paths 1, 2 and 3 respectively in the gravitational field of a point mass m,

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1 58. particles of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a c is varying with time t as a c = k 2 rt 2 where k is a constant. The power delivered to the particles by the force acting on it is: (a) 2π mk 2 r 2 t (b) mk 2 r 2 t (c) (mk 4 r 2 t 5 )/3 (d) zero Sol: (b) The centripetal acceleration, a c = k 2 r t 2 or ν 2 /r = k 2 rt 2, or ν = krt So, tangential acceleration, a t = dν/dt = kr, Work is done by tangential force. Power = F t.ν. cos 0 0, = (m a t ) (krt), = (mkr) (krt), = mk 2 r 2 t lternate Solution: The centripetal acceleration, a c = k 2 r t 2 ν 2 /r = k 2 rt 2, mν2 = k2 r 2 t 2 (i) K.E. = k2 r 2 t 2, d/dt (K. E) = mk 2 r 2 t, Power = mk 2 r 2 t 59. particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = - kx + ax 3. Here k and a are positive constants. For x 0, the functional form of the potential energy U(x) of the particle is: (a) (b) (c) (d) U(x) U(x) U(x) U(x) Sol: (d) du (x) = - Fdx, or U x =, = kx 2 /2 ax 4 /4 U = 0 at x = 0 and at x = ; we have potential energy zero twice (out of which one is at origin). lso, when we put x = 0 in the given function, we get F = 0. ut F = - t x = 0; du/dx = 0 i.e. the slope of the graph should be zero. These characteristics are represented by (d) 60. If W 1, W 2 and W 3 represent the work done in moving a particle from to along three different paths 1, 2 and 3 respectively in the gravitational field of a point mass m, m find the correct relation between W 1, W 2 and W 3. 2 (a) W 1 > W 3 > W 2 (b) W 1 = W 2 = W 1 3 (c) W 1 < W 2 < W 3 (d) W 2 < W 1 < W 3 Sol: (b) Note: In a conservative field work done does not depend on the path. 3 The gravitational field is a conservative field. r W 1 = W 2 = W Two small particles of equal masses start moving in opposite directions from a point in a horizontal circular orbit. Their tangential velocities are ν and, respectively, as shown in the figure. etween collisions, the particles move with constant speeds. fter making how many elastic collisions, other than that at, these two particles will again ν reach the point? (a) 4 (b) 3 (c) 2 (d) 1 ν Sol: (c) Let the radius of the circle be r. Then the two distance travelled by the two particles before first collision is 2πr. Therefore, X t + ν X t = 2πr where t is the time taken where t is the time taken for first collision to occur. r t = 2πr/3ν, or Distance travelled by particle with velocity ν is equal to ν X 2πr/3ν = 2πr/3. Therefore the collision occurs at. s the collision is elastic and the particles have equal masses, the velocities will interchanges as shown in the figure. ccording to the same reasoning as above, the 2 nd collision will take place at C and the velocities will again interchanges. With the same reasoning the 3 rd collision will occur at the point. Thus there will be zero elastic collisions before than particles again reach at. ν ν, 62. block of mass 2 kg is fee to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F (t) in the x direction. F(t) C 4N 3s s t

2 The force F (t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is (a) 4.50 J (b) 7.50 J (c) 5.06 J (d) J Sol: (c) rea under F = t graph gives the impulse or the change in the linear momentum of the body. s the initial velocity (and therefore the initial linear momentum) of the body is zero, the area under F t graph gives the final linear momentum of the body. F(t) rea of, = ½ X 3 X 4 = 6 N s 4N lso / = CD/C, 4/3 = CD/1.5 f, CD = 2 or rea of CD = - [1/2 X 1.5 X 2] = N s 4.5 s or The final linear momentum = = 4.5 N s 3s C or Kinetic energy of the block = p 2 /2m = (4.5) 2 /2 X 2 = 5.06 J 63. force F = -K (y + x) (where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0), and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particle is (a) 2Ka 2 (b) 2 Ka 2 (c) Ka 2 (d) Ka 2 Sol: (c) The expression of work done by the variable force F on the particle is given by, W = l In going from (0, 0) to (a, 0), the coordinate of x varies from 0 to a, while that of y remains zero. Hence, the work done along this path is: W 1 =. dx = 0, [or = 0] Y In going from (a, 0) to (a, a) the coordinate of x remains constant (= a) (a, a) while that of y changes from 0 to a. Hence, the work done along this path is W 2 = = ka = - Ka 2, Hence, W = W 1 + W 2 = - ka 2 dy D t (a, 0) dx 64. stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. t a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is (a) (b) (c) (d) ν Sol: (d) pplying the principle of conservation of energy (K.E.) + (P.E.) = (K.E.) + (P.E), we get ½ mν 2 + mgl = ½ mu 2, Hence, ν= (i) X Change in velocity = ν = ν, = l [From (i)] u 65. force = ( ) N is applied over a particle which displaces it from its origin to point = (2 - ) m. The work done on the particle in joules is (a) + 10 (b) + 7 (c) 7 (d) + 13 Sol: (b) Workdone in displacing the particle, W = = ( ). (2 - ), = 10 3 = 7 Joules 66. body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function of time is given by (a) ν. t2 (b) ν.t (c) Sol: (b) u = 0; ν = u + at; ν = at, ν ν.t2 (d).t Instantaneous power = F X ν = m. a. at = m. a 2. T or Instantaneous power = m (ν 2 /T 2 )t 67. The potential energy of a 1 kg particle free to move along the x-axis is given by V (x) = (x 4 /4 x 2 /2) J. The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is (a) 3/ (b) (c) 1/ (d) 2 Sol: (a) Velocity is maximum when K.E. is maximum for minimum, P.E., dv/dx = 0 x 3 x = 0 x = ± 1, Min. P.E. = - = - J K.E. (max) + P.E. (min) = 2 (Given), or K.E. (max. ) = 2 + = K.E. max = mν2 max, X 1 X ν2 max = 9/4 ν max. = 3/ 63

3 68. 2 kg block slides on a horizontal floor with a speed of 4m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000 N/m. The spring compresses by (a) 8.5 cm (b) 5.5 cm (c) 2.5 cm (d) 11.0 cm Sol: (b) Let the block compress the spring by x before stopping kinetic energy of the block = (P.E. of compressed spring) + Work done against friction. X 2 X (4)2 = X 10,000 X x X x, 10,000 x x 32 = 0, 5000x x 16 = 0 or x = - 15 ± / 2 X 5000, = 0.055m = 5.5 cm. 69. block of mass 0.50 kg is moving with a speed of 2.00 ms -1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is (a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J Sol: (c) Initial kinetic energy of the system, K.E i = ½ mu 2 + ½ M (0) 2 = ½ X 0.5 X 2 X = 1J For collision, applying conservation of linear momentum m X u = (m + M) X ν, or 0.5 X 2 = ( ) X ν ν= 2/3 m/s Final kinetic energy of the system is, K.E f = (m + M)ν2 = ( ) X X = J or Energy loss during collision = (1- ) J = 0.67 J 70. The potential energy function for the force between two U(x) = a/x 12 b/x 6, where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U (x = ) U at equilibrium ], D is (a) b 2 /2a (b) b 2 /12a (c) b 2 /4a (d) b 2 /6a Sol: (c) t equilibrium: du(x)/dx = 0, - 12a/x 11 = - 6a/x 5 x = (2a/b) 1/6 or U at equilibrium = - ) = - b2 /4a and U (x = ) = 0, or D = 0 (-b 2 /4a) = b 2 /4a 71. block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. n impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is (IIT-JEE 2011) (a) 4 (b) 3 (c) 2 (d) 1 Sol: (a) Let ν be the speed of the block just after impulse. t, the block comes to rest. Thereforeν ν Loss in K.E. of the block = Gain in P.E. of the spring + Work done against friction ½ mν 2 = ½ kx 2 + µg.x, or ν = µ or ν =, or ν = 4/10,or N = 4 SLUTINS 1. (c) Mass of the drop m = volume X density of water = r3, where is the density of water. Work done by gravitational force is W = mgh = r3 gh Thus W r (d) Density of brick = relative density X density of water = 3 X 1000 = 3000 kgm -3. Volume of the brick is V = mass/density = 6kg/3000kgm -3 = 1/500 m 3 Now, loss of weight of the brick in water = weight of 1/500 m 3 of water displaced by it = volume X density of water X acceleration due to gravity = 1/500 m 3 X 1000 kgm -3 X 10 ms -2 = 20 N. Therefore, the weight to be lifted = 6 X = 40 N. Thus the work done is W = 40 X 10 = 400 J 3. (b) Displacement along the z-axis is S = 4 metres. Therefore, work done is W 1 = F.S, = (- ). (4), = Now. = 0 and. = 0 because and are perpendicular to. ut. = 1. Therefore, W 1 = 12 J. 64

4 4. (a) Displacement along the y-axis is S = 3 metres. Therefore, the work done is W 2 = F.S = ( ). (3) = 6. = 6J 5. (c) Since work is a scalar, the total work done is just the algebraic sum of W 1 and W 2, i.e. W = W 1 + W 2 = = 18 J. 6 (a) Since the block is moved up the incline, the force applied on it has to over-come the component mg sin θ of its weight mg acting down the plane and the force of friction µmg cos θ also acting down the plane. Thus F = ms sin θ + µmg cos θ and Work done W = Fx = mgx (sin θ + µ cos θ). 7. (c) s shown in fig the height attained by the bob when the string subtends an angle α with the vertical is h = l l cos α = l (1 cos α) Its potential energy at the highest point = mgh, where m is the mass l of the bob. Let be the speed of the bob when it passes through. Its kinetic energy at = ½ m 2. From the principle of conservation of energy, we have ½ m 2 = mgh or = = h 8. (d) From the principle of conservation of energy, we have ½ m 2 + mgh = 0 + mgr, Which gives =. 9. (b) t the highest point, the energy of the ball is entirely potential = mgh and at the lowest point, the energy is entirely kinetic = ½ m 2. Since friction is absent, the principle of conservation of energy requires, ½ m 2 = mgh or = = = 2 ms (b) Let m be the mass of each bullet. Since the resistance offered by the plank is the same for the two bullets, the amount of work done by the plank is the same for the two bullets. From workenergy principle, the decrease in the kinetic energy is the same for the two bullets. Decrease in KE of first bullet= ½ mu 2 1 ½ m 2 1, = ½ m (200) 2 ½ m (180) 2 (i) If 2 is the speed for the second bullet after passing through the plank, then Decrease in KE of second bullet = ½ mu 2 2 ½ m 2 2, = ½ m (100) 2 ½ m 2 2 (ii) Equation (i) and (ii) we have, = ½ m (200) 2 ½ m (180) 2 = = ½ m (100) 2 ½ m 2 2 which gives 2 2 = 2400 or = 49 ms (b) Thickness of plank (S) = 1.0 m, Mass of bullet (m) = 10 g = 10-2 kg If F is the average resistive force exerted by the plank on the bullet, the work done by the plank on the bullet is, W = FS, Work done = decrease in KE of the bullet. Hence FS = ½ m (200) 2 ½ m (180) 2 = 3800 m, Putting S = 1 m and m = 10-2 kg, we get F = 38 N. 12. (b) Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to work only to overcome the frictional force (f). Since the distance moved in 1 second is metres, the work done per second or the power of the engine is P = f X, = 1500 X 5, = 7500 W = 7.5 kw 13. (c) When the car is being pulled along an inclined plane of gradient 1 in 10 (i.e. sin θ = 1/10), the engine has to be extra work against the component Mg sin θ of the weight Mg of the car. The extra work per second or extra power the engine has to develop to maintain the same speed is = Mg sin θ X, = 1500 X 10 X 1/10 X 5, = 7500 W = 7.5 kw 14. (b) Let and be the original speeds of the heavier and the lighter particles respectively. We then have ½ m 2 = ½ X {1/2 (m/2) 2 }, or 2 = ¼ 2 or = (a) When the heavier particle is speeded up by 2.0 m s -1, its kinetic energy becomes ½ m(+2) 2. Since this equals the original kinetic energy of the lighter particle, we have ½ m ( + 2) 2 = ½ (m/2) (4) 2, = 2 2 or = 0 = = = 22 The positive root is = = 2 (1 + ). 16. (b) Mass of block (M) = 0.9 kg, mass of bullet (m) = 0.1 kg, initial velocity of bullet (u) = 100 ms -1 and initial velocity of block (U) = 0. r Momentum of bullet and block before impact = mu + MU = mu. Let be the velocity of bullet + block after impact. Then, the momentum after impact = (m + M). From the principle of conservation of momentum, we have mu = (m + M) or = mu/(m + m) = 0.1 X 100/( ) = 10 ms -1, KE of bullet + block = ½ (m + M) 2 (i) 65

5 Let the bullet + block system rise to a height h. t this height, PE of bullet + block = (m + M) gh (ii) Equating (i) and (ii), we get, h = 2 /2g = (10) 2 /2 X 10 = 5 m 17. (a) Initial KE = ½ mu 2 Final KE = ½ (m + M) 2 or Loss in KE = ½ mu 2 ½ (m + M) 2, = ½ X 0.1 X (100) 2 ½ ( ) X (10) 2, = 450 J 18. (b) The system consists of three identical balls marked 1, 2 and 3. Let m be the mass of each ball. efore the collision, KE of the system = KE of 1 + KE of 2 + KE of 3 = ½ m = ½ m 2 case (a) fter the collision, KE of the system = KE of 1 + KE of 2 + KE of 3 = 0 + ½ m (/2) 2 + ½ m (/2) 2 = ¼ m 2 Case (b) KE of the system = KE of 1 + KE of 2 + KE of 3, = ½ m 2 = ½ m 2 Case (c) KE of the system = KE of 1 + KE of 2 + KE of 3 = ½ m (/3) 2 + ½ m (/3) 2 + ½ m (/3) 2, = 1/6 m 2, Case (d) KE of the system = 0 Now, in an elastic collision, the kinetic energy of the system remains unchanged. 19. (a) Suppose the bob acquires a velocity on reaching the bob. In a head on elastic collision between two bodies of the same mass when one of them is at rest, the velocities are exchanged after the collision. Hence the bob will come to rest at the lowermost position (occupied by before collision) and the bob will move to the left attaining a maximum height h 20. (a) The total momentum of the system must remain conserved after the collision. Since, at its highest point, the projectile has only a horizontal velocity, we must have (2 m) u cos θ = m X 0 + m X u or u = 2 u cos θ Thus the horizontal velocity of the other fragment is 2 u cos θ. The time taken to reach the highest point (as well as the time taken to fall down from this point) being, we have Total horizontal distance covered by the other fragment = u cos θ X u sin θ/g + 2u cos θ X u sin θ/g, = 21. (d) 22. (d) The velocity attained after a fall through a height h is given by 2 = 2gh Thus h 2. The velocity after first rebound is e. Therefore, the height attained after first rebound = e 2 h. Velocity after second rebound is e 2. Hence the height attained after second rebounds is e 4 h. 23. (a) From the principle of conservation it follows that the momentum of m kg piece has a magnitude p 3 = = 20 kg ms -1 or m = 20kg ms -1 /40 ms -1 = 0.5 kg or Total mass of the shell = = 3.5 kg. 24. (b) The first collision will be between balls 1 and 2. Since both have the same mass, after the collision ball 1 will come to rest and ball 2 will move with speed 1. This ball will collide with the stationary ball 3. fter this second collision, let 2 and 3 be the speeds of the balls 2 and 3 respectively. Since the collision is elastic, 2 and 3 are given by 2 = (m M/ m + M) 1 (i) and 3 = (2m/m + M) 1 (ii) If M < m, it follows from (i) and (ii) that 2 < 3 and both have the same direction. Therefore, ball 2 cannot collide with ball 3 again. Hence there are only two collisions. 25. (c) If M > m, we have from Eq. (i) 2 = - (M m/m + m) 1 The negative sign indicates that, after the second collision, ball 2 will move in opposite direction towards the ball 1 which is at rest after the first collision. Therefore, ball 2 will make another collision with ball 1. Hence, in this case, there are three collision in all between the balls. 26. (d) Mass of hanging part of chain is m = m/l X l/3 = m/3. This mass can be assumed to be located at the centre of mass of the hanging length of the chain which is at a distance x = l/6 from the edge of the table. Hence the work done is, W = m g x = m/3 X g X l/6 = mgl/ (a) The speed of the ball at the lowest point is =, where R is the radius of the bowl. Therefore KE = ½ m 2 = mgr = 0.05 kg X 10 ms -2 X 1.5 m = 2.25 J 66

6 28. (c) Velocity () = dx/dt = d/dt (3t 2 + 4t + 5) = 6t + 4. cceleration is a = d/dt = d/dt (6t + 4) =6 ms -2. Therefore, applied force is F = ma = 0.5 X 6 = 3 N. Now t = 2s, the distance moved is x = 3 X (2) X = 25 m or Work done W = Fx = 3 X 25 = 75 J. 29. (a) Initial PE = mgh. Now, gain in KE = loss in PE. Thus ½ m 2 = ½ mgh, or = 30. (d) Initial KE = ½ m 2. Now gain in PE = loss in KE. Thus mgh = ¼ m 2. r h = 2 /4g 31. (b) The weight of the rod acts at the centre of gravity which is at a distance of l/2 from the pivoted end. When the rod makes an angle θ with the horizontal, the vertical height of the centre of gravity is h = ½ sin θ i.e. the centre of gravity rises by an amount h. Therefore PE = mgh = l/2 mgl sin θ 32. (d) In a perfectly inelastic collision, two bodies stick together. fter, the collision, the speed of the ball and the bob (sticking together) is = /2. The height to which they will rise is given by = or h = 2 /2g = 2 /8g 33. (b) Mass of the ball and the bob sticking together is m = 2m. KE after collision = ½ m 2 = ½ X 2m X (/2) 2 = ¼ m 2. KE before collision = ½ m 2. Therefore, their ratio is 1: (b) In an elastic collision between two bodies of the same mass with one of them initially at rest, the moving body is brought to rest and the other moves in the same direction with the same speed. Thus the ball will come to rest and the bob of the pendulum acquires a speed. t this speed, it will rise to height h given by h = 2 /2g. 35. (a) Mass of neutron (m 1 ) = 1 unit. Mass of nucleus (m 2 ) = units. Here u 1 = u and u 2 = 0 Therefore the velocity of the neutron after the collision is 1 = (m 1 m 2 /m 1 + m 2 ) u = (1 /1 + ) u KE of neutron after collision = ½ m = ½ X 1 X (1 / 1 + ) 2 u 2 KE of neutron before collision = ½ m u 2 = ½ X 1 X u 2 = ½ u (c) The total number of nucleons (i.e. proton + neutrons) in a nucleus is called its mass number. n α-particle is a helium nucleus having 2 protons and 2 neutrons. So the mass number of an α- particle = 4. When a nucleus of mass number emits an α-particle, the mass number of the daughter nucleus reduces to ( 4). If V is the recoil speed of the daughter nucleus, we have, from the law of conservation of momentum, ( 4) V - 4 = 0, or V = 4/ (c) Refer to fig. Taking the components of the velocities along x and y-axes and using Y the law of conservation of momentum for x and y components we have mu = m cos θ 1 + m cos θ 2 (i) and 0 = m sin θ 1 m sin θ 2 (ii) From (ii) we get sin θ 1 = sin θ 2 or θ 1 = θ 2. Using θ 1 = θ 2 = θ in (i) we have, mu = 2 m cos θ, or cos θ = u/2 (iii) u Since the collision is elastic, kinetic energy is also conserved, i.e. m m ½ mu 2 = ½ m 2 + ½ m 2, or u 2 = 2 2 or u = (iv) Using (iv) and (iii) we have cos θ = 1/ or θ = Thus θ 1 + θ 2 = 2θ = (c) Let 1 and 2 be the speeds of the balls after the collision. Then Eq. (ii) above gives 1 sin θ 1 = 2 sin θ 2, (v) lso, conservation of kinetic energy gives, ½ mu 2 = ½ m ½ m 2 2, or u 2 = (vi) lso Eq. (i) above becomes, mu = m 1 cos θ 1 + m 2 cos θ 2, or 2 cos θ 2 = u 1 cos θ 1 (vii) Squaring (v) and (vii) and adding we get, 2 2 = u u cos θ 1 (viii) Eliminating 2 between (vi) and (viii), we get, cos θ 1 = 1 /u (ix) Eliminating 2 between (v) and (vii) we get, tan θ 2 = sin θ 1 /(u/ 1 cos θ 1 ) (x) Using (ix) in (x), we get, tan θ 2 = sin θ 1 /(1/cosθ 1 cosθ 1 ) = sin θ 1 cosθ 1 /(1 cos 2 θ 1 ) = cot θ 1 = tan (90 0 θ 1 ), or θ 2 = 90 0 θ 1 or θ 1 + θ 2 = (c) Power P = F = ma = m. or d = dt. Integrating, we have d = P/m dt (or P = constant) or 2 /2 = Pt/m, or = t 1/2, or dx/dt = t 1/2 or dx = t 1/2 dt Integrating again, we have, d x = t 1/2 dt, or x = 2/3 t 3/2 67 x

7 40. (b) Let be the speed of the bullet incident on the first plank. Its speed after it passes the plank = 9/10. If x is thickness of the plank, the deceleration a due to the resistance of the plank is given by 2ax = 2 (9/10) 2 = 19 2 /100 Suppose the bullet is stopped after passing through n such planks. Then the distance covered by the bullet is s = nx. Thus, we have 2 0 = 2as = 2anx or n = 2 /2ax = 2 X 100/19 2 [use Eq. (i) above], = 100/19 = 5.26 Thus the minimum number of planks required is (c) The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid-point of the rod is raised through a height h = l/2. Since the entire mass of the rod can be assumed to be concentrated at the mid-point (centre of gravity), D the work done = mgh = mgl/ (a) Refer to fig D = = l. In the inclined position, let the centre of gravity C of the rod be at a height h above the ground, so that C = l/2. In triangle C CE, we have h = C sin 30 0 = l/2 sin 30 0 = l/4, or PE = mgh = mgl/ (a) Let f be the force of friction and m be the mass of the body. The retardation a = f/m. h h 30 0 If is the initial speed of the body, then 2ax = 2 or max = ½ m 2 E = k ut ma = f. Therefore fx = k or f = k/x. 44. (a) Force of friction = µmg. Therefore, retardation a = µmg/m = µg. lso 2ax = 2 or 2am 2 x = m 2 2. ut P = m. Therefore, 2 am 2 x = P 2, ut a = µg. Therefore, 2 µg m 2 x = P 2 or µ = P 2 /2 gm 2 x. 45. (d) The mass per unit length of the chain m = M/L. The mass of the hanging portion of chain is m = ml/n. This mass can be assumed to be concentrated at the centre of the hanging portion of the chain which is a distance of x = L/2n from the edge of the table. Therefore, the work done in pulling the hanging portion of the chain on to the table top is W = m gx = ml/n X g X L/2n = mgl 2 /2n 2 = MgL/2n (c) Momentum of 5 kg mass (p 1 ) = 5 X 2 = 10 kg ms -1 along the x-axis. Momentum of 10 kg mass (p 2 ) = 10 kg ms -1 along the y-axis. These two momenta are perpendicular to each other. Therefore, the resultant initial momentum is, p = = = 20 kg ms -1 If ms -1 is the velocity of the combined mass, then the final momentum = (10 + 5) = 15 kg ms -1. Now, from the principle of conservation of momentum, we have 15 = 20 or = 4/3 ms (b) Mass of the second part = M m. If its velocity is, then we have MV = (M m) + m X 0, or = MV/(M m) 48. (a) Let V be the velocity of the block with the bullet embedded in it at the time of impact. Then from the principle of conservation of momentum, we have m = (M + m) V or V = m/(m + m) (i) If the block, with the bullet embedded in it, rises to a vertical height h, then from the principle of conservation of energy, we have ½ (M + m) V 2 = (M + m) gh, or V = (ii) Using (ii) in (i), we get, = m/(m + m) 49. (d) Refer to fig Here P = m and P = MV. The resultant of p and P is p r = = y, The angle which the resultant momentum p r subtends with the x-axis is given by tan θ = P/p = MV/m Loss of KE = (1/2 m 2 + ½ MV 2 ) ½ [m M 2 V 2 /(M + m)] Pr = ½ Mm/(M + m) (V ) P p 50. (a) P 1 = 2 kg ( + 2-3) ms -1 = ( ) kg ms -1 P 2 = 3 kg (2 + + ) ms -1 = ( ) kg ms -1 P Resultant momentum is P = P 1 + P 2, = ( ) + ( ), = ( )kg ms -1 Total mass (m) = = 5 kg. Therefore, the velocity of composite body is v = p/m = 1/5 ( ) ms -1 x 68

Chapter Work, Energy and Power. Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Ans: (a)

Chapter Work, Energy and Power Q1. The co-efficient of restitution e for a perfectly elastic collision is [1988] (a) 1 (b) 0 (c) (d) 1 Q2. A bullet of mass 10g leaves a rifle at an initial velocity of