6Class Exercise - 1In 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?
8Class Exercise - 2A 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 cannotmove on the plank
9Solutionx cm•x cm•Since there are no external forces acting on the plank + man system in x-direction, XCM will not change.
10Solution Let XCM be the position where the man is standing. The plank moves towards the west.
11Class Exercise - 3An 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,mM
12Class 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
13Solutionb, c, dSince 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.
14Class Exercise - 4 A hand ball falls on the ground and rebounds elastically along the sameline 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
15Solution(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.
16Class Exercise - 5Velocity 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 informationis insufficient
21Class Exercise - 7A machine gun fires 35.0 g bullets at a speed of 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?
22Solution Rate of change of momentum = mv × n, where n = Number of bullets fired per second.m = 35 g = kgv =750 m/s
23Class Exercise - 8An 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)1.5 msF = 18,000t (ms)N
24Solution (a) Impulse J = Area under F-t graph (b) Impulse J = force × t(c) Fmax = N
25Class Exercise - 9A small block of mass = 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 blockreaches the horizontal surface,(ii) what is the height (h) of the wedge?m1m24 m/sh
26SolutionTake 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
27Class Exercise - 10A 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?
28SolutionTo conserve the centre of mass the boat has to move 1 m away from the shore.So the position of the dog = 10 – = 7 m from the shore