Physics

Session Particle dynamics - 4

Session Objectives

Session Objective Motion of varying mass system (rockets) Problems

Session Opener Rocket moves when fuel is ejected as gases (carrying some momentum)  Rocket gets a momentum in opposite direction.  Mass of rocket reduces.

Motion of Varying System (Rockets)

Illustrative Problem A rocket moving in free space has a speed of 3.0  103 m/s relative to earth. Its engines are turned on and fuel is ejected in a direction opposite the rocket’s motion at a speed of 5.0  103 m/s relative to the rocket. (a) What is the speed of the rocket relative to Earth once its mass is reduced to one half its mass before ignition ? (b) What is the thrust on the rocket if it burns fuel at the rate of 50 kg/s ?

Solution (a) (b)

Class Exercise

Class Exercise - 1 A block of mass 2.5 kg rests on a smooth, horizontal surface. It is struck by a horizontal jet of water at the flow rate of 1kg/s and a speed of 10m/s. What will be the starting acceleration of the block? a Jet A

Solution Water jet supplies the force as mass changes over time and velocity is constant. = 1kg/s × 10m/s = 10N This provides the acceleration to the block.

Class Exercise - 2 Two forces f and F are acting on objects m and M respectively as shown. The horizontal surface is smooth and the spring joining the masses is light. What is the extension of the spring if its force constant is K? (F > f) T F f m M

Solution Total force providing the motion: F – f. Total mass accelerated: (m + M) For M: F – T = Ma For m: T – f = ma Substituting for ‘a’ in any of equations T F f m M

Class Exercise - 3 A light spring, hanging from a rigid support, supports a light pulley. A light and unstretchable string goes around the pulley. One end of the string is tied to the floor and the other end supports a hanging mass m. If the mass has dropped by a height h, express k in terms of m and h. m P k

Solution Let the spring stretch by x. Then the string moves by 2x. So when mass moves by h, the spring stretches by . T mg k h 2T

Solution Tension (force) on the spring = force on the pulley = 2T. T = mg as the mass is at rest  Spring force = 2T = 2mg = k(extension). 2mg = k.

Class Exercise - 4 An insect inside a hemispherical bowl is trying to climb out. The radius of the bowl is r and the coefficient of friction between the bowl and the insect is m . How high can the insect climb? • Insect r

Solution FBD of insect gives [at the point where it can reach. Beyond that point mgsinq > f and insect will slip down.] N = mg cosq, f = mgsinq f = mN = mmg cosq j h y q N r mg Tangent O  tan = m

Solution q f N mg cos mg sin

Class Exercise - 5 A mass m is sliding down an equally inclined right angled trough, inclined to the horizontal by an angle q. m is the coefficient of kinetic friction between the trough and m. What is the acceleration of the mass? 90o M q

Solution The block has two surfaces of contact. So mgsinq – mN1 – mN2 = ma ... (i) |N1| = |N2| (surfaces identical) x mg M N y q N = 2N1 cos45 =

Solution N N2 N1 But N = mgcosq So (i)

Class Exercise - 6 A block A of mass 2 kg rests on top of a block B of mass 3 kg which is on the floor of a lift. The lift accelerates upward with an acceleration of 2m/s2. Find the normal reaction on B by the floor. (g = 10m/s2) a A B

Solution Using FBD of A Using FBD of B B a NA mg mBg A NB

Class Exercise - 7 In the given figure the weight of A, hanging from a fixed, light pulley is 200 N and that of B, hanging from a free pulley is 300 N. Find the tensions T1 and T2 (g = 10m/s2) c P2 B 300 N 200 N N P1 T1 A T2

Solution From FBD of P2, 2T1 = T2 B will rise as upward force on it (T2) is more. Noting that when A falls by y, B rises by and in time t T1 T2 P2

Solution From FBD of A and B, T2 – mBg = mBaB mAg – T1 = mAaA

Solution Putting aB in (ii) = 163.6N T2 = 2T1 = 327.2N

Class Exercise - 8 Three blocks A, B and C, weighing 3kg, 4kg and 8kg respectively are arranged one on top of other as shown. Top block A is attached to a rigid wall by a rigid, light rod. Blocks B and C are connected to each other by a light, unstretchable string passing around a light pulley. Find the force F to drag block C with constant speed. All surfaces have coefficient of kinetic friction equal to 0.25(g = 10m/s2) F B C A rod

Solution MAg NA R f1 MBg NB NA T f1 f2 For B:

Solution MCg NC NB T F f2 f3 For C: F – T – f3 – f2 = 0 = 0.25 × 10 × 32 = 80.0N

Class Exercise - 9 Block A has a mass of 200 gm and block B has a mass of 700 gm in the arrangement shown. Coefficient of friction is 0.2. How large should be the value of force F so that block B has an acceleration of 0.5m/s2? (g = 10m/s2). Friction is present at all the contact surfaces. F A B

Solution For mA: T – f2 = mAa MAg f2 T a NA T f1 f2 NB NA a F MBg

Solution F = = (0.2 + 0.7)0.5 + 0.2 (0.7 + 0.6)10 = 0.45 + 2.60 = 3.05N

Class Exercise - 10 In the arrangement shown, when m is 3.0 kg its acceleration is 0.6m/s2 and when m is 4.0kg its acceleration is 1.6 m/s2. Find the frictional force on M and the mass of M. Assume the pulleys to be massless and strings to be light (g = 10m/s2) m M

Solution mg – 2T1 = ma ... (i) 2T1 m m T1 – f = M.2a 2T1 – 2f = 4Ma ... (ii) Using (i) and (ii) mg – 2f = (m + 4M)a m T2 = 2T1 T1

Solution as given for m = 3kg a = 0.6m/s2 3 × 10 – 2f = (3 + 4M)(0.6) ... (iii) and for m = 4kg a = 1.6 m/s2 Using (iv) and (iii) 40 – 30 = 6.4 + 6.4M – 1.8 – 2.4M  4M = 5.4 M = 1.35 kg Substituting M in (iii), 2f = 30 – (3 × 4 × 1.35) (0.6)= 12.48N

Thank you