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101. A frame of reference in which Newton’s first law holds without adding any extra force is called
ⓐ. an inertial frame
ⓑ. a heating frame
ⓒ. a magnetic frame
ⓓ. a deforming frame
Correct Answer: an inertial frame
Explanation: An inertial frame is a frame in which a body with zero net external force remains at rest or continues in uniform straight-line motion. In such a frame, Newton’s first law is valid in its usual form. The frame itself is not accelerating in a way that creates apparent changes in motion. For many problems, the ground is treated as approximately inertial when Earth’s rotation and other small effects are ignored. A frame name is not decided by temperature, magnetism, or deformation. The key test is whether the first-law behaviour is observed without introducing an extra apparent force.
102. A bus suddenly accelerates forward, and a standing passenger seems to fall backward relative to the bus. The bus frame during this acceleration is best treated as
ⓐ. an inertial frame because the passenger appears to move
ⓑ. a non-inertial frame because the bus accelerates
ⓒ. an inertial frame because Earth attracts the passenger
ⓓ. a frame with no possible laws of motion
Correct Answer: a non-inertial frame because the bus accelerates
Explanation: The bus frame is accelerating forward, so it is not an inertial frame during that interval. A passenger who was initially at rest tends to remain in the earlier state of motion because of inertia. Relative to the accelerating bus, the passenger appears to move backward. This apparent backward tendency is not caused by a real backward contact force on the whole body. In a ground frame, the explanation can be given using inertia and contact forces from the bus floor. The same event can look different in different frames, so the nature of the chosen frame matters.
103. In many ordinary problems on blocks, carts, and projectiles near Earth, the ground frame is approximately treated as inertial because
ⓐ. Earth has no gravity near its surface
ⓑ. all forces vanish in the ground frame
ⓒ. velocity cannot be measured from the ground
ⓓ. its acceleration effects are negligible
Correct Answer: its acceleration effects are negligible
Explanation: A perfectly inertial frame is an idealisation, but many frames are close enough for ordinary calculations. The ground frame is commonly treated as approximately inertial in mechanics because effects such as Earth’s rotation are often small for the situation being studied. This does not mean gravity vanishes near Earth. Forces such as weight, normal reaction, friction, and tension may still act. The approximation only means Newton’s laws can be applied in their usual form with good accuracy. The validity of the approximation depends on the scale and precision of the problem.
104. Study the table and identify the row that is most suitable.
| Row | Frame description | Classification |
|---|---|---|
| P | A frame moving with constant velocity relative to an inertial frame | Inertial |
| Q | A frame accelerating forward relative to the ground | Inertial |
| R | A frame rotating steadily with a turntable | Strictly inertial |
| S | A frame in a suddenly braking bus | Inertial |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row S
ⓓ. Row P
Correct Answer: Row P
Explanation: A frame moving with constant velocity relative to an inertial frame is also treated as inertial. In such a frame, Newton’s first law can hold in the usual form. A frame that is accelerating forward is non-inertial because the frame itself is changing velocity. A rotating frame is also non-inertial because its direction of motion changes continuously. A suddenly braking bus is an accelerating frame, so it is not inertial during braking. The difference is not whether objects are moving in the frame, but whether the frame itself has acceleration.
105. Assertion: A frame attached to an accelerating lift is a non-inertial frame.
Reason: The lift frame has acceleration relative to an approximately inertial ground frame.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A lift that accelerates upward or downward carries its frame along with that acceleration. A frame with acceleration relative to an inertial or approximately inertial frame is non-inertial. In such a frame, bodies may appear to have accelerations that are not explained by the usual real forces alone unless an apparent force is introduced. The Reason gives the actual basis for calling the lift frame non-inertial. A lift moving with constant velocity would be a different case from an accelerating lift. The classification depends on acceleration of the frame, not simply on whether the lift is high, low, or moving.
106. A coin lies on a smooth tray inside a train. The train moves with constant velocity along a straight track, and the coin remains at rest relative to the tray. This observation is consistent with treating the train frame as
ⓐ. approximately inertial
ⓑ. necessarily non-inertial
ⓒ. a frame where force has no direction
ⓓ. a frame where mass becomes zero
Correct Answer: approximately inertial
Explanation: If the train moves with constant velocity in a straight line, its acceleration is zero relative to the ground. A frame moving uniformly relative to an inertial frame can also be treated as inertial. The coin remaining at rest relative to the tray is consistent with zero net horizontal force in that frame. This would not be the same if the train suddenly accelerated, braked, or turned. The frame does not make mass vanish or remove the vector nature of force. The important condition is uniform straight-line motion of the frame itself.
107. A frame fixed to a car moving around a circular path at constant speed is not inertial because
ⓐ. the car’s velocity direction changes
ⓑ. the car has constant speed during the turn
ⓒ. the car has non-zero mass only
ⓓ. the road has a normal reaction
Correct Answer: the car’s velocity direction changes
Explanation: A car moving around a circular path at constant speed still has changing velocity because velocity includes direction. Since the car frame changes direction continuously, it is an accelerating frame. An accelerating frame is non-inertial. Constant speed alone is not enough to make a frame inertial. The car’s mass and the road’s normal reaction are real physical features, but they do not decide the inertial character by themselves. The frame classification depends on whether the frame has acceleration, including acceleration due to direction change.
108. Read the case below and answer the question.
Two observers describe the same ball. Observer P stands on the ground beside a straight road. Observer Q sits inside a car that is speeding up along the road. The ball is momentarily free from horizontal contact forces.
Which description about the frames is most suitable?ⓐ. Both frames must be non-inertial
ⓑ. Observer Q’s frame is inertial because the car has speed
ⓒ. P is approximately inertial; Q is non-inertial
ⓓ. Observer P’s frame is impossible because the ground has gravity
Correct Answer: P is approximately inertial; Q is non-inertial
Explanation: The ground frame is often treated as approximately inertial for ordinary mechanics near Earth. The car frame is speeding up, so it is accelerating relative to the ground. Therefore Observer Q’s frame is non-inertial during that speeding-up interval. Having speed does not by itself make a frame non-inertial; acceleration is the deciding point. Gravity near the ground does not make the ground frame impossible for applying Newton’s laws in usual situations. The comparison shows why the same physical event must be described carefully after choosing the frame.
109. In an accelerating frame, an apparent force may be introduced only to
ⓐ. replace every real interaction in all frames
ⓑ. prove that Newton’s first law is false in inertial frames
ⓒ. change the unit of mass from \( \text{kg} \) to \( \text{N} \)
ⓓ. describe motion from that non-inertial frame
Correct Answer: describe motion from that non-inertial frame
Explanation: In a non-inertial frame, the frame itself is accelerating. To use a Newton-like equation from that frame, an apparent or pseudo force is sometimes introduced. This is a description tool for the accelerating frame, not a new real interaction between two bodies. In an inertial frame, the same event can usually be explained using only real forces and acceleration. The idea does not change the SI unit of mass, which remains \( \text{kg} \). At this level, the main boundary is to recognise non-inertial frames without turning pseudo force into an ordinary contact or field force.
110. A frame moving along a straight line has velocity \(12\,\text{m s}^{-1}\) for \(5\,\text{s}\) without changing speed or direction. Relative to an inertial frame, it is best described as
ⓐ. inertial during this interval
ⓑ. non-inertial because its velocity is non-zero
ⓒ. non-inertial because time is passing
ⓓ. inertial only if no body is observed in it
Correct Answer: inertial during this interval
Explanation: \( \textbf{Given frame motion:} \) The frame has constant velocity \(12\,\text{m s}^{-1}\) along a straight line.
\( \textbf{Acceleration check:} \) Since speed and direction do not change, the frame acceleration is zero.
\( \textbf{Condition for inertial treatment:} \) A frame with no acceleration relative to an inertial frame can be treated as inertial.
The non-zero velocity of the frame does not make it non-inertial.
\( \textbf{Why the time interval matters:} \) The given \(5\,\text{s}\) only says how long the constant-velocity condition lasts.
If the frame started accelerating after this interval, the classification would need to be reconsidered then.
\( \textbf{Final answer:} \) During the stated interval, the frame can be treated as inertial.
111. Newton’s second law in momentum form relates net external force to
ⓐ. rate of change of momentum
ⓑ. the total distance travelled
ⓒ. the colour of the moving body
ⓓ. the mass of Earth only
Correct Answer: rate of change of momentum
Explanation: Newton’s second law in its more general form states that the net external force is equal to the rate of change of momentum. Symbolically, it is written as \(\vec{F}=\frac{d\vec{p}}{dt}\). This form emphasizes that force is connected with how quickly momentum changes. Momentum can change because speed changes, direction changes, mass changes in special systems, or more than one of these occurs. In the usual constant-mass case, this law leads to \(\sum\vec{F}=m\vec{a}\). Distance travelled alone does not decide the force unless it is connected with a change of momentum or energy in a specific situation.
112. The vector direction of the net force in the momentum form of Newton’s second law is the same as the direction of
ⓐ. change of displacement only
ⓑ. rate of change of momentum
ⓒ. mass only
ⓓ. time only
Correct Answer: rate of change of momentum
Explanation: The momentum form of Newton’s second law is \(\vec{F}=\frac{d\vec{p}}{dt}\). This is a vector equation, so the direction of \(\vec{F}\) is the direction of \(\frac{d\vec{p}}{dt}\). It is not necessarily the direction of velocity or displacement. For example, a body slowing down has a force opposite to its velocity because its momentum is decreasing in that direction. Mass and time are scalar quantities in this context and do not provide a spatial direction. The force direction is decided by how the momentum vector is changing.
113. A force acting for a short time changes a body’s momentum from \(6\,\text{kg m s}^{-1}\) east to \(14\,\text{kg m s}^{-1}\) east in \(2\,\text{s}\). The average force is
ⓐ. \(10\,\text{N}\) east
ⓑ. \(4\,\text{N}\) west
ⓒ. \(4\,\text{N}\) east
ⓓ. \(20\,\text{N}\) east
Correct Answer: \(4\,\text{N}\) east
Explanation: \( \textbf{Initial momentum:} \) \(\vec{p}_1=6\,\text{kg m s}^{-1}\) east.
\( \textbf{Final momentum:} \) \(\vec{p}_2=14\,\text{kg m s}^{-1}\) east.
\( \textbf{Time interval:} \) \(\Delta t=2\,\text{s}\).
Take east as positive.
\( \textbf{Change in momentum:} \)
\[\Delta\vec{p}=\vec{p}_2-\vec{p}_1\]
\[\Delta\vec{p}=14-6=8\,\text{kg m s}^{-1}\]
\( \textbf{Average-force relation:} \)
\[\vec{F}_{\text{avg}}=\frac{\Delta\vec{p}}{\Delta t}\]
\( \textbf{Substitution:} \)
\[\vec{F}_{\text{avg}}=\frac{8\,\text{kg m s}^{-1}}{2\,\text{s}}\]
\[\vec{F}_{\text{avg}}=4\,\text{kg m s}^{-2}=4\,\text{N}\]
The positive sign gives an eastward average force.
\( \textbf{Final answer:} \) The average force is \(4\,\text{N}\) east.
114. A body’s momentum-time graph is described below.
The graph of momentum \(p\) versus time \(t\) is a straight line rising uniformly from left to right.
For motion along one straight line, the slope of this graph representsⓐ. force
ⓑ. velocity
ⓒ. displacement
ⓓ. mass
Correct Answer: force
Explanation: In one-dimensional motion, the slope of a momentum-time graph gives \(\frac{\Delta p}{\Delta t}\). Newton’s second law in momentum form connects this rate of change of momentum with force. For an instantaneous graph, the slope represents \(\frac{dp}{dt}\). If the graph is a straight rising line, the force is constant and in the positive direction chosen for momentum. Velocity is related to momentum through \(p=mv\) only when mass is known. The graph directly gives force through its slope, not displacement or mass.
115. A ball initially moving east is struck so that its momentum changes toward west. The average force during the strike is directed
ⓐ. east
ⓑ. vertically upward
ⓒ. nowhere, because force is scalar
ⓓ. west
Correct Answer: west
Explanation: Force in the momentum form follows the direction of change in momentum. If the ball’s momentum changes toward west, then \(\Delta\vec{p}\) is westward. The average force is \(\vec{F}_{\text{avg}}=\frac{\Delta\vec{p}}{\Delta t}\), and \(\Delta t\) is a positive scalar. Therefore the force has the same direction as \(\Delta\vec{p}\). It need not be in the direction of the initial velocity. Force is a vector quantity, so its direction is essential in this situation.
116. Consider the statements below.
I. Newton’s second law in momentum form is \(\vec{F}=\frac{d\vec{p}}{dt}\).
II. The force in this form refers to net external force.
III. The force must always be in the direction of velocity.
The suitable set is
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I, II, and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I gives the standard momentum form of Newton’s second law. Statement II is also important because the law relates the rate of change of momentum of a body or system to the net external force. Statement III is not generally true. A force can act opposite to velocity when a body slows down, or perpendicular to velocity when the direction changes. The force direction follows the change in momentum, not necessarily the velocity itself. This distinction is especially useful in collisions, braking, and circular motion.
117. A \(0.40\,\text{kg}\) ball moving east at \(10\,\text{m s}^{-1}\) is brought to rest in \(0.20\,\text{s}\). Taking east as positive, the average force on the ball is
ⓐ. \(-20\,\text{N}\)
ⓑ. \(+20\,\text{N}\)
ⓒ. \(-2.0\,\text{N}\)
ⓓ. \(+2.0\,\text{N}\)
Correct Answer: \(-20\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(m=0.40\,\text{kg}\), \(v_1=+10\,\text{m s}^{-1}\), \(v_2=0\), and \(\Delta t=0.20\,\text{s}\).
East is taken as positive.
\( \textbf{Initial momentum:} \)
\[p_1=mv_1=(0.40)(+10)=+4.0\,\text{kg m s}^{-1}\]
\( \textbf{Final momentum:} \)
\[p_2=mv_2=(0.40)(0)=0\]
\( \textbf{Change in momentum:} \)
\[\Delta p=p_2-p_1=0-(+4.0)\]
\[\Delta p=-4.0\,\text{kg m s}^{-1}\]
\( \textbf{Average force:} \)
\[F_{\text{avg}}=\frac{\Delta p}{\Delta t}\]
\[F_{\text{avg}}=\frac{-4.0}{0.20}\,\text{N}\]
\[F_{\text{avg}}=-20\,\text{N}\]
The negative sign shows that the force is opposite to the initial eastward motion.
\( \textbf{Final answer:} \) The average force is \(-20\,\text{N}\).
118. In the equation \(\vec{F}=\frac{d\vec{p}}{dt}\), the term \(\frac{d\vec{p}}{dt}\) means
ⓐ. momentum multiplied by time
ⓑ. momentum divided by mass only
ⓒ. distance travelled per unit time
ⓓ. rate of change of momentum
Correct Answer: rate of change of momentum
Explanation: The notation \(\frac{d\vec{p}}{dt}\) represents the instantaneous rate at which momentum changes with time. Since \(\vec{p}\) is a vector, its rate of change is also a vector quantity. Newton’s second law identifies this vector rate with the net external force. It is not the same as multiplying momentum by time. Distance per unit time is speed, not rate of change of momentum. This notation becomes especially useful when force or momentum changes continuously.
119. A body has momentum \(p\) along a straight line. The graph of \(p\) against \(t\) is a horizontal line. According to Newton’s second law in momentum form, the net force is
ⓐ. zero
ⓑ. equal to \(p\)
ⓒ. equal to \(pt\)
ⓓ. always increasing
Correct Answer: zero
Explanation: A horizontal momentum-time graph means momentum is constant with time. Therefore the rate of change of momentum is zero. In symbols, \(\frac{d\vec{p}}{dt}=0\). Newton’s second law in momentum form gives \(\vec{F}=\frac{d\vec{p}}{dt}\). Hence the net force is zero in the direction represented by the graph. A body may still have non-zero momentum, but a constant non-zero momentum does not imply a non-zero net force.
120. A constant net force of \(5\,\text{N}\) acts on a body along a straight line for \(3\,\text{s}\). The change in momentum of the body is
ⓐ. \(15\,\text{kg m s}^{-1}\)
ⓑ. \(5\,\text{kg m s}^{-1}\)
ⓒ. \(3\,\text{kg m s}^{-1}\)
ⓓ. \(1.67\,\text{kg m s}^{-1}\)
Correct Answer: \(15\,\text{kg m s}^{-1}\)
Explanation: \( \textbf{Given:} \) Constant force \(F=5\,\text{N}\) and time interval \(\Delta t=3\,\text{s}\).
\( \textbf{Required:} \) Change in momentum \(\Delta p\).
From Newton’s second law in average form:
\[F=\frac{\Delta p}{\Delta t}\]
\( \textbf{Rearranging:} \)
\[\Delta p=F\Delta t\]
\( \textbf{Substitution:} \)
\[\Delta p=(5\,\text{N})(3\,\text{s})\]
\( \textbf{Calculation:} \)
\[\Delta p=15\,\text{N s}\]
\( \textbf{Unit equivalence:} \)
\[1\,\text{N s}=1\,\text{kg m s}^{-1}\]
So,
\[\Delta p=15\,\text{kg m s}^{-1}\]
The direction of \(\Delta\vec{p}\) is the direction of the applied net force.
\( \textbf{Final answer:} \) The change in momentum is \(15\,\text{kg m s}^{-1}\).