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

Consider a gas endosed in a box. A molecule of mass m, having a velocity −2i^+3j^+4k^-2\hat{i}+3\hat{j}+4\hat{k} mlides with a wall parallel to the xzxz plane. What will be its velocity after collision and its change in momentum? j^\hat{j} & k^\hat{k} are unit vectors along the xyxy & zz axis.

Answer

Velocity after collision: v⃗f=−2i^−3j^+4k^\vec{v}_f = -2\hat{i} - 3\hat{j} + 4\hat{k}

Change in momentum: Δp⃗=−6mj^\Delta \vec{p} = -6m\hat{j}

Explanation

Solution

The problem describes the elastic collision of a gas molecule with a wall. In an elastic collision with a stationary wall, the component of the molecule's velocity perpendicular to the wall reverses 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: 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: The change in momentum, Δp⃗\Delta \vec{p}, 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.