# COLM-03- Subjective Solved

26 May 2023

### COLM-03- Subjective Solved

• 1. SUBJECTIVE SOLVED PROBLEMS 1. Locate the centre of mass of a uniform semicircular rod of radius R and linear density λ kg/m. Solution: From the symmetry of the body we see that the CM must lie along the y axis, so xCM = 0. In this case it is convenient to express the mass element in terms of the angle θ , measured in radians. The element, which subtends an angle dθ at the origin, has a length R dθ and a mass dm = λ R dθ . Its y coordinate is y = R sin θ . Therefore, yCM =  M ydm yCM = M R M R d R M 2 0 2 0 2 2 ] cos [ sin 1             y x  R d dm The total mass of the ring is M = π R λ ; Therefore, yCM =  R 2 . 2. Find the centre of mass of a uniform solid hemisphere of radius R and mass M with centre of sphere at origin and the flat of the hemisphere in the x, y plane. Solution: Let the center of the sphere be the origin and let the flat of the hemisphere lie in the x, y plane as shown. By symmetry . 0   y x Consider the hemisphere divided into a series of slices parallel to x, y plane. Each slice is of thickness dz The slice between z and (z + dz) is a disk of radius, r = 2 2 z R  . Let r be the constant density of the uniform sphere. Mass of the slice, dm =     2 2 2 r dz R z dz      The z value is obtained by z = M dm z R  0 = M dz z z R R   0 3 2 ) (  = R z z z z R M                  0 4 2 2 4 2  y z x R r  4 4 R R 2 4 z M         
• 2.  4 R z 4M   Since M = 3 2 V R 3            4 3 R 3 z R 2 8 4 R 3            Hence center of mass has positive coordinates as       R 8 3 , 0 , 0 . 3. A projectile is fired at a speed of 100 m/s at an angle of 370 above horizontal. At the highest point the projectile breaks into two parts of mass ratio 1 : 3. Find the distance from the launching point to the point where the heavier piece lands. The smaller mass has zero velocity with respect to the earth immediately after explosion. Solution: Refer to the figure. At the highest point, the projectile has horizontal velocity. The lighter part comes to rest. Hence the heavier part will move with increased velocity in the horizontal direc- tion. In the vertical direction both parts have zero velocity and undergo same acceleration. Hence they will cover equal vertical displacements in a given time. Thus both will hit the ground together. As internal forces do not affect the motion of the center of mass, the center of mass hits the ground at the position where the original projectile would have landed. The range of the original projectileis Xcm = g u   cos sin 2 2 = m 960 10 5 4 5 3 ) 100 ( 2 2     370 u x1 xcm x2 where sin θ = 3/5, cos θ = 4/5 and g = 10 m/s2 The center of mass will hit the ground at this position. As the smaller mass comes to rest after breaking it falls down vertically and hits the ground at half the range = 480m. If the heavier block hits the ground at x2 , xcm = 2 1 2 2 1 1 m m x m x m   960 = M x M M 2 4 3 480 4    Solving, x2 = 1120 m M k m v0 4. A bullet of mass m strikes a block of mass M connected to a light spring of stiffness k, with a speed v0 and gets embedded into mass M. Find the loss of K.E. of the system just after
• 3. impact. Solution: The process of impact of bullet and block is transient. Within a very short time of impact, the compression of the spring is negligible. Therefore the corresponding spring force is negligible. Even though it is external to the system (M + m), we can conserve its momentum just before and after the impact (impact force is internal). Conservation of linear momentum of bullet plus block just after and before impact yields (M + m)V = mV0  V = m M mV  0 where V = common velocity of block and bullet. M k m v Therefore the loss of K. E. of the system  KE = 2 2 0 ) ( 2 1 2 1 V m M mV   Putting V = M m mV  0 we obtain,  KE = ) ( 2 2 0 m M MmV  . 5. A shell flying with a velocity u = 500 m/s bursts into three identical fragments so that the kinetic energy of the system increases k times. What maximum velocity can one of the fragments obtain if k = 1.5 ? Solution: Let the mass of the shell be 3m. The mass of each fragment is m. The particle with maximum velocity must be in the forward direction. By law of conservation of momentum, 3mu = mv1 - mv2 cos  2 - mv3 cos  3 3u = v1 - v2 cos  2 - v3 cos  3 v1 = 3u + v2 cos  2 + v3 cos  3 ...(i) Also mv2 sin 2 = mv3 sin 3 ...(ii) If v1 is to be maximum  2 =  3 = 0 From (ii), if  2 =  3 mv2 mv1 mv3 3mu 2 3 v2 = v3 = v (say) Equation (i) becomes v1 = 3u + 2v v = (v1 - 3u)/2 ...(iii) from question,            2 2 1 2 mv 2 1 2 mv 2 1 u m 3 2 1 k 3ku2 = 2 2 1 2v v  ...(iv) Substituting for v from(iii) 3ku2 = ) 6 9 ( 2 1 1 2 2 1 2 1 u v u v v    Solving for v1
• 4. v1 = u 1 2(k 1)       For u = 500 m/s and k = 1.5 v1 = 500 1 2(1.5 1)       = 1000 m/s. 6. A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same K.E. Find the energy of explosion. Solution: Let the three fragments move along X, Y and Z axes. Therefore their velocities can be given as j V V i V V ˆ , ˆ 2 1   & k V V ˆ 3  , where V = speed of each of the three fragments. Let the velocity of the fourth fragment be 4 V  . Since, in explosion no net external force is involved, the net momentum of the system remains conserved just before and after the explosion.     i f P P   0 4 3 2 1     V m V m V m V m (Pi = 0 because the body was stationary), putting the values of 2 1,V V and , 3 V we obtain,   k j i V V ˆ ˆ ˆ 4     Therefore, V V 3 4  The energy of explosion E = KEf - KEi = ) 0 ( 2 1 2 1 2 1 2 1 2 4 2 3 2 2 2 1           mV mV mV mV Putting V1 = V2 = V3 = V & setting 2 1 mV2 = E0 , we obtain, E = 6E0 . M u m 7. A cannon with shots of total mass M0 is kept on a rough horizontal surface. The coefficient of friction between the cannon and the horizontal surface is  . If the cannon fires the shots with a velocity u relative to it, find the veloc- ity of the cannon when it possesses a total mass M with the remaining shots, after time t from starting. Assume that the cannon fires shots at the same fequency. Solution: Because each shot of mass m (say) leaves the cannon with a relative velocity u with a frequency n, the rate of loss of mass of the system is given as r = mn dt dm  f= N  R N a M m (At time t) R v  the impact force dm R u mnu dt    The net force acting on the system after a time t  F = R - f  Ma = mnu -  N where f is the force of friction
• 5. where N = Mg  Ma = mnu -  Mg ...(i) M = mass of the cannon with the shots remaining inside it after a time t given as M = M0 - rt  M = M0 - mnt ...(ii) using (i) and (ii) a = 0 mnu g M mnt   Integrating both sides v t t 0 0 0 0 dt dv mnu g dt M mnt       v = mnu t 0 0 1 ln | M mnt | gt mn           v = -u ln 0 0 0 0 M mnt M gt uln gt M M mnt                  Since after a time t, the cannon + remaining shots has mass M  M0 - mnt = M  v = u ln (M0 /M) -  gt. C B v A 8. Two blocks B and C of mass m each connected by a spring of natural length l and spring constant k rest on an absolutely smooth horizontal surface as shown in figure. A third block A of same mass collides elasti- cally block B with velocity v. Calculate the velocities of blocks, when the spring is compressed as much as possible and also the maximum compression. Solution: Let A be the moving block and B and C the stationary blocks. SinceAand B are of equal mass, Ais stopped dead and B takes off with its velocity. Now B and C move under their mutual action and reaction and so their momentum is conserved. Let v1 and v2 be their instantaneous velocities when the compression of spring is x. By the principle of conservation of momentum, mv = m(v1 + v2 ) v1 + v2 = v (a constant) By the principle of conservation of energy. C B v A m m m 2 2 2 2 1 2 2 1 2 1 2 1 2 1 kx mv mv mv     v2 = 2 2 2 2 1 x m k v v    v2 = (v1 + v2 )2 - 2v1 v2 + 2 .x m k  v2 = v2 - 2v1 v2 + 2 .x m k  v1 v2 = 2 . 2 x m k Obviously compression (x) is maximum, when v1 v2 is maximum under the condition that their
• 6. sum (v1 + v2 ) is constant. We have, (v1 + v2 )2 = (v1 - v2 )2 + 4v1 v2  v2 = (v1 - v2 )2 + 4v1 v2  4v1 v2 = v2 - (v1 - v2 )2 Obviously v1 v2 is maximum when (v1 - v2 )2 is minimum. But it is a real positive quantity. Its minimum value is zero. (v1 v2 )max = 2 v 4 when v1 = v2 2 2 max 1 2 max 2m 2m v x (v v ) . k k 4   xmax = m .v 2k 9. A ball of mass m is projected with speed u into the barrel of spring gun of mass M initially at rest on a frictionless surface. The mass m sticks in the barrel at the point of maximum compression of the spring. What fraction of the initial kinetic energy of the ball is stored in the spring ? Neglect the friction. Solution: Let v be the velocity of system after the ball of mass m sticks in the barrel. Applying law of conservation of linear momentum, we have mu = (m + M)v ...(i) The initial K. E.= 2 2 1 mu of the ball is converted into elastic potential energy = 2 2 1 kx of the spring and kinetic energy of the whole system = 2 ) ( 2 1 v M m  . That is 2 2 2 ) ( 2 1 2 1 2 1 v M m kx mu    ...(ii) where k is the spring constant and x is its maximum compression. Dividing equation (ii) by 2 2 1 mu u m 1 = 2 2 2 2 2 1 ) ( 2 1 2 1 2 1 mu v M m mu kx   ...(iii) 1 = 2 2 2 2 ) ( mu v M m mu kx   ...(iv) From equation (i), ) ( m M m u v   Substituting this value in equation (iv) 1 = 2 2 2 2 2 2 2 2 kx (m M) m kx m kx m M 1 mu m (m M) mu m M mu m M (m M)            
• 7. The energy stored in spring = 2 2 1 kx Initial K.E. of the ball = 2 2 1 mu Hence, 2 2 mu kx represents the fraction of initial energy, which is stored in the spring.  fraction = M m M  v0 v1 v2 v3 v3 v2 v1 v0 10. A ballof mass mis projected vertically up froma smooth horizontal floor with a speed V. Find the total momen- tum delivered by the ball to the surface, assuming e as the coefficient of restitution of impact. Solution: Referring the figure, the momentum delivered by the ball at first, second, third impact etc. can be given as the corresponding change, in its momenta (  P). (  P )1 = j V m j mV ˆ ) ( ˆ ( 0 1     P1 = m(V1 + V0 ) Similarly  P2 = m(V1 + V2 ),  The total momentum transferred  P =  P1 +  P2 +  P3 + . . . . Putting the values of  P1 ,  P2 etc., we obtain,  P = m[V0 + 2(V1 + V2 + V3 + . . . . .)] Putting V1 = eV0 , V2 = e2 V0 , V3 = e3 V0 we obtain,  P = mV0 [1 + 2(e + e2 + e3 + . . . . . )]   P = mV0                  e e mV e e 1 1 1 2 1 0 M m R A B 11. A block of mass M with a semicircular track of radius R rests on a horizontal frictionless surface. A uniform cylin- der of radius r and mass m is released from rest at the top point A (see figure). The cylinder slips on the semicircular frictionless track. How far has the block moved when the cylinder reached the bottom (point B) of the track? How fast is the block moving when the cylinder reaches the bottom of the track? Solution: The horizontal component of forces acting on M-m system is zero and the center of mass of the system cannot have any horizontal displacement. When the cylinder is at B its displacement relative to the block in the horizontal direction is (R - r). Let the consequent displacement of the block to the left be x. The displacement of the cylinder relative to the ground is (R - r - x) Since the center of mass has no horizontal displacement M.x = m(R - r - x) x(M + m) = (R - r)m
• 8. x = ) ( ) ( m M m r R   When the cylinder is at A, the total momentum of the system in the horizontal direction is zero. If v is the velocity of the cylinder at B and V, the velocity of the block at the same instant, then mv + MV = 0, by principle of conservation of momentum. Potential energy of the system at A = mg(R - r) Kinetic energy of the cylinder at B = 2 2 1 mv The kinetic energy of the block at that instant = 2 2 1 MV By principle of conservation of energy, mg(R - r) = 2 2 2 1 2 1 MV mv  since v = - m MV mg(R - r) =                   M m M V MV m MV m 2 2 2 2 2 2 1 2 1 mg(R - r) = ) ( 2 2 2 Mm M m V  V2 = ) ( ) ( 2 2 2 Mm M r R g m   V = ) ( ) ( 2 2 m M M r R g m   . 12: A wagon of mass M can move without friction along horizontal rails. A simple pendulum con- sisting of a bob of mass m is suspended from the ceiling by a string of length l. At the initial moment, the wagon and pendulum are at rest and the string is deflected through an angle α from the vertical. Find :(i) the velocity of wagon, when the string forms an angle  ( < α ) with vertical. (ii) the velocity of wagon, when the pendulum crosses its mean position. Solution: (i) Let v be the leftward velocity of wagon (absolute that is relative to earth). Let u be the velocity of pendulum in a frame fixed to the wagon. Then u cos is the relative horizontal velocity of the bob and u sinβ is its vertical velocity. Let vx and vy be the absolute horizontal and vertical downward velocities of the bob.  vx = u cos - v and u sin = vy There is no external force on the system in the horizontal direction. Therefore, by the principle of conservation of momentum to the right, 0 = m(u cos  - v) - Mv
• 9.  u cos  - v = v m M .  u =  cos ) ( m v m M   Before releasing the bob Kinetic energy of bob = ) ( 2 1 2 2 y x v v m  By the conservation of energy, mgl(1 - cos  ) = mgl(1 - cos ) + ] sin ) cos [( 2 1 2 1 2 2 2 2   u v u m Mv    or, 2mgl (cos - cos  ) = Mv2 + m   2 2 2 2 2 2 2 2 cos ) ( sin m v m M m m v M   = Mv2     2 2 2 2 2 2 2 2 2 cos sin ) ( ) ( sin cos ) ( 1 m v m M v m m M M m v m M m M              or 2m2 gl (cos - cos  )cos2  = M(M + m)v2 cos2  + (M + m)2 v2 sin2  = (M + m)v2 [M cos2 + (M + m)sin2 ]  v2 =              2 2 2 sin cos ) cos (cos 2 m M m M gl m  v u After releasing the bob  v =              2 2 2 sin cos ) cos (cos 2 m M m M gl m (ii) In this particular case when  = 0, M m M gl m M m M gl m 2 sin 2 2 ) cos 1 ( 2 2 2 2       v = 2 m sin M m M gl ) ( 2   . 13. Two balls of masses m and 2m are suspended by two threads of same length l from the same point on the ceiling. The ball m is pulled aside through an angle α and released after giving a tangential velocity v0 towards the other stationary ball is imparted to it. To what heights will the balls rise after collision, if the collision is perfectly elastic? Solution: The velocity acquired by m on reaching the lowest position is v (say). Then, 2 2 0 2 1 ) cos 1 ( 2 1 mv mgl mv     v2 = ) cos 1 ( 2 2 0    gl v
• 10. by conservation of momentum, mv = mv1 + 2mv2 v = v1 + 2v2 or v - v1 = 2v2 ...(i) By conservation of kinetic energy, 2 2 2 1 2 2 2 1 2 1 2 1 mv mv mv   v2 = 2 2 2 1 2v v    2 2 2 1 2 2v v v   (v - v1 ) (v1 + v) = 2 2 2v ...(ii) Using (i) in (ii), v1 + v = v2 ...(iii) Solving (i) and (iii), v2 = v 3 2 and v1 = 3 v  Let m rise by h1 and 2m by h2 , then 1 2 1 2 1 mgh mv  or gh1 =   ) cos 1 ( 2 18 1 9 2 1 2 0 2      gl v v h1 =   ) cos 1 ( 2 18 1 2 0    gl v g 2 2 2 2 2 2 1 mgh mv   gh2 =   ) cos 1 ( 2 18 4 9 4 2 1 2 0 2      gl v v h2 =   ) cos 1 ( 2 18 4 2 0    gl v g h v0 =45º  M m 14. A triangular wedge of mass M is moving with uniform velocity v0 along a smooth horizontal surface in the leftward direction. A particle of mass m falls from rest from h on to the inclined face, colliding elastically with it. Find velocity of the ball and wedge after the impact taking M = 2m. Q R u 450  N N 450 O v1 Solution: Impact incline is a straight line perpendicular to the incline. Normal reaction force between the body and the wedge acts along the impact line. This normal force becomes an internal force when we consider (wedge + body) as a total system. But, normal reaction and considerable in magnitude because the impact forceduring collision has contribution towards mak- ing of the normal force.  Momentum of the system is conserved along a line perpendicular to this normal force and momentum of the system is conserved along horizontal. ...(i)
• 11.  Momentum of the body particle is conserved along the common tangent at the point of impact. ...(ii)  As this is an elastic collision, relative velocity of separation along the impact line = relative velocity of approach. ...(iii)  If the wedge were at rest then the body /particle would deflect in the horizontal direction after collision because it is an elastic collision ...(iv)  As, in this case, wedge is moving, the object would not be deflected horizontally, but at an angle 'α ' to the impact ...(v) Velocity of body particle before impact = u = gh 2 along PO Velocity of body particle after collision = v1 along OQ Velocity of wedge before collision = v0 along RO Velocity of wedge after collision = v2 along RO v0 cos 450 - u cos 450 = -v1 cos  - v2 cos 450  v1 cos  + 0 2 v v u 2 2 2   2 v1 cos  + v2 = u - v0 ...(A) According to logic (i) Mv0 = Mv2 - mv1 cos (450 -  )  2v0 = 2v2 - v1 cos (450 -  )  2v2 - v1 cos (450 -  ) = 2v0 ...(B) According to logic (ii) mu sin 450 = mv1 sin  v1 sin  = u solving v1 =            20 6 ) 2 4 ( 0 2 v u u v2 = 5 ) 1 2 ( 0 u v   .  usin v u ucos 15. Afreight car is moving on smooth horizontal track with- out any external force. Rain is falling with a velocity u m/s at an angle  with the vertical. Rain drops are collected in the car at the rate of m kg/s. If initial mass of the car is m0 and velocity v0 then find its velocity after time t. Solution: After time t mass of the car with water is mt = (m0 + mt)kg. Let at that momentum speed of the car be v.  mt dt dm v F dt dv rel ext   (m0 + mt) dv dt        = 0 + (u sin  + v)m
• 12.                t v v t m dt v u dv 0 0 0 ) sin (    ln                                        / ln sin sin 0 0 0 m t m v u v u  ) ( sin sin 0 0 0 t m m v u v u         v = u sin θ ) ( 1 0 0 0 0 0 t m v m t m m                0 0 0 0 t m v v usin m t (m t)              . A B m u 3m m 16. A system of two blocks A and B are connected by an inextensible massless string as shown. The pulley is massless and frictionless. A bullet of mass 'm' moving with a velocity 'u' as shown hits the block 'B' and gets embedded into it. Find the impulse imparted by tension force to the system. Solution: Let velocity of B and A after collision has magnitude v. At the time of collision, tension = T Impulse provided by tension = Tdt  Consider change of momentum of (B + bullet) mu - Tdt  = 2mv ...(i) Consider change of momentum A Tdt  = 3mv ...(ii) from (i) and (ii), mu = 5mv v = u/5 Hence, Impulse T dt  = 3mv = 3m 5 3 5 mu u        . 17. A gun is mounted on a gun carriage movable on a smooth horizontal plane and the gun is elevated at an angle 450 to the horizon. A shot is fired and leaves the gun inclined at an angle  to the horizontal. If the mass of gun and carriage is n times that of the shot, find the value of  . Solution: Let m be the mass of shot. mn = mass of gun w = velocity of shot relative to gun v = velocity of recoil of gun
• 13. Since the gun is inclined at an angle  to horizontal, the direction of w makes an angle  with horizontal. The horizontal and vertical components are w cos and w sin . When the shot leaves the muzzle the horizontal velocity relative to ground (w cos - v). The vertical component of shot relative to ground is the same as relative to gun since the gun moves horizontal. If the shot leaves at an angle  to horizontal, tan  = shot of velocity of component Horizontal shot of velocity of component Vertical = v w w    cos sin ...(i) by conservation of momentum in horizontal direction, mnv = m(w cos  - v) v = ) 1 ( cos  n w  Substituting in (i), tan = 1 cos cos sin   n w w w    tan =    tan 1 1 cos sin ) 1 (          n n n  = tan-1        n n 1 ) 1 45 tan ( 0   18. A glass ball collides with a smooth horizontal surface with a velocity ˆ ˆ V ai bj    . If the coeffi- cient of restitution of collision be e, find the velocity of the ball just after the collision. Solution: Collision takes place along the normal. Therefore the magnitude of normal component (Vy ) of the velocity of the glass ball is changed to y Vy eV   just after the collision whereas the horizon- tal component (Vx ) of its velocity remains constant due to the absence of any horizontal force.  The velocity of the ball just after the impact = V  = ' ' y x V V   j V i V V y x ˆ ˆ ' ' '   where, x ' V = a and y ' V = eb ( j b i a V ˆ ˆ    )  j eb i a V x ˆ ˆ '   Therefore the magnitude of the velocity ' V V  = 2 2 2 b e a  and the direction is given as  = tan-1                 eb a tan ' V ' V 1 y x to the normal (vertical).
• 14. 19. A ball with a velocity of 4 ms-1 impinges at an angle of 30º with the vertical on a smooth horizon- tal fixed plane. If the coefficient of restitution is 0.5, find its velocity and direction of motion after the first impact. Solution: Let the ball be reflected at angle  with the vertical with speed v. Since the plane is smooth, there is no interaction along the horizontal and so that momentum of the ball along this direction remains unchanged.  4cos60º vcos(90º )   or 2 = v sin ...(i) The velocity of the ball relative to the plane before collision 4cos30º 0 4cos30º    The velocity of the ball relative to the plane after collision ( vcos ) 0 vcos        By Newton’s law of collision 30º  O vcos 0.5 4cos30º      or 3 vcos 2 3 2     ...(ii) Squaring and adding (i) and (ii) 2 v 4 3 7    or 1 v 7ms  Dividing, tan 2/ 3   . Thus the ball is reflected with velocity 1 7ms at 1 tan 2/ 3  with the vertical. 20. Consider a one-dimensional elastic collision between an incoming body A and a body B initially at rest. How would you choose the mass of B, in comparison to the mass of A in order that B should recoil with (a) the greatest speed, (b) the greatest momentum, the (c) the greatest kinetic energy? Solution: By the conservation of momentum 1 1 1 1 2 2 m v m v m v    or 1 1 1 2 2 m (v v ) m v    By the conservation of kinetic energy 2 2 2 1 1 1 1 2 2 1 1 1 m v m v m v 2 2 2    or 2 1 1 1 1 1 2 2 m (v v )(v v ) m v       1 1 2 v v v    1 1 ( v v )    2 1 1 2 1 m v v v m    Adding 2 1 2 1 m 2v 1 v m         or 1 2 2 1 2v v 1 (m / m )    (a) v2 is maximum when m2 is negligible in comparison to m1 . (b) 2 1 1 2 1 1 1 2 2 2 2 1 1 2 1 2 2m v 2m m v 2m v p r v 1 (m / m ) m m 1 (m / m )        Obviously p2 is maximum when m2 is large in comparison to m1 .
• 15. (c) 2 2 1 2 2 2 2 2 2 1 4v 1 1 T m v m 2 2 (1 (m / m ))    2 2 2 2 2 1 1 1 1 2 2 1 2 1 2 2 2m v m 2m v (m m ) m m m             Now 2 2 1 1 2 2 1 2 2 m m m m 4m m m                     Obviously T2 is maximum when 2 1 2 2 m m m          is minimum but it is minimum when 2 1 2 2 m m m          is minimum. This a positive quantity. Its minimum value is zero. T2 is maximum when 1 2 2 m m 0 m   , or 1 2 m m  that is, when the masses are equal.