Velocity Of Approach

Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m//s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s. When the relative velocity of approach becomes 3m/s, the velocity of the centre of mass is 0.75 m/s.

Two particles of mass m and M are initially at rest and infinitely separated from each other. Due to mutual interaction, they approach each other. Their relative velocity of approach at a separation d between them, is

Assertion :- velocity time graph of two paticles undergoing head - on collison is shown in the figure. If collision is inelastic then value of y must be less than x . Reason :- Coefficient of restitution (e)=(|"velocity of separation"|)/(|"velocity of approach"|)

These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five responses. Reason: For inelastic collision, 0leelt1 . Hence, the magnitude of relative velocity of separation after collision is less than relative velocity of approach before collision.

Two particles of masses m_(1) and m_(2) initially at rest a infinite distance from each other, move under the action of mutual gravitational pull. Show that at any instant therir relative velocity of approach is sqrt(2G(m_(1)+m_(2))//R) where R is their separation at that instant.

Two bodies of mass m_(1) and m_(2) are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attaction. Show that their relative velocity of approach at separation r betweeen them is v=sqrt(2G(m_(1)+m_(2)))/(r)

Two particles of mass 'm' and 3 m are initially at rest an infinite distance apart. Both the particles start moving due to gravitational attraction. At any instant their relative velocity of approach is sqrt((ne Gm)/d where 'd' is their separation at that instant. Find ne.

Comprehension # 2 When two bodies collide normally they exert equal and opposite impulses on each other. Impulse = change in linear momentum. Coefficient of restitution between two bodies is given by :- e = |"Relative velocity of separation"|/|"Relative velocity of approach"| = 1 , for elastic collision Two bodies collide as shown in figure. During collision they exert impulse of magnitude J on each other. For what value of J (in N-s ) the 2 kg block will change its direction of velocity :-

Comprehension # 2 When two bodies collide normally they exert equal and opposite impulses on each other. Impulse = change in linear momentum. Coefficient of restitution between two bodies is given by :- e = |"Relative velocity of separation"|/|"Relative velocity of approach"| = 1 , for elastic collision Two bodies collide as shown in figure. During collision they exert impulse of magnitude J on each other. If the collision is elastic, the value of J is ............. N-s :-

A body of mass m moving with a velocity v is approaching a second object of same mass but at rest. The kinetic energy of the two objects as viewed from the centre of mass is