Question

A particle of mass 2m moving with velocity v strikes a stationary particle of mass 3m and sticks to it. The speed of the system will be xv/10. Calculate x

Answer 1

Given:-

→ Mass of the 1st particle = 2m

→ Velocity of the 1st particle = v

→ Mass of the 2nd particle = 3m

→ Speed of the system after collision = xv/10

To find:-

→ Value of 'x'.

Solution:-

In this case :-

• Velocity of the 2nd particle before collision will be zero as it was stationary.

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By using Law of Conservation of momentum, we get :-

m₁u₁ + m₂u₂ = (m₁ + m₂)v₁

Where :-

• m₁ is the mass of the 1st particle.

• u₁ is the velocity of the 1st particle.

• m₂ is the mass of the 2nd particle.

• u₂ is the velocity of the 2nd particle.

• v₁ is the velocity of the system after collision.

Substituting values, we get :

⇒ 2m × v + 3m × 0 = (2m + 3m)xv/10

⇒ 2mv + 0 = 5mxv/10

⇒ 2mv = mxv/2

⇒ 2(2mv) = mxv

⇒ 4mv = mxv

⇒ x = 4mv/mv

⇒ x = 4

Thus, value of 'x' is 4 .

Answer 2

Answer:

Given :-

• A particle of mass is 2m moving with velocity v strikes a stationery particle of mass 3m and strikes to it. The speed will be xv/10.

To Find :-

• What is the value of x.

Formula Used :-

$\bigstar \: \boxed{\sf{m_1u_1 + m_2u_2 =\: (m_1 + m_2)v_1}}$

where,

• m₁ = Mass of particle A
• m₂ = Mass of particle B
• u₁ = Velocity of particle A before collision
• u₂ = Velocity of particle B before collision
• v₁ = Velocity of particle A after collision

Solution :-

Given :

• m₁ = 2m
• m₂ = 3m
• u₁ = v
• u₂ = 0
• v₁ = xv/10

According to the question by using the formula we get,

↦ $\sf \dfrac{2m \times v + 3m \times 0 =\: (2m + 3m) \times xv}{10}$

↦ $\sf \dfrac{2mv + 0 =\: (5m) \times xv}{10}$

↦ $\sf \dfrac{2mv =\: \cancel{5}mxv}{\cancel{10}}$

↦ $\sf \dfrac{2mv =\: mvx}{2}$

↦ $\sf 4mv =\: mvx$

↦ $\sf \dfrac{4\cancel{mv}}{\cancel{mv}} =\: x$

↦ $\sf 4 =\: x$

➠ $\sf\bold{\red{x =\: 4}}$

$\therefore$ The value of x is 4 .