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(C) The average force $\vec{F}_{avg}$ is defined as the change in momentum divided by the time interval: $\vec{F}_{avg} = \frac{\Delta \vec{p}}{\Delta

Vedclass · Accepted Answer

$\frac{2mv^2 \sin(	heta/2)}{R	heta}$

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(C) The average force $\vec{F}_{avg}$ is defined as the change in momentum divided by the time interval: $\vec{F}_{avg} = \frac{\Delta \vec{p}}{\Delta t}$.Let the velocity at point $1$ be $\vec{v}_1$ and at point $2$ be $\vec{v}_2$. Since the speed is constant,$|\vec{v}_1| = |\vec{v}_2| = v$.The angle between the velocity vectors is $	heta$. The change in momentum is $\Delta \vec{p} = m(\vec{v}_2 - \vec{v}_1)$.The magnitude of the change in momentum is $|\Delta \vec{p}| = m \sqrt{v^2 + v^2 - 2v^2 \cos 	heta} = m \sqrt{2v^2(1 - \cos 	heta)} = m \sqrt{4v^2 \sin^2(	heta/2)} = 2mv \sin(	heta/2)$.The distance traveled along the arc is $s = R	heta$. Since the speed $v$ is constant,the time taken is $\Delta t = \frac{s}{v} = \frac{R	heta}{v}$.Therefore,the average force is $F_{avg} = \frac{|\Delta \vec{p}|}{\Delta t} = \frac{2mv \sin(	heta/2)}{R	heta/v} = \frac{2mv^2 \sin(	heta/2)}{R	heta}$.

$A$ ring of mass $m$ moves from point $1$ to point $2$ along a smooth rigid horizontal wire with a constant speed $v$. The average force acting on the ring over the time of its motion from $1$ to $2$ is

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