Question

Two objects of mass 100 gm and 200 gm are moving along the same line and same direction with

velocities of 2 ms-1 and 1 ms-1

respectively. They collide and after the collision the first object moves with the

velocity of 1.5 ms-1

. Determine the velocity of the second object.​

Answer 1

Answer:

Explanation:

Given,

Mass of one of the objects, m₁ = 100 g = 0.1 kg  

Mass of the other object, m₂ = 200 g = 0.2 kg  

Velocity of m₁ before collision, v₁ = 2 m/s  

Velocity of m₂ before collision, v₂ = 1 m/s  

Velocity of m₁ after collision, v₃ = 1.5 m/s

To Find,

Velocity of the second object

Formula to be used,

According to law of conservation of momentum.

Total momentum before collision = Total momentum after collision  

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

Solution,

Putting all the values, we get

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

⇒ 0.1 × 2 + 0.2 × 1 = 0.1 × 1.5 + 0.2 × v₄

⇒ 0.4 = 0.15 + 0.2v₄

⇒ 0.2v₄ = 0.15 - 0.4

⇒ 0.2v₄ = 0.25

⇒ v₄ = 0.25/0.2

⇒ v₄ = 1.25 m/s,

Hence, the velocity of the second object is 1.25 m/s.

Answer 2

Answer:

Given :-

• Mass of object 1 = 0.1 kg
• Mass of Object 2 = 0.2 kg
• Velocity of Object 1 before Collision = 2 m/s
• Velocity of Object 2 before Collision = 1 m/s
• Velocity of Object 1 after before Collision = 1.5 m/s

To Find :-

Velocity of Object 2

Solution :-

According to the law of conversion of momentum

$\large \bf \: m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$

$\large \sf \: 0.1 \times 2 + 0.2 \times 1 = 0.1 \times 1.5 + 0.2 \times v$

$\large \sf \: 0.2 + 0.2 = 1.5 + 0.2v$

$\large \sf \: 0.4 \: = 0.15 + 0.2v$

$\large \sf \: 0.2v = 0.4 - 0.15$

$\large \sf \: 0.2v = 0.25$

$\large \sf \: v \: = \dfrac{0.25}{0.2}$

${ \boxed {\frak{ \red{v \: = 1.25 \: {ms}^{ - 1}} }}}$