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Question: Consider a gas enclosed in a box. A molecule of mass m, having a velocity $2\hat{i} + 3\hat{j} + 4\h...

Consider a gas enclosed in a box. A molecule of mass m, having a velocity 2i^+3j^+4k^2\hat{i} + 3\hat{j} + 4\hat{k} collides with a wall parallel to the xz plane. What will be its velocity after collision and its change in momentum? i^\hat{i}, j^\hat{j} & k^\hat{k} are unit vectors along the x,y & z axis.

Answer

Velocity after collision: 2i^−3j^+4k^2\hat{i} - 3\hat{j} + 4\hat{k}

Change in momentum: −6mj^-6m\hat{j}

Explanation

Solution

The problem describes an elastic collision of a gas molecule with a stationary wall. In such a collision, the component of the molecule's velocity perpendicular to the wall reverses its direction, while the components parallel to the wall remain unchanged.

Given:

  • Initial velocity of the molecule, v⃗i=2i^+3j^+4k^\vec{v}_i = 2\hat{i} + 3\hat{j} + 4\hat{k}.
  • Mass of the molecule = mm.
  • The wall is parallel to the xzxz plane. This means the normal to the wall is along the y-axis.

1. Velocity after collision (v⃗f\vec{v}_f):

The components of velocity parallel to the xzxz plane are the x-component (2i^2\hat{i}) and the z-component (4k^4\hat{k}). These components remain unchanged. The component of velocity perpendicular to the xzxz plane is the y-component (3j^3\hat{j}). This component reverses its direction. So, 3j^3\hat{j} becomes −3j^-3\hat{j}.

Therefore, the velocity after collision, v⃗f\vec{v}_f, will be: v⃗f=2i^−3j^+4k^\vec{v}_f = 2\hat{i} - 3\hat{j} + 4\hat{k}

2. Change in momentum (Δp⃗\Delta \vec{p}):

The change in momentum is given by the difference between the final momentum and the initial momentum: Δp⃗=p⃗f−p⃗i=mv⃗f−mv⃗i=m(v⃗f−v⃗i)\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i = m(\vec{v}_f - \vec{v}_i)

Substitute the initial and final velocities: Δp⃗=m[(2i^−3j^+4k^)−(2i^+3j^+4k^)]\Delta \vec{p} = m[(2\hat{i} - 3\hat{j} + 4\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})] Δp⃗=m[2i^−3j^+4k^−2i^−3j^−4k^]\Delta \vec{p} = m[2\hat{i} - 3\hat{j} + 4\hat{k} - 2\hat{i} - 3\hat{j} - 4\hat{k}]

Combine the components: Δp⃗=m[(2−2)i^+(−3−3)j^+(4−4)k^]\Delta \vec{p} = m[(2-2)\hat{i} + (-3-3)\hat{j} + (4-4)\hat{k}] Δp⃗=m[0i^−6j^+0k^]\Delta \vec{p} = m[0\hat{i} - 6\hat{j} + 0\hat{k}] Δp⃗=−6mj^\Delta \vec{p} = -6m\hat{j}

The change in momentum is entirely along the negative y-axis, which is the direction perpendicular to the wall.