CBSE Class 11 Physics Newton laws of Motion Advanced Problems

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

A uniform chain of length \(l\) lies on a smooth horizontal table with its length perpendicular to the edge of the table and a small portion of the chain hanging over the edge. The chain starts sliding from rest due to the weight of the hanging part. Find the acceleration and speed of the chain when the length of the hanging part is \(x\).

Answer: Weight of the hanging part \(=\dfrac{M}{l}xg\) moves the whole mass \(M\), so

\[ a=\frac{gx}{l},\qquad v=\sqrt{\frac gl}\,x \]

(from \( v\,dv=a\,dx \), starting from \(x\approx0\)).

Figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is \(\mu_1\), and that between the block of mass 4.0 kg and the incline is \(\mu_2\). Calculate the acceleration of the 2.0 kg block if

(a) \(\mu_1=0.20\) and \(\mu_2=0.30\)

(b) \(\mu_1=0.30\) and \(\mu_2=0.20\). Take \(g=10\ \text{m/s}^2\).

Answer: (a) \(2.7\ \text{m/s}^2\) (the blocks stay in contact and move together)

(b) \(2.4\ \text{m/s}^2\) (the blocks separate)

A light pulley is attached to the ceiling of a lift. A massless string passing over the pulley has masses \(m_1\) and \(m_2\) (\(m_1>m_2\)) attached to its two ends. If the lift starts moving up with an acceleration \(a\) then find

(a) the acceleration of \(m_1\) relative to the lift and the shaft of the lift;

(b) the force exerted by the pulley on the ceiling of the lift.

Answer: (a) Relative to the lift: \( \dfrac{(m_1-m_2)(g+a)}{m_1+m_2} \). Relative to the shaft (ground): \( \dfrac{(m_1-m_2)g-2m_2a}{m_1+m_2} \).

(b) \( \dfrac{4m_1m_2(g+a)}{m_1+m_2} \)

In the arrangement shown in the figure, the mass of ball B is equal to that of rod A and the length of the rod is \(l\). The pulleys and strings are massless and smooth. Initially the ball is set at the same level as the lower end of the rod and then released. Find the time taken by the ball to cross the upper end of the rod.

Answer: \[ t=\sqrt{\frac{10\,l}{3g}} \]

Block A of mass \(m\) and block B of mass \(2m\) are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in the figure. The wedge is inclined at 45° to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the system of A and B is released from rest, find

(a) the acceleration of A,

(b) the tension in the string, and

(c) the magnitude and direction of the force of friction acting on A. (\(g=10\ \text{m/s}^2\))

Answer: (a) Zero (the system stays at rest).

(b) \( T=\dfrac{50\,m}{3\sqrt2}\ \text{N}\approx11.8\,m\ \text{N} \) (\(m\) in kg)

(c) \( \dfrac{10\sqrt2\,m}{3}\ \text{N}\approx4.7\,m\ \text{N} \), directed down the incline.

Note: the friction forces here are not fully fixed by statics; the printed answers correspond to friction on A at its limiting value \( \tfrac23 mg\cos45^\circ \). The key omits the factor \(m\) in (c).

Two masses \(m\) and \(2m\) are connected by a massless string which passes over a light frictionless pulley as shown in figure. The masses are initially held with equal lengths of the strings on either side of the pulley. Find the velocity of the masses at the instant the lighter mass \(m\) moves up a distance of 6.54 metre. This string is suddenly cut at that instant. Calculate the time taken by each mass to reach the ground. Take \(g=9.81\ \text{m/s}^2\).

Answer: Acceleration \( a=\dfrac{(2m-m)g}{3m}=\dfrac g3=3.27\ \text{m/s}^2 \), so \( v=\sqrt{2\times3.27\times6.54}=6.54\ \text{m/s} \). The heavier mass (2 m) is then 6.54 m above the ground and moving down at 6.54 m/s: \( 6.54=6.54t+4.905t^2 \Rightarrow t\approx0.67 \) s. The lighter mass is 19.62 m above the ground and moving up at 6.54 m/s: \( -19.62=6.54t-4.905t^2 \Rightarrow t\approx2.78 \) s.

Notes: the question text says masses \(m\) and \(3m\), but the figure shows \(m\) and \(2m\) and the printed 0.67 s matches \(2m\) (with \(3m\) the heavier mass would take 0.60 s), so \(2m\) is used. The key gives 2.88 s for the lighter mass; the working above gives 2.78 s.

A chain of length \(l\) is placed on a smooth spherical surface of radius \(R\) with one of its ends fixed at the top of the sphere. What will be the acceleration \(a\) of each element of the chain when its upper end is released? It is assumed that the length of the chain \(l<\pi R/2\).

Answer: \[ a=\frac{Rg}{l}\left[1-\cos\left(\frac lR\right)\right] \]

Note: the source writes the chain length as \(\lambda\) in the condition; it is the same length \(l\).

A smooth semicircular wire-track of radius \(R\) is fixed in a vertical plane as given in the figure. One end of a massless spring of natural length \(3R/4\) is attached to the lower point O of the wire track. A small ring of mass \(m\), which can slide on the track, is attached to the other end of the spring. The ring is held stationary at point P such that the spring makes an angle of 60° with the vertical. The spring constant \(K=mg/R\). Consider the instant when the ring is released, determine the tangential acceleration of the ring and the normal reaction.

Answer: Tangential acceleration \( =\dfrac{5\sqrt3}{8}\,g \); normal reaction \(=\dfrac{3mg}{8}=3.75\,m\) newton (with \(g=10\ \text{m/s}^2\), \(m\) in kg).

Calculate the force \(F\) required to cause the block of weight \(W_1=200\) N just to slide the block of weight \(W_2=100\) N. What is the tension in the string AB? Coefficient of friction \(\mu=0.25\) for all surfaces in contact.

Answer: For \(W_2\): \(T\cos45^\circ=\mu N_2\) and \(N_2+T\sin45^\circ=100\), giving \( T=20\sqrt2\approx28.3\ \text{N} \) and friction on \(W_2\) \(=20\) N, \(N_2=80\) N. For \(W_1\): \( F=20+0.25\times(200+80)=90\ \text{N} \).

Tension in AB \(=20\sqrt2\) N; \(F=90\) N.

Note: the printed key gives "20 N, \(20\sqrt2\) N"; 20 N is only the friction between the two blocks. Since \(\mu=0.25\) applies to all surfaces, the floor friction (70 N) must be added, giving 90 N.

A very small cube of mass \(m\) is placed on the inside of a funnel rotating about a vertical axis at a constant rate of \(f\) rev/s. The wall of the funnel makes an angle \(\theta\) with the horizontal as shown in the figure. If the coefficient of static friction between the cube and the funnel is \(\mu\) and the centre of the cube is at a distance \(r\) from the axis of rotation, what are the largest and the smallest values of \(f\) for which the block will not move with respect to the funnel?

Answer: \[ f_{max}=\frac1{2\pi}\sqrt{\frac{g(\sin\theta+\mu\cos\theta)}{r(\cos\theta-\mu\sin\theta)}},\qquad f_{min}=\frac1{2\pi}\sqrt{\frac{g(\sin\theta-\mu\cos\theta)}{r(\cos\theta+\mu\sin\theta)}} \]

Note: the printed key leaves out \(r\) inside the square roots; it is needed (from \(\omega^2r\)) for the units to be those of frequency.

In the figure, a block of mass \(m=5\) kg is on the smooth inclined face of the wedge M of mass 10 kg, the inclination to the horizontal being \(\theta=37^\circ\). The wedge is resting on a smooth horizontal plane. Assuming the pulley P to be smooth and the string is light and inextensible, find the acceleration of M, when M and m are always in contact.

Answer: \( 2.5\ \text{m/s}^2 \) (the wedge moves towards the wall).

A plank of mass \(m_1\) with a bar of mass \(m_2\) placed on it lies on a smooth horizontal plane. A horizontal force growing with time \(t\) as \(F=bt\) (\(b\) is constant) is applied to the bar. Find how the accelerations of the plank \(a_1\) and of the bar \(a_2\) depend on \(t\), if the coefficient of friction between the plank and the bar is equal to \(\mu\). Draw the approximate plots of these dependences.

Answer: For \( t\le t_0 \): \( a_1=a_2=\dfrac{bt}{m_1+m_2} \), where \( t_0=\dfrac{\mu m_2 g\,(m_1+m_2)}{b\,m_1} \).

For \( t\ge t_0 \): \( a_1=\dfrac{\mu g\,m_2}{m_1} \), \( a_2=\dfrac{bt-\mu m_2g}{m_2} \).

Find the acceleration \(a\) of body 2 in the arrangement shown in Fig, if its mass is \(\eta\) times as great as the mass of bar 1 and the angle that inclined plane forms with the horizontal is equal to \(\alpha\). The masses of the pulleys and the threads, as well as the friction, are assumed to be negligible. Look into possible cases.

Answer: \[ a=\frac{2g\,(2\eta-\sin\alpha)}{4\eta+1} \]

Possible cases: if \(2\eta>\sin\alpha\), body 2 goes down with this acceleration (and bar 1 goes up the plane with \(a/2\)); if \(2\eta<\sin\alpha\), the expression is negative, so body 2 goes up and bar 1 slides down; if \(2\eta=\sin\alpha\), the system stays at rest.

In the arrangement shown in Fig. the mass of the rod \(M\) exceeds the mass \(m\) of the ball. The ball has an opening permitting it to slide along the thread with some friction. At the initial moment the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if \(t\) seconds after the beginning of motion the ball got opposite the upper end of the rod. The rod length equals \(l\).

Answer: \[ F=\frac{2l\,mM}{(M-m)\,t^2} \]

Figure shows a small block of mass \(m\) kept at the left end of a larger block of mass \(M\) and length \(l\). The system can slide on a horizontal road. The system is started towards right with an initial velocity \(v\). The friction coefficient between the road and the bigger block is \(\mu\) and that between the blocks is \(\mu/2\). Find the time elapsed before the smaller block separates from the bigger block.

Answer: \[ t=\sqrt{\frac{4Ml}{(M+m)\,\mu g}} \]

Find the acceleration of the block of mass \(M\). The coefficient of friction between the two blocks is \(\mu_1\) and that between the bigger block and the ground is \(\mu_2\).

Answer: \[ a=\frac{\left[2m-\mu_2(M+m)\right]g}{M+m\left[5+2(\mu_1-\mu_2)\right]} \]

Note: in the printed key the factor \(g\) is misplaced inside the bracket; it multiplies the whole numerator.

Find the acceleration of \(M\). The pulleys are light and the strings are massless. The pulleys and masses have negligible size.

Answer: \[ a=\frac{12m_1m_2-m_1M-m_2M}{36m_1m_2+m_1M+m_2M}\,g \]

Direction: downward (as given in the key).

Note: the direction is given as downward in the printed key.

A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30°. The friction coefficient between the road and the tyre is \( \dfrac{1}{2\sqrt3} \). Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take \(g=10\ \text{m/s}^2\).

Answer: Braking can at most give the full friction force \(\mu mg\cos30^\circ\). The smallest acceleration down the slope is then \( a=g\sin30^\circ-\mu g\cos30^\circ=5-2.5=2.5\ \text{m/s}^2 \). With \(u=21.6\ \text{km/h}=6\) m/s, \( v^2=6^2+2(2.5)(12.8)=100 \), so \(v=10\) m/s \(=36\) km/h at least.

Note: with the numbers given, the minimum speed is exactly 36 km/h, so "greater than" should read "not less than".

A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Find the frictional force exerted by the wall on the block.

Answer: 20 N. The weight (20 N) and the 15 N force together need 25 N of friction, but the largest friction available is \(0.5\times40=20\) N, so the block slips and the friction is 20 N (taking \(g=10\ \text{m/s}^2\)).

Further entries in the answer key:

Key entry (its question number is not legible in the PDF): \( \dfrac{19g}{29} \) (upward), \( \dfrac{17g}{29} \) (downward).

20. \( 17:15 \)

21. (a) \( m_1(\sin\alpha+\mu\cos\alpha) \)

(b) \( m_1(\sin\alpha-\mu\cos\alpha) \)

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