Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:
(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?
(a) 9 m
(b) 6 m
(c) 3 m
(d) 2 m
(e) m/3
Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.
Since the momentum is to be conserved, we have
p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.
[The final momentum of particle A is negative since its direction is reversed].
Therefore, p’ = 4p/3
The kinetic energy gained by particle B due to the collision is p’2/2M where M is the mass of particle B.
The kinetic energy lost by particle A due to the collision is (8/9)×p2/2m.
[Note that the initial kinetic energy of particle A is p2/2m and its final kinetic energy is (1/9) p2/2m].
Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have
p’2/2M = (8/9) p2/2m
Substituting for p’ = 4p/3, we have
(16/9) (p2/2M) = (8/9) p2/2m from which M = 2 m.
(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?
(a) 4v
(b) 2√2 v
(c) 2v
(d) √2 v
(e) v/√2
The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).
The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.
Since the momentum is conserved, we have
2m V = √2 mv so that V = v/√2
beautiful job sir ji..........
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