Question

A large box slides across a frictionless
surface with a velocity of 12 m/s and a
mass of 4 kg, collides with a smaller
box with a mass of 2 kg that is
stationary. The boxes stick together.
What is the velocity of the two
combined masses after collision?

Answer 1

Given:-

For larger box

• Mass of large box (m¹) = 4kg

• Initial Velocity of the large box (v¹) = 12m/s

For smaller box

• Mass of small box (m²) = 2kg

• Initial Velocity (v²) = 0m/s (At rest)

To Find:-

• The Velocity of two combined masses after Collision.

Formulae used:-

• Conservation of momentum

Accordung to Conservation of momentum total Momentum of system is conserved.

→ m¹u¹ + m²u² = m¹v¹ + m²v²

Now,

→ m¹u¹ + m²u² = m¹v¹ + m²v²

Atq.

→ m¹u¹ + m²u² = (m¹ + m²)v

→ (4)(12) + (2)(0) = (4 + 2)v

→ 48 + 0 = 6v

→ 48 = 6v

→ v = 48/6

→ v = 8m/s

Hence, The Velocity of the two combined masses after Collision is 8m/s.

Answer 2

Answer:

VELOCITY OF COMBINED MASSES = 8m/s

Explanation:

GIVEN :  

Mass of surface (m₁) = 4kg

Initial velocity of surface (u₁) = 12m/s

Mass of box (m₂) = 2kg

Initial velocity of box (u₂) = 0m/s

TO FIND : Velocity of combined masses after collision

STEP BY STEP EXPLANATION :

This problem is based on conservation of linear momentum ,

 Initial momentum = final momentum

 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

4 x 12 + 2 x 0 = 4v + 2v      ( Since , v₁ = v₂ = v )

48 + 0 = 6v

v = 48/6

v = 8m/s

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