Question

two objects each of mass 1.5 kg are moving in the same straight line But in opposite directions the velocity of an object is 2.5 M per second before the collision during which they stick together what will be the velocity of the combined object after collision

Answer 1

Let us consider that the velocity of the combination becomes v.

Then, according to the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

So, m1v1 + m2v2 = (m1+m2)v

That is, (1.5 * 2.5) + (1.5 * -2.5) = 3 * v

3.75 – 3.75 = 3v

Therefore, v = 0 m/s

Thus the velocity of the objects after sticking is 0 m/s.

Answer 2

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,

Given, let us consider,

Mass of the first object as variable "$m_1$" = 1.5 kg

Mass of the second object as variable "$m_2$" = 1.5 kg

Before colliding the velocity of first object that is "$m_1$", considered as variable "$v_1$" = 2.5 m/s

Before colliding and moving into an opposite direction the velocity of second object that is "$m_2$", considered as variable "$v_2$" = - 2.5 m/s

As, the second object is moving in a opposite direction it'll be taken as negative. And, after the collision has occurred, they're sticking together (as said in question).

Therefore, total mass of this newly formed combined object will be;

$m_1 v_1 + m_2 + v_2 = (m_1 + m_2) \times v$

$m_1 \: \: and \: \: m_2 = 1.5 \: kg$, $v_1 = 2.5 \: m/s\: \: and \: \: v_2 = - 2.5 \: m/s$

Now, Substituting the values given above into this equation :

$\bf{1.5 (2.5) + 1.5 (- 2.5) = (1.5 + 1.5) \times v}$

$\bf{3.75 - 3.75 = 3 \times v}$

Subtract the values and divided by "3".

$\bf{\frac{3.75 - 3.75}{3} = \frac{3}{3} \times v}$

$\therefore \: \: v = 0 \: m/s$

Therefore, the total combined object, after colliding should've a velocity of $\bf{0 \: m/s}$

HOPE THIS HELPS YOU AND CLEARS YOUR DOUBTS FOR FINDING VELOCITY AFTER THE COLLISION IS DONE!!!!!!!!