A 300-g ball is kicked with a velocity of vA=25m/s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction θ of the velocity of the rebounding ball at B.

Question

A 300-g ball is kicked with a velocity of { v }_{ A } = 25 m/s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction heta of the velocity of the rebounding ball at B.

Accepted Answer

Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the symmetrical properties of the trajectory, [latex]v_B = v_A = 25 m/s[/latex] and [latex]\phi = 30°[/latex].

Conservation of Linear Momentum: Since no impulsive force acts on the football along the x axis, the linear momentum of the football is conserved along the x axis.

[latex]\begin{aligned} (\underleftarrow{+}) \quad\quad & m (v_B)_x = m(v_B^{'})_x \ & 0.3(25 \cos 30°) = 0.3 (v_B^{'})_x \ & (v_B^{'})_x = 21.65 m/s \leftarrow \end{aligned}[/latex]

Coefficient of Restitution: Since the ground does not move during the impact, the coefficient of restitution can be written as

[latex]\begin{aligned} (+\uparrow) \quad\quad & e = \frac { 0 - (v_B^{ ' })_y } { (v_B)_y - 0 } \ & 0.4 = \frac { -(v_B^{ ' })_y } { -25 \sin 30° } \ & (v_B^{ ' })_y = 5 m/s \uparrow \end{aligned}[/latex]

Thus, the magnitude of [latex] extbf{v}_B^{'}[/latex] is

[latex]v_B^{'} = \sqrt{(v_B^{'})_x + (v_B^{'})_y} = \sqrt{21.65^2 + 5^2} = 22.2 m/s[/latex]

and the angle of [latex] extbf{v}_B^{'}[/latex] is

[latex] heta = an^{ -1 } [\frac { (v_B^{'})_y } { (v_B^{'})_x }] = an^{ -1 } (\frac 5 { 21.65 }) = 13.0°[/latex]

Question : A 300-g ball is kicked with a velocity of vA=25m/s at point ...

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