CBSE Class 11 Physics Newton laws of Motion Basic Problems

Class 12 Physics: Newton's Laws of Motion (Basic Level Assignment)

A mass \(M\) placed on a smooth table is connected to two light, inextensible strings which pass over opposite edges of the table to hang vertically. Masses \(m_1, m_2\) \((m_1>m_2)\) are attached to the other ends of the strings. Show that the acceleration of the system when released is \( \dfrac{m_1-m_2}{m_1+m_2+M}\,g \). Find the tensions in the two strings.

Answer: For \(m_1\): \( m_1g-T_1=m_1a \). For \(m_2\): \( T_2-m_2g=m_2a \). For \(M\): \( T_1-T_2=Ma \). Adding the three equations, \( (m_1-m_2)g=(m_1+m_2+M)a \), so

\[ a=\frac{m_1-m_2}{m_1+m_2+M}\,g \]

\[ T_1=m_1g\left[\frac{M+2m_2}{m_1+m_2+M}\right],\qquad T_2=m_2g\left[\frac{M+2m_1}{m_1+m_2+M}\right] \]

Three blocks A, B and C having masses 2 kg, 1 kg and 3 kg respectively, are hanging over a fixed pulley as shown. Find the accelerations of the masses and the tensions in the two strings. (\(g=10\ \text{m/s}^2\))

Answer: \( a=\dfrac{10}{3}\approx3.3\ \text{m/s}^2 \) (A moves up, B and C move down); \( T_1=26.7\ \text{N} \) (string over the pulley); \( T_2=20\ \text{N} \) (string between B and C).

One end of a light string is attached to a 10 kg mass placed on a smooth table. The string passes over the edge of the table and to its other end is attached a light pulley. Over this pulley passes another string having masses 6 kg and 3 kg attached to its ends. Find the acceleration of the 10 kg mass. Friction is absent everywhere.

Answer: \( \dfrac{4g}{9} \)

In the figure, block B weighs 10 N. The coefficient of friction between B and the table is 0.25. Find the maximum weight of block A for which the system will be in equilibrium.

Answer: Maximum friction on B \(=0.25\times10=2.5\) N, so the horizontal string can have tension \(T_1=2.5\) N at most. At the knot, \(T_2\cos45^\circ=T_1\) and \(T_2\sin45^\circ=W_A\), so \( W_A=T_1=2.5\ \text{N} \).

Note: the printed key says 25 N, which is the answer if B weighed 100 N. With the 10 N given in the question the answer is 2.5 N.

In the figure, A is a 10 kg block and B is a 5 kg block.

(a) Find the minimum weight (block C) which must be placed on A to keep it from sliding if \(\mu_s\) between A and the table is 0.20.

(b) The block C is suddenly lifted off A. What is the acceleration of A if \(\mu_k\) between A and the table is 0.15?

Answer: (a) 15 kg (i.e. a weight of 15 kg-wt)

(b) \( 2.3\ \text{m/s}^2 \)

In the given set up, find the acceleration of the blocks and the tensions in the strings. The strings and pulleys \(P_1\) and \(P_2\) are massless and the inclined plane is fixed. Neglect friction.

Answer: \( a_1=\dfrac{8}{3}\ \text{m/s}^2 \) (2 kg block, up the incline), \( a_2=\dfrac{4}{3}\ \text{m/s}^2 \) (4 kg block, downward), \( T_1=\dfrac{52}{3}\ \text{N} \), \( T_2=\dfrac{104}{3}\ \text{N} \) (taking \(g=10\ \text{m/s}^2\)).

Note: the printed key shows \(a_1=8\) and \(a_2=4\) m/s², with the "/3" missing; the tensions given in the key (and the constraint \(a_1=2a_2\)) fit \(8/3\) and \(4/3\).

Two particles of each equal masses \(m\) are connected by a light string of length \(2l\) and are placed on a smooth horizontal table as shown in figure. A constant force \(F\) is applied continuously at the mid-point of the string, always along the perpendicular bisector of the straight line joining the two particles. When the distance between the two particles is \(2x\), find the acceleration of the particle.

Answer: If the tension is \(T\) and each half of the string makes angle \(\theta\) with the line of \(F\), then \( 2T\cos\theta=F \) with \( \cos\theta=\dfrac{\sqrt{l^2-x^2}}{l} \). The force on each particle along the table is \( T\sin\theta=T\dfrac{x}{l} \), so

\[ a=\frac{F}{2m}\cdot\frac{x}{\sqrt{l^2-x^2}} \]

Note: the printed key has \(l\) (not \(x\)) in the numerator, which does not follow from the working above and looks like a misprint.

Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length one metre, as given in figure. The upper wire has negligible mass and the lower wire has a uniform mass of 0.2 kg/m. The whole system of blocks, wires and support have an upward acceleration of 0.2 m/s². Acceleration due to gravity is 9.8 m/s². Find the tension

(a) at the middle point of the lower wire.

(b) at the middle point of the upper wire.

Answer: (a) 20 N

(b) 50 N

A block of mass 2 kg slides on an inclined plane which makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surfaces is \( \sqrt{3/2} \). (\(g=9.8\ \text{m/s}^2\))

(a) What force should be applied to the block so that the block moves down without any acceleration?

(b) What force should be applied to the block so that it moves up without any acceleration?

(c) Calculate the ratio of power needed in the above two cases, if the block moves with the same speed in both the cases.

Answer: (a) 11 N (along the incline, up the plane)

(b) 30.6 N (along the incline, up the plane)

(c) \( \dfrac{P_1}{P_2}=\dfrac{F_1}{F_2}\approx0.36 \)

Note: the printed key gives 0.3610; the exact ratio is \(10.99/30.59=0.359\).

A bob hanging from the ceiling of a car acts as an accelerometer. Derive a relation for the horizontal acceleration \(a\) of the car to the angle \(\theta\) made by the bob with the vertical. If the brakes are applied, the speed of the car changes from 5 m/s to \( \dfrac53 \) m/s during 3 seconds. Find the angle through which the thread will deflect from the vertical direction. (\(g=10\ \text{m/s}^2\)).

Answer: (a) The bob moves with the car, so the horizontal component of the tension gives the acceleration and the vertical component balances the weight: \( T\sin\theta=ma \) and \( T\cos\theta=mg \). Hence \( a=g\tan\theta \).

(b) The retardation is \( a=\dfrac{5-5/3}{3}=\dfrac{10}{9}\ \text{m/s}^2 \), so \( \tan\theta=\dfrac{a}{g}=\dfrac19 \) and \( \theta\approx6^\circ20' \).

Two touching bars 1 and 2 are placed on an inclined plane forming an angle \(\alpha\) with the horizontal. The masses of the bars are equal to \(m_1\) and \(m_2\), and the coefficients of friction between the inclined plane and these bars are equal to \(\mu_1\) and \(\mu_2\) respectively, with \(\mu_1>\mu_2\). Find:

(a) the force of interaction of the bars in the process of motion;

(b) the minimum value of the angle \(\alpha\) at which the bars start sliding down.

Answer: (a) \[ F=\frac{(\mu_1-\mu_2)\,m_1m_2\,g\cos\alpha}{m_1+m_2} \]

(b) \[ \tan\alpha_{min}=\frac{\mu_1m_1+\mu_2m_2}{m_1+m_2} \]

The friction coefficient between the board and the floor shown in figure is \(\mu\). Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Answer: \[ T_{max}=\frac{\mu(M+m)g}{1+\mu} \]

Block A in figure weighs 2.70 N, and block B weighs 5.40 N. The coefficient of kinetic friction between all surfaces is 0.25. Find the magnitude of the horizontal force \(\mathbf F\) necessary to drag block B to the left at constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

Answer: 3.38 N

The monkey B shown in figure is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in the tail, what force should it apply on the rope in order to carry the monkey B with it? Take \(g=10\ \text{m/s}^2\).

Answer: Between 70 N and 105 N.

Find the mass M of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

Answer: \[ M=\frac{M^\prime+m}{\cot\theta-1} \]

Find the acceleration of the 500 g block in figure.

Answer: \( \dfrac{8}{13}\,g \) downwards

In a simple Atwood machine two unequal masses \(m_1\) and \(m_2\) are connected by a string going over a clamped light smooth pulley. In a typical arrangement \(m_1=300\) g and \(m_2=600\) g. The system is released from rest. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.

Answer: \( \dfrac23 \) s

The elevator shown in figure is descending with an acceleration of 2 m/s². The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B?

Answer: 4 N

Note: the question does not state \(g\); the key's value uses \(g=10\ \text{m/s}^2\) (\(N=m(g-a)=0.5\times8\)).

Let \(m_1=1\) kg, \(m_2=2\) kg and \(m_3=3\) kg in figure. Find the accelerations of \(m_1\), \(m_2\) and \(m_3\). The string from the upper pulley to \(m_1\) is 20 cm when the system is released from rest. How long will it take before \(m_1\) strikes the pulley?

Answer: \( \dfrac{19}{29}g \) (up) for \(m_1\), \( \dfrac{17}{29}g \) (down) for \(m_2\), \( \dfrac{21}{29}g \) (down) for \(m_3\); time taken \(=0.25\) s.

A particle is attached by means of two equal strings, each of length \(l\) m, to two points A and B in the same vertical line and describes a horizontal circle with a uniform speed as shown in figure. If the number of revolutions per second is 2, find the ratio of the tensions in the upper and the lower strings. Assume \(\theta\) to be small.

Answer: \( T_1:T_2=17:15 \)

Note: this follows from \( T_1+T_2=m\omega^2l \) and \( T_1-T_2\approx mg \), with \(\omega=4\pi\) rad/s, \(l=1\) m and \(g=9.8\ \text{m/s}^2\) (the length is not given numerically in the question).

A block with mass \(m_1\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block that has mass \(m_2\) by a cord passing over a small, frictionless pulley. The coefficient of static friction is \(\mu_s\), and the coefficient of kinetic friction is \(\mu_k\).

(a) Find the mass \(m_2\) for which block \(m_1\) moves up the plane at constant speed once it has been set in motion.

(b) Find the mass \(m_2\) for which block \(m_1\) moves down the plane at constant speed once it has been set in motion.

(c) For what range of values of \(m_2\) will the blocks remain at rest if they are released from rest?

Answer: (a) \( m_2=m_1(\sin\alpha+\mu_k\cos\alpha) \)

(b) \( m_2=m_1(\sin\alpha-\mu_k\cos\alpha) \)

(c) \( m_1(\sin\alpha-\mu_s\cos\alpha)\le m_2\le m_1(\sin\alpha+\mu_s\cos\alpha) \)