IIT JEE Advanced 2010 Paper 1

From principle of conservation of linear momentum, we have
1 kg × u1 + 0 = - 1 kg ×

2m

s

+ 5 kg × v2
From conservation of kinetic energy, we have

1

2

1 kg × u12 + 0 =

1

2

1 kg × 22 +

1

5

5 kg × v22
Solving the above two equations, we get v2 = 1 m/s. Thus, the total momentum (by solving for total momentum after collision) is obtained as follows:
Ptotal = 1 kg × - 2 + 5 kg × 1 = 3 kg/s
The velocity of center of mass is
Vcm =

(5kg×1m∕s)−(1kg×2m∕s)

6

=

1

2

m/s
The kinetic energy of the center of mass is
KEcm =

1

2

6kg×(

1

2

)2 = 0.75 J