CBSE Class 12 Physics Centre of Mass Advanced Assignment.

CBSE Class 12 Physics Centre of Mass Advanced Assignment. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Class 12 Physics: Centre of Mass – Advanced Level Assignment

A plastic ball is dropped from a height of 1 m and rebounds several times from the floor. If 1.3 seconds elapse from the moment it is dropped to the second impact with the floor, what is the coefficient of restitution between the ball and the floor?

Answer: \( e=0.94 \)

A ball is dropped on the ground from a height \(h\). If the coefficient of restitution is \(e\), then find the total distance travelled by the ball before coming to rest and the total time elapsed.

Answer: \[ \frac{h(1+e^2)}{1-e^2},\qquad \sqrt{\frac{2h}{g}}\left[\frac{1+e}{1-e}\right] \]

A block of mass \(m_1 = 150\) kg is at rest on a very long frictionless table, one end of which is terminated in a wall. Another block of mass \(m_2\) is placed between the first block and the wall, and set in motion towards \(m_1\) with constant speed \(u_2\). Assuming that all collisions are completely elastic, find the value of \(m_2\) for which both blocks move with the same velocity after \(m_2\) has collided once with \(m_1\) and once with the wall. The wall has effectively infinite mass.

Answer: 50 kg

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position and released. The ball hits the wall, the coefficient of restitution being \( 2/\sqrt5 \). What is the minimum number of collisions after which the amplitude of oscillation becomes less than \(60^\circ\)?

Answer: 4

A block A of mass \(2m\) is placed on another block B of mass \(4m\) which in turn is placed on a fixed table. The two blocks have the same length \(4d\) and they are placed as shown in the figure. The coefficient of friction (both static and kinetic) between the block B and the table is \(\mu\). There is no friction between the two blocks. A small object of mass \(m\) moving horizontally along a line passing through the centre of mass of the block B and perpendicular to its face with a speed \(v\) collides elastically with the block B at a height \(d\) above the table. (a) What is the minimum value of \(v\) (call it \(v_0\)) required to make the block A topple? (b) If \(v=2v_0\), find the distance (from the point P) at which the mass \(m\) falls on the table after collision.

Answer: (a) \( v_0=\dfrac52\sqrt{6\mu g d} \)

(b) \( 6d\sqrt{3\mu} \)

Note: the printed key gives \(6d\sqrt{4\mu}\) for part (b). Working: after the elastic collision the small mass rebounds with speed \( \tfrac35 v=3\sqrt{6\mu g d} \) and falls through a height \(d\) in time \( \sqrt{2d/g} \), so the distance is \( 3\sqrt{12\mu}\,d=6d\sqrt{3\mu} \). Part (a) agrees with the key.

A 60 kg man and a 50 kg woman are standing on opposite ends of a platform of mass 20 kg. The platform is placed on smooth horizontal ground. The man and the woman begin to approach each other. Find the displacement of the platform when the two meet, in terms of the displacement \(x_0\) of the man relative to the platform. The length of the platform is 6 m.

Answer: \[ \frac{11x_0-30}{13}\ \text{m} \]

Note: this is the displacement of the platform in the direction opposite to the man's motion (towards the woman's side). Taking the man's direction as positive, the displacement is \( (30-11x_0)/13 \).

A rope thrown over a pulley has a ladder with a man A on one of its ends and a counterbalancing mass \(M\) on its other end. The man, whose mass is \(m\), climbs upwards by \(\Delta\vec r\) relative to the ladder and then stops. Ignoring the masses of the pulley and the rope, as well as the friction in the pulley axis, find the displacement of the centre of mass of this system.

Answer: \[ \frac{m}{2M}\,\Delta\vec r \]

A drinking straw of length \( \dfrac{3a}{2} \) and mass \(2m\) is placed on a square table of side \(a\), parallel to one of its sides, such that one third of its length extends beyond the table. An insect of mass \(m/2\) lands on the inner end of the straw (i.e. the end which lies on the table) and walks along the straw until it reaches the outer end. It does not topple even when another insect lands on top of the first one. Find the largest mass of the second insect that can be there without toppling the straw. Neglect friction.

Answer: \( m/2 \)

Working: the straw's centre of mass is \( 3a/4 \) from the inner end, i.e. \( a/4 \) inside the table edge (the edge is \(a\) from the inner end). Taking torques about the edge, the weight of the straw gives a restoring torque \( 2m g\cdot\frac a4=\frac{mga}{2} \). The two insects at the outer end (\(a/2\) beyond the edge) give \( \left(\frac m2+x\right) g\cdot\frac a2 \). For no toppling \( \frac{mga}{2}\ge\left(\frac m2+x\right)\frac{ga}{2} \), so \( x\le \frac m2 \).

Note: the printed key gives "5m", which does not agree with this working.

A boy throws a ball with initial speed \( 2\sqrt{ag} \) at an angle \(\theta\) to the horizontal. It strikes a smooth vertical wall and returns to his hand. Show that if the boy is standing at a distance \(a\) from the wall, the coefficient of restitution between the ball and the wall equals \( \dfrac{1}{4\sin2\theta-1} \). Also show that \(\theta\) cannot be less than \(15^\circ\).

Answer: Let \(u=2\sqrt{ag}\). Time to reach the wall is \( \dfrac{a}{u\cos\theta} \); after the impact the horizontal speed is \( e\,u\cos\theta \), so the return time is \( \dfrac{a}{e\,u\cos\theta} \). The total time equals the time of flight, \( \dfrac{2u\sin\theta}{g} \):

\[ \frac{a}{u\cos\theta}\left(1+\frac1e\right)=\frac{2u\sin\theta}{g}\ \Rightarrow\ 1+\frac1e=\frac{u^2\sin2\theta}{ag}=4\sin2\theta\ \Rightarrow\ e=\frac{1}{4\sin2\theta-1} \]

Since \(e\le1\), \(4\sin2\theta-1\ge1\), i.e. \( \sin2\theta\ge\tfrac12 \), so \(2\theta\ge30^\circ\) and \( \theta\ge15^\circ \).

A ball is projected from a point A on a smooth inclined plane which makes an angle \(\alpha\) to the horizontal. The velocity of projection makes an angle \(\theta\) with the plane upwards. If on the second bounce the ball is moving perpendicular to the plane, find \(e\) in terms of \(\alpha\) and \(\theta\). Here \(e\) is the coefficient of restitution between the ball and the plane.

Answer: \[ e=\frac12\cot\alpha\cot\theta-1 \]

Note: the printed key shows "cos \(\theta\)" in place of "cot \(\theta\)". Working: the time of the first flight is \( \dfrac{2u\sin\theta}{g\cos\alpha} \) and of the second \( e \) times that. The component of velocity along the plane vanishes at the second impact: \( u\cos\theta=g\sin\alpha\,\dfrac{2u\sin\theta(1+e)}{g\cos\alpha} \), which gives \(1+e=\tfrac12\cot\alpha\cot\theta\).

Two identical smooth balls are projected towards each other from points A and B on the horizontal ground with the same speed of projection. The angle of projection in each case is \(30^\circ\). The distance between A and B is 100 m. The balls collide in air and return to their respective points of projection. If the coefficient of restitution is \(e=0.7\), find (a) the speed of projection of either ball, (b) the coordinates of the point, with respect to A, where the balls collide. (Take \(g=10\ \text{m/s}^2\).)

Answer: (a) 37.5 m/s (b) (50 m, 17 m)

Three identical particles A, B and C lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC, and \(\angle ABC\) is \(135^\circ\). An impulse \(J\) is applied to the particle C in the direction BC. Find the initial speed of each particle. The mass of each particle is \(m\).

Answer: \[ v_A=\frac{\sqrt2\,J}{7m},\qquad v_B=\frac{\sqrt{10}\,J}{7m},\qquad v_C=\frac{3J}{7m} \]

A 2 kg sphere A is connected to a fixed point O by an inextensible cord of length 1.2 m. The sphere is resting on a frictionless horizontal surface at a distance of 0.5 m from O when it is given a velocity \(v_0\) in a direction perpendicular to the line OA. It moves freely until it reaches position A′ when the cord becomes taut. Determine (a) the maximum allowable velocity \(v_0\) if the impulse of the force exerted on the cord is not to exceed 3 N s, (b) the loss of energy as the cord becomes taut, if the sphere is given the maximum allowable velocity \(v_0\).

Answer: (a) 1.65 m/s (b) 2.25 J

An open car of mass 1000 kg and running at 25 m/s holds three men each of mass 75 kg. Each man runs with a speed of 5 m/s relative to the car and jumps off from the back end. Find the speed of the car if the three men jump off (a) in succession, (b) all together. Neglect friction between the car and the ground.

Answer: (a) 25.981 m/s (b) 25.918 m/s

Note: the printed key gives 25.981 m/s for both parts. For (b): \( 1225\times25=1000v+225(v-5)\Rightarrow v=\dfrac{31750}{1225}=25.918 \) m/s. Part (a) (successive jumps: 25.306, 25.632, 25.981 m/s) agrees with the key.

Ball B is hanging from an inextensible cord BC. An identical ball A is released from rest when it is just touching the cord and acquires a velocity \(v_0\) before striking ball B. Assuming perfectly elastic impact (\(e=1\)) and no friction, determine the velocity of each ball immediately after impact.

Answer: \( |\vec v_A'|=0.721\,v_0 \) directed at \(16.1^\circ\) to the horizontal, as shown; \( |\vec v_B'|=0.693\,v_0 \) (horizontal).

A particle whose initial mass is \(m_0\) is projected vertically upwards at time \(t=0\) with speed \(gT\), where \(T\) is a constant. The particle gradually acquires mass on its way up and at time \(t\) the mass of the particle has increased to \(m_0e^{t/T}\). If the added mass is at rest relative to the particle when it is acquired, find the time when the particle is at its highest point and its mass at that instant.

Answer: \[ T\ln 2,\qquad 2m_0 \]

Two blocks of mass 2 kg and \(M\) are at rest on an inclined plane and are separated by a distance of 6.0 m as shown. The coefficient of friction between each block and the inclined plane is 0.25. The 2 kg block is given a velocity of 10.0 m/s up the inclined plane. It collides with \(M\), comes back and has a velocity of 1.0 m/s when it reaches its initial position. The other block \(M\) after the collision moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block \(M\). [Take \(\sin\theta\approx\tan\theta=0.05\) and \(g=10\ \text{m/s}^2\).]

Answer: \( e=0.84 \), \( M=15.12 \) kg

Working: up the incline the retardation is \(g(\sin\theta+\mu\cos\theta)=3\) m/s\(^2\); the 2 kg block reaches \(M\) with \(v=\sqrt{100-2(3)(6)}=8\) m/s. Coming down, friction exceeds the gravity component, so the retardation is \(2\) m/s\(^2\) and \(1=v'^2-2(2)(6)\), giving \(v'=5\) m/s down just after the collision. \(M\) rises 0.5 m with retardation 3 m/s\(^2\), so its speed after the collision is \(\sqrt3\) m/s. Then \(e=\dfrac{\sqrt3+5}{8}=0.84\), and momentum conservation gives \( 2(8)=-2(5)+M v_M \), with \(v_M=0.84\times8-5=1.72\) m/s (using the rounded \(e=0.84\)), gives \( M=\dfrac{26}{1.72}=15.12 \) kg (the unrounded \(v_M=\sqrt3\) would give 15.01 kg).

A small sphere of radius \(R\) is held against the inner surface of a larger sphere of radius \(6R\) (as shown in the figure). The masses of the large and small spheres are \(4M\) and \(M\) respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position.

Answer: \( (L+2R,\ 0) \)

A block of mass \(m\) rests on a wedge of mass \(M\) which in turn rests on a horizontal table (see figure). All the surfaces are smooth and the system is at rest initially. Find the velocity of the wedge at the instant when the block, after sliding on the wedge, touches the table. (\(h\) is the height of the wedge, from which the block starts.)

Answer: \[ V=\left[\frac{2m^2gh\cos^2\theta}{(M+m)(M+m\sin^2\theta)}\right]^{1/2} \]

Two smooth spheres A and B, of equal radius but of masses \(m\) and \(M\), are free to move on a horizontal table. A is projected with speed \(u\) towards B which is at rest. After impact, the line joining their centres is inclined at an angle \(\theta\) to the direction of velocity of A before impact. (a) If \(e\) is the coefficient of restitution between the spheres, find the speed with which B begins to move. (b) If A's path after impact is perpendicular to its path before impact, show that \( \tan^2\theta=\dfrac{eM-m}{M+m} \).

Answer: (a) \[ v_B=\frac{(1+e)\,m\,u\cos\theta}{m+M} \]

(b) With the velocity of A along the line of centres \( \dfrac{(m-eM)}{m+M}u\cos\theta \) and perpendicular component \( u\sin\theta \) (unchanged), the component of A's velocity along the original direction is zero when \( \dfrac{m-eM}{m+M}\cos^2\theta+\sin^2\theta=0 \), i.e. \( \tan^2\theta=\dfrac{eM-m}{M+m} \).