The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is _____

Answer :

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Where, u = initial velocity

v = final velocity

Put the value into the formula

[tex]1\times3+2\times0=(1+2)v[/tex]

[tex]v=\dfrac{3}{3}[/tex]

[tex]v=1\ m/s[/tex]

Hence, The speed of the boxes are 1 m/s.

After the collision, the speed of the two boxes will reduce due to lost in kinetic energy associated with inelastic collision.

The given parameters;

  • mass of the box, m = 1 kg
  • mass of the second box, = 2 kg

Apply the principle of conservation of linear momentum to determine the final velocity of the two boxes after collision.

m₁u₁  +  m₂u₂  = v(m₁  +  m₂)

1(u₁) + 2u₂ = v(3)

u₁ + 2u₂ = 3v

Where;

u₁ and u₂ are the initial velocity of the first and second box respectively

v is the final velocity of the two boxes

Thus, we can conclude that after the collision, the speed of the two boxes will reduce due to lost in kinetic energy associated with inelastic collision.

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