bodies need not be at rest; a moving isolated system can have constant total momentum. Mass does not become zero in an isolated system. The key boundary is between forces from inside the system and forces from outside it.
207. A cart of mass \(2\,\text{kg}\) moves east at \(3\,\text{m s}^{-1}\), and another cart of mass \(1\,\text{kg}\) moves west at \(4\,\text{m s}^{-1}\). Taking east as positive, the total momentum is
ⓐ. \(+2\,\text{kg m s}^{-1}\)
ⓑ. \(+10\,\text{kg m s}^{-1}\)
ⓒ. \(-2\,\text{kg m s}^{-1}\)
ⓓ. \(-10\,\text{kg m s}^{-1}\)
Correct Answer: \(+2\,\text{kg m s}^{-1}\)
Explanation: \( \textbf{Sign convention:} \) East is positive, so west is negative.
\( \textbf{Cart 1 data:} \) \(m_1=2\,\text{kg}\), \(v_1=+3\,\text{m s}^{-1}\).
\( \textbf{Cart 2 data:} \) \(m_2=1\,\text{kg}\), \(v_2=-4\,\text{m s}^{-1}\).
\( \textbf{Momentum of cart 1:} \)
\[p_1=m_1v_1=(2)(+3)=+6\,\text{kg m s}^{-1}\]
\( \textbf{Momentum of cart 2:} \)
\[p_2=m_2v_2=(1)(-4)=-4\,\text{kg m s}^{-1}\]
\( \textbf{Total momentum:} \)
\[P=p_1+p_2=+6+(-4)\]
\[P=+2\,\text{kg m s}^{-1}\]
The positive sign means the total momentum is eastward.
\( \textbf{Final answer:} \) The total momentum is \(+2\,\text{kg m s}^{-1}\).
208. A system has total momentum changing with time. The most direct conclusion is that
ⓐ. internal forces are absent
ⓑ. all particles must have equal masses
ⓒ. the system must be at rest
ⓓ. a non-zero external net force acts
Correct Answer: a non-zero external net force acts
Explanation: For a system, Newton’s second law can be written in the form \(\sum\vec{F}_{\text{ext}}=\frac{d\vec{P}}{dt}\). If the total momentum \(\vec{P}\) changes with time, then \(\frac{d\vec{P}}{dt}\) is not zero. This means the net external force is not zero. Internal forces can change the momenta of individual parts, but they do not change the total momentum of the whole system. Equal masses are not required for conservation. A system can be moving and still have constant total momentum if the net external force is zero.
209. Consider the statements below.
I. Momentum conservation is applied to a chosen system.
II. Internal forces can change individual momenta inside the system.
III. Internal forces alone can change the total momentum of an isolated system.
The suitable set is
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is suitable because momentum conservation always refers to a selected system. Statement II is also suitable because internal forces can transfer momentum from one part of the system to another. Statement III is not suitable because internal forces occur in equal and opposite pairs within the system. Their total contribution to the rate of change of system momentum cancels. For an isolated system, total momentum remains constant even though individual momenta may change. The distinction between individual momentum and total system momentum is essential here.
210. A shell explodes into two parts while floating freely in space, with no external force during the explosion. If the shell was initially at rest, the total momentum of the two parts just after explosion is
ⓐ. equal to the larger part’s momentum only
ⓑ. equal to the smaller part’s momentum only
ⓒ. always vertically downward
ⓓ. zero
Correct Answer: zero
Explanation: The shell is the chosen system, and the explosion forces between its parts are internal. Since no external force is acting during the explosion, the total momentum of the system is conserved. The initial momentum of the shell was zero because it was at rest. Therefore the total momentum after explosion must also be zero. The two parts can move in opposite directions with equal and opposite momenta. The total momentum is not the momentum of one part alone, because the vector sum of both parts must be considered.
211. A \(3\,\text{kg}\) body moving east at \(2\,\text{m s}^{-1}\) and a \(1\,\text{kg}\) body moving east at \(5\,\text{m s}^{-1}\) form a system. The system’s total momentum is
ⓐ. \(7\,\text{kg m s}^{-1}\) east
ⓑ. \(3\,\text{kg m s}^{-1}\) east
ⓒ. \(15\,\text{kg m s}^{-1}\) east
ⓓ. \(11\,\text{kg m s}^{-1}\) east
Correct Answer: \(11\,\text{kg m s}^{-1}\) east
Explanation: \( \textbf{Body 1 data:} \) \(m_1=3\,\text{kg}\), \(v_1=2\,\text{m s}^{-1}\) east.
\( \textbf{Body 2 data:} \) \(m_2=1\,\text{kg}\), \(v_2=5\,\text{m s}^{-1}\) east.
Both momenta point east, so their magnitudes add.
\( \textbf{Momentum of body 1:} \)
\[p_1=m_1v_1=(3)(2)=6\,\text{kg m s}^{-1}\]
\( \textbf{Momentum of body 2:} \)
\[p_2=m_2v_2=(1)(5)=5\,\text{kg m s}^{-1}\]
\( \textbf{Total momentum:} \)
\[P=p_1+p_2=6+5\]
\[P=11\,\text{kg m s}^{-1}\]
The direction remains east because both momenta are eastward.
\( \textbf{Final answer:} \) The total momentum is \(11\,\text{kg m s}^{-1}\) east.
212. Two bodies interact only with each other in deep space. During the interaction, one body gains momentum \(+8\,\text{kg m s}^{-1}\). The momentum change of the other body is
ⓐ. \(+8\,\text{kg m s}^{-1}\)
ⓑ. \(-8\,\text{kg m s}^{-1}\)
ⓒ. \(0\)
ⓓ. \(+16\,\text{kg m s}^{-1}\)
Correct Answer: \(-8\,\text{kg m s}^{-1}\)
Explanation: The two bodies form an isolated system because they interact only with each other. Total momentum of the system must remain constant. If one body gains \(+8\,\text{kg m s}^{-1}\), the other must gain an equal change in the opposite direction. This keeps the total change in system momentum equal to zero. In symbols, \(\Delta p_1+\Delta p_2=0\). Therefore \(\Delta p_2=-\Delta p_1=-8\,\text{kg m s}^{-1}\). Internal forces change individual momenta in opposite ways.
213. A system contains two carts connected by a compressed spring. The spring is released on a smooth horizontal track. Treating both carts as one system, the spring force is
ⓐ. internal to the two-cart system
ⓑ. external, so it must change the total momentum of the two-cart system
ⓒ. gravitational, so it acts vertically only
ⓓ. absent because the track is smooth
Correct Answer: internal to the two-cart system
Explanation: If both carts and the spring interaction between them are considered within the same system, the spring force is internal. The spring can push the carts apart and change each cart’s momentum separately. However, the forces on the two carts are equal and opposite, so their contributions to total system momentum cancel. A smooth track reduces external horizontal forces such as friction. The spring force is not absent; it is simply internal to the chosen system. This is why the total horizontal momentum can remain conserved while the carts move apart.
214. A momentum-conservation equation is written for two bodies before and after interaction in one dimension. The correct general form is
ⓐ. \(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)
ⓑ. \(m_1u_1-m_2u_2=m_1v_1-m_2v_2\)
ⓒ. \(m_1u_1+m_2u_2=0\) always
ⓓ. \(u_1+u_2=v_1+v_2\) always
Correct Answer: \(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)
Explanation: In one-dimensional motion, momentum conservation is written by equating total initial momentum to total final momentum. The initial momenta are \(m_1u_1\) and \(m_2u_2\). The final momenta are \(m_1v_1\) and \(m_2v_2\). Therefore the suitable equation is \(m_1u_1+m_2u_2=m_1v_1+m_2v_2\), with signs chosen according to direction. The total final momentum is not always zero; it equals the initial total momentum. Adding velocities alone is not enough because mass also affects momentum.
215. The derivation of momentum conservation for two interacting bodies uses Newton’s third law mainly to show that
ⓐ. external force must always be large
ⓑ. both bodies must have the same acceleration
ⓒ. both bodies must stop after interaction
ⓓ. internal force changes cancel in total
Correct Answer: internal force changes cancel in total
Explanation: During an interaction, body \(1\) exerts a force on body \(2\), and body \(2\) exerts an equal and opposite force on body \(1\). By Newton’s second law in momentum form, each force equals the rate of change of momentum of the body on which it acts. Because the internal forces are equal and opposite, the rates of change of the two momenta are equal and opposite. Adding them gives zero rate of change of total momentum when no net external force acts. This gives conservation of total momentum. The derivation does not require equal accelerations, because the two masses may be different.
216. Assertion: Internal forces cannot change the total momentum of an isolated system.
Reason: Internal forces between two parts of the system occur in equal and opposite pairs.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Internal forces act between parts of the same chosen system. Newton’s third law says that each such interaction produces equal and opposite forces on the two interacting parts. These forces can change the momentum of each part separately. However, their contributions to the total momentum change of the whole system cancel. Therefore internal forces cannot change total momentum of an isolated system. The Reason gives the mechanism by which the Assertion becomes true.
217. A two-body system has total initial momentum \(18\,\text{kg m s}^{-1}\) east. During a short interaction, the net external impulse is negligible. The total final momentum is
ⓐ. \(0\)
ⓑ. \(18\,\text{kg m s}^{-1}\) west
ⓒ. \(18\,\text{kg m s}^{-1}\) east
ⓓ. impossible to relate to the initial momentum
Correct Answer: \(18\,\text{kg m s}^{-1}\) east
Explanation: \( \textbf{Initial total momentum:} \) \(\vec{P}_i=18\,\text{kg m s}^{-1}\) east.
\( \textbf{Condition:} \) Net external impulse is negligible.
Negligible external impulse means the change in total system momentum is negligible.
\( \textbf{Conservation statement:} \)
\[\vec{P}_f=\vec{P}_i\]
\( \textbf{Substitution:} \)
\[\vec{P}_f=18\,\text{kg m s}^{-1}\text{ east}\]
Internal forces during the interaction may redistribute momentum between the two bodies.
They do not change the total momentum of the system under the stated condition.
\( \textbf{Final answer:} \) The total final momentum is \(18\,\text{kg m s}^{-1}\) east.
218. Read the case below and answer the question.
Two carts collide on a horizontal track. In Case P, the track is smooth and the collision time is very short. In Case Q, a strong external brake acts on one cart during the collision.
Momentum conservation for the two-cart system is more reliable in
ⓐ. Case Q
ⓑ. Case P
ⓒ. both cases equally, regardless of the brake
ⓓ. neither case because collisions never conserve momentum
Correct Answer: Case P
Explanation: Momentum conservation for a chosen system requires the net external force or external impulse to be zero or negligible during the interaction. In Case P, the smooth track and short collision time make external horizontal impulse small, so conservation is more reliable. In Case Q, the external brake applies an outside force to one cart during the collision. That external impulse can change the total momentum of the two-cart system. Collisions can conserve momentum when the proper system is chosen and external effects are negligible. The brake case shows why the conservation condition must be checked before writing the equation.
219. Two bodies of masses \(m\) and \(2m\) are initially at rest on a smooth surface. They push apart due only to an internal spring. If the body of mass \(m\) moves right with speed \(2v\), the body of mass \(2m\) moves
ⓐ. right with speed \(v\)
ⓑ. left with speed \(2v\)
ⓒ. left with speed \(v\)
ⓓ. right with speed \(4v\)
Correct Answer: left with speed \(v\)
Explanation: \( \textbf{Initial condition:} \) Both bodies are at rest, so total initial momentum is zero.
Since the spring force is internal and the surface is smooth, total momentum remains zero.
Take right as positive.
\( \textbf{Momentum of mass \(m\):} \)
\[p_1=m(2v)=2mv\]
Let the velocity of mass \(2m\) be \(V\).
\( \textbf{Conservation of momentum:} \)
\[0=2mv+(2m)V\]
\( \textbf{Solving:} \)
\[(2m)V=-2mv\]
\[V=-v\]
The negative sign means the body of mass \(2m\) moves left.
\( \textbf{Final answer:} \) The \(2m\) body moves left with speed \(v\).
220. In a derivation using \(\sum\vec{F}_{\text{ext}}=\frac{d\vec{P}}{dt}\), if \(\sum\vec{F}_{\text{ext}}=0\), then
ⓐ. every particle in the system is at rest
ⓑ. every internal force becomes zero
ⓒ. total kinetic energy must be zero
ⓓ. \(\vec{P}\) is constant
Correct Answer: \(\vec{P}\) is constant
Explanation: The equation \(\sum\vec{F}_{\text{ext}}=\frac{d\vec{P}}{dt}\) relates the net external force on a system to the rate of change of total momentum. If \(\sum\vec{F}_{\text{ext}}=0\), then \(\frac{d\vec{P}}{dt}=0\). This means the total momentum \(\vec{P}\) does not change with time. It does not require every particle to be at rest. Internal forces may still act and may change the momenta of individual particles. Momentum conservation is a statement about the total vector momentum of the system, not about the absence of all motion.
