Question

two objects of mass 100 g and 200g are moving along the same line and direction with velocity is 2 metre per second and 1 metre per second respectively after the collision the first object moves with a velocity of 1.67m/s determine the velocity of second object

Answer 1

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


We are provided the data of the following the masses and the velocity;

Mass of the first object is variable "$m_1$" = 100 g = $\frac{100}{1000} \: kg$ = 0.1 kg.

Mass of the second object is variable "$m_2$" = 200 g = $\frac{200}{1000} \: kg$ = 0.2 kg.

$m_1's$ Velocity before it's due collision as variable $v_1$ = 2 m/s.

$m_2's$ Velocity before it's due collision as variable $v_2$= 1 m/s.

$m_1's$ Velocity after it's due collision as variable $v_3$ = 1.67 m/s.

$m_2's$ Velocity after it's due collision as variable $v_4$ = ? m/s.

Now, Applying the principles of this famous equation for momentum that is, Law of conversation of momentum, it's stated as;

The total Momentum before it's due collision = The total Momentum after it's due collision .

This gives us,,,,

$\boxed{\bf{m_1 v_1 + m_2 v_2 = m_3 v_3 + m_4 v_4}}$

Substituting the above given values into this equation , we get ,,,

$\bf{0.1 \times 2 + 0.2 \times 1 = 0.1 \times 1.67 + 0.2 \times v_4}$

$\bf{0.2 + 0.2 = 0.167 + 0.2v_4}$

$\bf{0.4 = 0.167 + 0.2v_4}$

$\therefore \: \: 1.165 \: ms^{- 1} = v_4$

Therefore, second object's Velocity or $m_2's$ Velocity after colliding will be $\bf{1.165 \: ms^{- 1}}$.


HOPE THIS HELPS YOU AND CLEARS THE DOUBTS FOR FINDING VELOCITY BY LAW OF CONVERSATION OF MOMENTUM!!!!!!!!!!