1) Two bodies each of mass 0.5 kg are moving in a straight line but opposite in direction with the same velocity of 2m/s. They collide with each other and stick to each other after collision. What is the common velocity of these bodies after collision ?

2) An object of mass 1.5 kg travelling in a straight line with a velocity of 5 m/s collides with a wooden block of mass 5kg resting on the floor. This object sticks with the wooden block after collision and both move together in a straight line. Calculate a) the total momentum before collision b) the total momentum after collision and c) the velocity of th e combination of these objects after collision.

Please answer fast and explain me.

(1)

This is a case of perfectly inelastic collision and the final velocity of the system is given as

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

here

m₁ = m₂ = 0.5 kg

and as they move in opposite directions

u₁  =  2 m/s

u₂ = -2 m/s

thus,

v = (0.5x2 - 0.5x2) / (0.5 + 0.5)

so,

v = 0

..

..

(2)

let us assume

Mass of the object, m₁ = 1 kg
Velocity of the object before collision, u₁ = 10 m/s

Mass of the wooden block = 5kg
Velocity of the wooden block before collision, u₂ = 0 m/s

..

now, the total momentum before collision will be

p_initial  = m₁ u₁ + m₂u₂
= 1 (10) + 5 (0)

so,

p_initial = 10 kg m/s

now
as the object and the wooden block stick together we shall consider the case of perfectly inelastic collision. 
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v

...

Thus, according to the law of conservation of momentum
Total momentum before collision = Total momentum after collision

thus, we get

 m₁ u₁ + m₂u₂ = (m₁ + m₂) v

or
1 (10) + 5 (0) = (1 + 5) v
10 = 6v 

so, final velocity

v = 5/3 m/s = 1.67 m/s
...

thus,
Total momentum just before the impact = 10 kg m/s
Total momentum just after the impact = (m₁+m₂)v = 6×5/3 = 10kg m/s
and

Velocity of the combined object after collision = 1.67 m/s