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Session. System of Particle - 3. Session Objectives. Session Objective. Problems. Questions. Class Exercise - 1.

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**Session**System of Particle - 3**Session Objective**• Problems**Class Exercise - 1**In a gravity-free space, a rifle fires a bullet of mass ‘m’ at a stationary block of mass M, a distance D away from it. When the bullet has moved a distance d towards the block, where does the centre of mass of bullet + block system lie?**d**M D Solution Let the origin be at the bullet. Then**A man of mass 2m stands on a plank of mass m. The plank is**lying on a smooth horizontal floor. Initially, both are at rest. The man starts walking on the plank towards east and stops after moving a distance l on the plank. Then (a) centre of mass of the system always remains stationary (b) the plank will slide to the west by a distance (c) the plank will slide to the west by a distance (d) since the floor is smooth, therefore, man cannot move on the plank Class Exercise - 2**x cm**• x cm • • Solution Since there are no external forces acting on the plank + man system in x-direction, XCM will not change.**The plank moves towards the west.**Solution Let XCM be the position where the man is standing.**An ideal spring is permanently connected to blocks of masses**M and m. Block-spring system can move over a smooth horizontal table in a straight line along the length of the spring as shown in the figure. The blocks are brought nearer to compress the spring and then released. In the subsequent motion, M m Class Exercise - 3**Class Exercise - 3**(a) initially they move in opposite directions with velocities inversely propotional to their masses (b) the ratio of their velocities remains constant (c) linear momentum and energy of the system remains conserved (d) the two blocks will oscillate about their centre of mass**Solution**b, c, d Since the system is on a smooth horizontatal floor, no external forces act on it. Initially, cm is stationary so the block will oscillate about the same point, i.e. their centre of mass = (d) is correct. Since masses are oscillating without encountring any resistance (friction), no loss in energy = (c) is also correct.**A hand ball falls on the ground and**rebounds elastically along the same line of motion. Then (a) the linear momentum of the universe remains conserved (b) the linear momentum of the ball is not conserved (c) during the collision the whole of the kinetic energy of the ball is converted into potential energy and then completely vanishes (d) during the collision the kinetic energy remains constant Class Exercise - 4**Solution**(a, b) In this collision, if we take universe as a system, then there is no external force getting on the system. Hence, linear momentum of universe is conserved = (a) is correct. When it collides with the ground a reaction force acts on the ball and since it is an external force linear momentum of ball changes.**Velocity of centre of mass of system of two particles is VC**and total mass of particles is equal to m. The kinetic energy of the system (a) may be equal to (d) cannot be calculated because this information is insufficient Class Exercise - 5**Case - i**VC VC M1 M2 Solution a, c**Case - ii**V1 V2 M1 M2 = (c) is correct while (b) is wrong Solution**Two particles of masses 100 g and 900 g have velocity**vectors and respectively at a given time. Find the velocity vector of the centre of mass. Class Exercise - 6**Class Exercise - 7**A machine gun fires 35.0 g bullets at a speed of 750.0 m/s. If the gun can fire 200 bullets per minute, what is the average force the shooter must exert to keep the gun from moving?**Solution**Rate of change of momentum = mv × n, where n = Number of bullets fired per second. m = 35 g = 0.035 kg v =750 m/s**An estimated force time curve for a base ball struck by a**bat is shown in the figure. From this curve determine (i) the impulse delivered to the ball, (ii) the average force exerted on the ball, and (iii) the peak force exerted on the ball. F (N) F = 18,000 N t (ms) 1.5 ms Class Exercise - 8**(b) Impulse J = force × t**Solution (a) Impulse J = Area under F-t graph (c) Fmax = 18000 N**A small block of mass = 0.500 kg is released from rest at**the top of a curved frictionless wedge of mass = 3 kg, which sits on a frictionless horizontal surface. When it leaves the wedge the block’s velocity is 4 m/s to the right, • what is the velocity of the wedge after the block • reaches the horizontal surface, m1 m1 (ii) what is the height (h) of the wedge? m2 h m2 4 m/s Class Exercise - 9**Solution**Take wedge and the block as a system. Since there are no external forces acting on the system in x-direction, therefore linear momentum is conserved. Height h = 0.8 m**Class Exercise - 10**A dog weighing 5 kg is standing on a flat boat so that he is 10 m from the shore. He walks 4 m on the flat boat towards the shore and then halts. The boat weighs 20 kg and one can assume that there is no friction between it and the water. How far is the dog from the shore at the end of this time?**Solution**To conserve the centre of mass the boat has to move 1 m away from the shore. So the position of the dog = 10 – 4 + 1 = 7 m from the shore

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