Question

A 2 kg mass with a speed of 0.5m/s collides head on with a 1.5kg mass moving with a speed of 0.3m/s toward the first mass. What is the speed of the 2nd mass after collision? (a) 0.75m/s (b) 0.37m/s (c) 0.23m/s (d) 0.88m/s

Answer 1

Given :
m1=2kg
u1=0.5m/s
m2=1.5kg
u2= -0.3m/s( moving in opposite direction)
v1=0(suppose it comes to rest)
v2=?
according to law of conservation of momentum
momentum before collision = momentum after collision
m1u1+m2u2=m1v1+m2v2
(2x0.5) -(1.5 x 0.3)= 0 + 1.5 xv2
v2=0.37m/s
so correct option is b

Answer 2

It is not specified that  the second mass comes to rest after collision.

m1 = 2 kg          u1 = 0.5 m/s         m2 = 1.5 kg         u2 = - 0.3 m/s

Assume an elastic collision.  So Kinetic energy is conserved.   Apply the conservation of linear momentum principle.  Let the velocities of the bodies be v1 and v2 respectively.

m1 u1 + m2 u2 = m1 v1 + m2 u2
    => 2 * 0.5 - 1.5 * 0.3 = 2 v1 + 1.5 v2
    => 2 v1 + 1.5 v2 = 0.55  __ (1)

1/2  m1 v1² + 1/2 m2 v2²  =  1/2   m1 u1² + 1/2  m2  u2² 
 =>  2  v1²  + 1.5 v2² = 2 * 0.5² + 1.5 * 0.3²  = 0.635   -- (2)

This equation is too cumbersome to solve.  Instead we could use the principle that in an elastic collision, velocity of separation is equal to the velocity of approach.

So    v2 - v1 = (u1 - u2) = 0.8      --- (3)

Solving (1) and (3) , we get    v2 = 2.15/3.5 = 43/70  m/sec
so v1 = -13/70  m/sec

The answers do not match any of the options given.
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If the second mass comes to rest after collision then   we get,   using  equation (1)
    v2 = 0.55/1.5 = 0.37 (rounded)      then   option  (b) is the answer