In a direct impact loss in kinetic energy is given by
`Delta K = (M_(1)M_(2))/(2(M_(1) + M_(2)))(V_(1) - V_(2))^(2) (1- k^(2))`
with usual notations (except k) The quantity k will have dimensional formula

Use of dilution formula (M_(1)V_(1) = M_(2) V_(2))

Use of dilution formula (M_(1)V_(1) = M_(2) V_(2))

Two particles having masses m_(1) and m_(2) are moving with velocities vec(V)_(1) and vec(V)_(2) respectively. vec(V)_(0) is velocity of centre of mass of the system. (a) Prove that the kinetic energy of the system in a reference frame attached to the centre of mass of the system is KE_(cm) = (1)/(2)mu V_(rel)^(2) . Where mu=(m_(1)m_(2))/(m_(1)+m_(2)) and V_(rel) is the relative speed of the two particles. (b) Prove that the kinetic energy of the system in ground frame is given by KE=KE_(cm)+(1)/(2)(m_(1)+m_(2))V_(0)^(2) (c) If the two particles collide head on find the minimum kinetic energy that the system has during collision.

If v=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))

If v=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))

What is r.m.s speed of O_(2) molecule if its kinetic energy is 2 k cal "mol"^(-1) ?

What is r.m.s speed of O_(2) molecule if its kinetic energy is 2 k cal "mol"^(-1) ?

The kinetic energy of a moving body is given by k = 2v^(2), k being in joules and v in m//s . It's momentum when travelling with a velocity of 2m//s will be (in kg ms^(-1) )

The gravitational force between two point masses m_(1) and m_(2) at separation r is given by F = k(m_(1) m_(2))/(r^(2)) The constant k doesn't

According to Maxwell - distribution law, the probability function representing the ratio of molecules at a particular velocity to the total number of molecules is given by f(v)=k_(1)sqrt(((m)/(2piKT^(2))))4piv^(2)e^(-(mv^(2))/(2KT)) Where m is the mass of the molecule, v is the velocity of the molecule, T is the temperature k and k_(1) are constant. The dimensional formulae of k_(1) is