Question

a particle of mass 'm' is projected with velocity 'u' at angle 'x' with horizontal. During the period when the particle descends from highest point to the position where its velocity vector makes an angle x/2 with horizontal. work done by gravity force is

Answer 1

Answer on Question #58487, Physics / Mechanics | Relativity

A particle of mass $m$ is projected with velocity $u$ at angle $x$ with horizontal. During the period when the particle descends from highest point to the position where its velocity vector makes an angle $x/2$ with horizontal. Work done by gravity force is -?

Solution:

Initial condition:

A mass of particle $m$, and its initial velocity $\vec{u}$ and initial angle between $\vec{u}$ and horizontal equal $x$. In finish this angle is equal $\frac{x}{2}$.

$W = \vec{F} \cdot \vec{S}$, where $\vec{F} = m\vec{g}$ - gravitational force, where $\vec{g}$ - gravitational acceleration, and $\vec{S}$ - displacement vector.

$W - ?$

$$W = \vec{F} \cdot \vec{S} = F \cdot S \cdot \cos \theta = F \cdot \Delta h.$$

$\Delta h$ - displacement of the particle parallel vertical.

First velocity equal $\vec{u} = (v_{x0}; v_{y0}) = \vec{i} v_{x0} + \vec{j} v_{y0}$,

Where $v_{x0} = u \cos x$ and $v_{y0} = u \sin x$ projection vector on the x-axis and y-axis, respectively. $\vec{v}(t) = (v_x; v_y)$ velocity of the particle at time $t$.

We have only one force, which parallel y-axis, then $v_{x} = u \cos x = \text{const}$

$$v_y = u \sin x - g t.$$

We should be obtain work the particle descends from highest point to the position where its velocity vector makes an angle $x/2$ with horizontal

i. e. at $t = 0$ $v_{x}(0) = u \cos x$ and $v_{y}(0) = 0$ and

at moment $t$ $v_{x}(t) = u \cos x$ and $v_{y} = -gt$. When the angle between velocity of particle and horizontal increased to $\frac{x}{2}$, we defined $t_f$ - finish time.

After defined projection of displacement on y-axis $\Delta h$, from motion equation.

$\frac{v_y}{v_x} = \tan \theta$ - relation between the components of the velocity vector,

When $t = t_f, \quad \theta = -\frac{x}{2}$, because y-component of particle position decreasing

$$\tan \theta = \tan \left(- \frac {x}{2}\right) = \frac {v _ {y} (t _ {f})}{v _ {x} (t _ {f})} = \frac {- g t _ {f}}{u \cos x}$$ $$- \tan {\frac {x}{2}} = - \frac {g t _ {f}}{u \cos x}$$ $$\tan {\frac {x}{2}} = \frac {g t _ {f}}{u \cos x}$$ $$t _ {f} = \frac {u \cos x \tan {\frac {x}{2}}}{g}$$ $$\tan {\frac {x}{2}} = \frac {\sin x}{1 + \cos x}$$ $$t _ {f} = \frac {u \cos x \tan {\frac {x}{2}}}{g} = \frac {u \cos x \sin x}{g (1 + \cos x)}$$ $$\Delta h = v _ {y} (0) \cdot t _ {f} - \frac {g t _ {f} ^ {2}}{2}, v _ {y} (0) = 0$$ $$\Delta h = - \frac {g t _ {f} ^ {2}}{2} = - \frac {g u ^ {2} \cos^ {2} x \sin^ {2} x}{g ^ {2} (1 + \cos x) ^ {2}} = - \frac {u ^ {2} \cos^ {2} x \sin^ {2} x}{g (1 + \cos x) ^ {2}}$$

"minus" sign means that displacement of the particle along the y-axis (height variation) directed downward, and the gravitational force acting on the particle, also aims to down.

Gravity $F = mg$ does work $W = mgh$ along any descending path

$$W = F _ {g} \cdot \Delta h = (- m g) \cdot \left(- \frac {u ^ {2} \cos^ {2} x \sin^ {2} x}{g (1 + \cos x) ^ {2}}\right) = m g \cdot \frac {u ^ {2} \cos^ {2} x \sin^ {2} x}{g (1 + \cos x) ^ {2}} =$$ $$= m \frac {u ^ {2} \cos^ {2} x \sin^ {2} x}{(1 + \cos x) ^ {2}}$$ $$\frac {\sin^ {2} x}{(1 + \cos x) ^ {2}} = \tan^ {2} \frac {x}{2}:$$ $$W = m u ^ {2} \cos^ {2} x \tan^ {2} \frac {x}{2} = m \frac {u ^ {2} \cos^ {2} x \sin^ {2} x}{(1 + \cos x) ^ {2}} = m \frac {u ^ {2} \sin^ {2} 2 x}{4 (1 + \cos x) ^ {2}}$$

These three answers are equal.

Answer: $W = m u^{2} \cos^{2} x \tan^{2} \frac{x}{2}$ or $W = m \frac{u^{2} \cos^{2} x \sin^{2} x}{(1 + \cos x)^{2}}$ or $W = m \frac{u^{2} \sin^{2} 2x}{4(1 + \cos x)^{2}}$

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