Question

Consider an isolated system of n interacting particles. The mutual forces between pairs of particles in the system cause changes in momenta of the individual particles. By third law, the mutual forces between any pair of particles are equal and opposite. By second law, the change in momenta for any pair of particles are Ft and -Ft. Thus, the momentum changes cancel in pairs and total momentum of the system remains constant. This leads to a fundamental principle of physics called the law of conservation of linear momentum. This law states that the total linear momentum of an isolated system of interacting particles is conserved. The recoil of a gun on firing, explosion of a bomb into different fragments due to internal forces, the working of rockets and jet planes, etc.; can be explained on basis of momentum conservation.

A rubber ball falls from a height h and rebounds to a height h/3. A rubber ball of double the mass falling from the same height h rebounds to a height

A bomb is thrown in a horizontal direction with a velocity of 50 ms⁻¹. It explodes into two part of masses 5 kg and 03 kg. The heavier fragment continuous to move in the horizontal direction with a velocity twice the initial velocity of lighter particle. If the final velocity of lighter fragment is 60 m/s then calculate the final velocity of the heavier fragment.

A hunter with a machine gun can fire 20g bullets with a velocity of 800ms⁻¹. A 40 kg tiger springs at him with a velocity of 10ms⁻¹. How many bullets must he fire onto the tiger in order to stop it in his track?