Class 11 Physics | Laws of Motion MCQs with Answers – Part 6

Explanation: An inertial frame is one in which Newton’s first law holds without adding any fictitious force. A bus accelerating forward is itself an accelerating frame. A passenger who is not firmly supported may appear to move backward relative to the bus, even when no real backward interaction is acting on the passenger in the ground frame. To describe motion from the accelerating bus frame, a pseudo force may be introduced opposite to the frame’s acceleration. This does not remove real forces such as weight, normal reaction, or friction. The need for such an extra frame-dependent force is the sign that the bus frame is non-inertial.

Explanation: Newton’s first law states that a body continues in its state of rest or uniform straight-line motion unless acted upon by a non-zero net external force. If the puck moves with constant velocity, its acceleration is zero. Therefore the net force along the horizontal direction is nearly zero. A continuous forward force is not required to maintain uniform motion in the absence of friction. In ordinary situations, moving objects slow down because resistive forces such as friction act on them. The frictionless example separates the idea of motion from the idea of force.

Explanation: \( \textbf{Given:} \) \(m=2\,\text{kg}\), initial velocity \(6\,\text{m s}^{-1}\) east, final velocity \(6\,\text{m s}^{-1}\) north. \( \textbf{Initial momentum magnitude:} \) \[p_i=mv=(2)(6)=12\,\text{kg m s}^{-1}\] \( \textbf{Final momentum magnitude:} \) \[p_f=12\,\text{kg m s}^{-1}\] The two momentum vectors are perpendicular. \( \textbf{Magnitude of change in momentum:} \) \[|\Delta \vec{p}|=\sqrt{p_i^2+p_f^2}\] \[|\Delta \vec{p}|=\sqrt{12^2+12^2}\] \[|\Delta \vec{p}|=12\sqrt{2}\,\text{kg m s}^{-1}\] The speed is unchanged, but momentum changes because momentum is a vector. \( \textbf{Final answer:} \) \(12\sqrt{2}\,\text{kg m s}^{-1}\).

Explanation: Newton’s second law in its general form is \(\vec{F}=\frac{d\vec{p}}{dt}\). It says that the net external force is equal to the rate of change of momentum. For constant mass, \(\vec{p}=m\vec{v}\), so \(\frac{d\vec{p}}{dt}=m\frac{d\vec{v}}{dt}=m\vec{a}\). This gives the familiar form \(\vec{F}=m\vec{a}\). The momentum form is broader in meaning because it begins with momentum change rather than only acceleration. It also makes impulse and momentum-change ideas fit naturally with Newton’s laws.

Explanation: \( \textbf{Given relation:} \) \[p=4t+3\] Newton’s second law in momentum form is: \[F=\frac{dp}{dt}\] The slope of the \(p-t\) relation gives the force. \( \textbf{Differentiate momentum with respect to time:} \) \[\frac{dp}{dt}=4\] So: \[F=4\,\text{N}\] The constant term \(3\) represents momentum at \(t=0\), not force. \( \textbf{Final answer:} \) The net force is \(4\,\text{N}\).

Explanation: For constant mass, Newton’s second law gives \(a=\frac{F}{m}\). If the original acceleration is \(a=\frac{F}{m}\), then after doubling both force and mass the new acceleration is: \[a'=\frac{2F}{2m}\] \[a'=\frac{F}{m}=a\] The acceleration depends on the ratio of net force to mass. Doubling only the force would double the acceleration, and doubling only the mass would halve it. Changing both by the same factor leaves the ratio unchanged. \( \textbf{Final answer:} \) The acceleration is unchanged.

Explanation: Mass is a measure of inertia and does not change merely because the body is taken to another planet. The mass remains \(5\,\text{kg}\). Weight is the gravitational force and is given by \(W=mg\). On the new planet: \[W=(5)(4)\] \[W=20\,\text{N}\] The unit \( \text{kg} \) belongs to mass, while \( \text{N} \) belongs to weight. The distinction matters because the same object can have different weights in different gravitational fields. \( \textbf{Final answer:} \) Mass \(=5\,\text{kg}\), weight \(=20\,\text{N}\).

Explanation: \( \textbf{Graph meaning:} \) Impulse is the area under the force-time graph. The graph is a trapezium with parallel sides \(4\,\text{N}\) and \(10\,\text{N}\). The time interval is \(3\,\text{s}\). \( \textbf{Trapezium area:} \) \[J=\frac{1}{2}(F_1+F_2)\Delta t\] \( \textbf{Substitution:} \) \[J=\frac{1}{2}(4+10)(3)\] \[J=\frac{1}{2}(14)(3)\] \[J=21\,\text{N s}\] This is also the change in momentum along the force direction. \( \textbf{Final answer:} \) The impulse is \(21\,\text{N s}\).

Explanation: \( \textbf{Given:} \) Impulse \(J=24\,\text{N s}\), contact time \(\Delta t=0.006\,\text{s}\). Average force is related to impulse by: \[J=F_{\text{avg}}\Delta t\] So: \[F_{\text{avg}}=\frac{J}{\Delta t}\] \( \textbf{Substitution:} \) \[F_{\text{avg}}=\frac{24}{0.006}\] \[F_{\text{avg}}=4000\,\text{N}\] The large value occurs because the same impulse is delivered in a very short time. \( \textbf{Final answer:} \) The average force is \(4000\,\text{N}\).

Explanation: \( \textbf{Given:} \) \(m=6\,\text{kg}\), \(F=50\,\text{N}\), \(\theta=37^\circ\), \(\mu_k=0.25\), and \(g=10\,\text{m s}^{-2}\). \( \textbf{Horizontal component of pull:} \) \[F_x=F\cos37^\circ=(50)(0.8)=40\,\text{N}\] \( \textbf{Upward component of pull:} \) \[F_y=F\sin37^\circ=(50)(0.6)=30\,\text{N}\] \( \textbf{Weight:} \) \[mg=(6)(10)=60\,\text{N}\] \( \textbf{Normal reaction:} \) \[N=mg-F_y=60-30=30\,\text{N}\] \( \textbf{Kinetic friction:} \) \[f_k=\mu_kN=(0.25)(30)=7.5\,\text{N}\] \( \textbf{Net horizontal force:} \) \[\sum F_x=40-7.5=32.5\,\text{N}\] \( \textbf{Acceleration:} \) \[a=\frac{32.5}{6}\approx5.42\,\text{m s}^{-2}\] The closest listed value is \(5.0\,\text{m s}^{-2}\), but the exact calculation gives approximately \(5.42\,\text{m s}^{-2}\). \( \textbf{Final answer:} \) The acceleration is closest to \(5.0\,\text{m s}^{-2}\).

Explanation: A push directed below the horizontal has a downward vertical component. This component adds to the weight in the vertical force balance. Therefore the normal reaction becomes \(N=mg+F\sin\theta\) when there is no vertical acceleration. Since friction in the elementary model is proportional to normal reaction, \(f=\mu N\), the frictional force or frictional limit increases. This is different from pulling upward at an angle, which reduces \(N\). The direction of friction remains along the surface, not perpendicular to it.

Explanation: The expression for sliding down a rough inclined plane is: \[a=g(\sin\theta-\mu_k\cos\theta)\] If \(\mu_k=0\), the surface is frictionless in this model. Substitute \(\mu_k=0\): \[a=g(\sin\theta-0\cdot\cos\theta)\] \[a=g\sin\theta\] This matches the standard result for a smooth inclined plane. The limiting case is useful because it checks whether the rough-plane equation reduces to the known frictionless form. \( \textbf{Final answer:} \) \(a=g\sin\theta\).

Explanation: The block moves up the plane with constant velocity, so acceleration along the plane is zero. The applied force \(P\) acts up the plane. The component of weight \(mg\sin\theta\) acts down the plane. Since the block slides up, kinetic friction also acts down the plane. The normal reaction is: \[N=mg\cos\theta\] So kinetic friction is: \[f_k=\mu_kmg\cos\theta\] For zero acceleration: \[P=mg\sin\theta+\mu_kmg\cos\theta\] Both opposing terms must be balanced by the applied force. \( \textbf{Final answer:} \) \(P=mg\sin\theta+\mu_kmg\cos\theta\).

Explanation: For a fixed mass with no opposing force, \(F=ma\), so an \(F-a\) graph passes through the origin. If a constant opposing force such as kinetic friction must first be overcome, the applied force may be written as \(F_{\text{applied}}=ma+f\). This gives a straight line with a non-zero intercept on the force axis. The slope can still represent mass, but the intercept represents the extra force term. Such a graph should not be read as a failure of Newton’s second law. It shows that the plotted applied force is not the same as the net force.

Explanation: The centripetal force can be written as: \[F_c=\frac{mv^2}{r}\] Using \(v=\omega r\), we also have: \[\omega=\frac{v}{r}\] So: \[mv\omega=mv\left(\frac{v}{r}\right)\] \[mv\omega=\frac{mv^2}{r}\] Therefore \(F_c=mv\omega\). This form is useful when both linear speed and angular speed are known. \( \textbf{Final answer:} \) \(F_c=mv\omega\).

Explanation: At the no-friction design speed, the normal reaction components alone provide the required centripetal force and vertical balance. If the speed is increased, the vehicle needs a larger inward centripetal force. The tendency is to move up the bank relative to the road surface. Static friction opposes this tendency, so it acts down the bank. This down-bank friction has a component toward the centre and helps supply the extra inward force. At lower than design speed, the friction direction would reverse.

Explanation: In a conical pendulum, the bob moves in a horizontal circle. The string length \(L\) is the slant length from the support to the bob. If the string makes angle \(\theta\) with the vertical, the horizontal distance from the vertical axis to the bob is the radius \(r\). In the right triangle formed, this horizontal side is opposite the angle \(\theta\). Therefore: \[r=L\sin\theta\] Using \(L\cos\theta\) would give the vertical projection of the string, not the circle radius. \( \textbf{Final answer:} \) \(r=L\sin\theta\).

Explanation: In the ideal pulley model, the pulley is smooth and the string is light, so the same tension is used on both sides. If the pulley has friction, this simplification may fail. The tension on one side of the pulley can differ from the tension on the other side. This does not mean tension becomes zero. It also does not remove the weights of the masses or force the acceleration to vanish. The equality of tension is a modelling assumption, so it must be checked from the problem statement.

Explanation: \( \textbf{Given:} \) \(m=0.10\,\text{kg}\), initial speed \(20\,\text{m s}^{-1}\), rebound speed \(20\,\text{m s}^{-1}\). Take the initial direction as positive. Then: \[u=+20\,\text{m s}^{-1}\] \[v=-20\,\text{m s}^{-1}\] \( \textbf{Change in momentum:} \) \[\Delta p=m(v-u)\] \[\Delta p=0.10[-20-(+20)]\] \[\Delta p=0.10(-40)\] \[\Delta p=-4.0\,\text{kg m s}^{-1}\] The magnitude of impulse is \(|\Delta p|=4.0\,\text{N s}\). The answer is not zero because the direction of momentum reverses. \( \textbf{Final answer:} \) The impulse magnitude is \(4.0\,\text{N s}\).

Explanation: \( \textbf{Given:} \) \(m=3\,\text{kg}\), applied force \(F=20\,\text{N}\), maximum static friction \(18\,\text{N}\), and kinetic friction \(12\,\text{N}\). Since: \[20\,\text{N}\gt18\,\text{N}\] the block starts moving. After sliding begins, kinetic friction is used, not maximum static friction. \( \textbf{Net force during sliding:} \) \[\sum F=20-12=8\,\text{N}\] \( \textbf{Acceleration:} \) \[a=\frac{\sum F}{m}\] \[a=\frac{8}{3}\,\text{m s}^{-2}\] The friction value changes when the block changes from rest to sliding. \( \textbf{Final answer:} \) \(a=\frac{8}{3}\,\text{m s}^{-2}\).