Question 13.199: A 2-kg ball B is traveling horizontally at 10 m/s when it st...

Velocity components: \quad \quad \quad \nu_{0}=10 \mathrm{~m} / \mathrm{s}

\begin{aligned}& \left(\nu_{0}\right)_{x}=\nu_{0} \quad\left(\nu_{0}\right)_{n}=\nu_{0} \cos 40^{\circ} \quad\left(\nu_{0}\right)_{t}=\nu_{0} \sin 40^{\circ} \\& \left(\nu_{A}\right)_{x}=\nu_{A} \quad\left(\nu_{A}\right)_{n}=\nu_{A} \cos 40^{\circ} \\& \left(\nu_{B}\right)_{x}=\left(\nu_{B}\right)_{n} \cos 40^{\circ}+\left(\nu_{B}\right)_{t} \sin 40^{\circ}\end{aligned}

Impulse-momentum for ball B alone.

t-direction:

\begin{aligned}m_{B}\left(\nu_{0}\right)_{t} & =m_{B}\left(\nu_{B}\right)_{t} \\\left(\nu_{B}\right)_{t} & =\left(\nu_{0}\right)_{t}=10 \sin 40^{\circ}=6.4279 \mathrm{~m} / \mathrm{s} \quad \quad \text{(1)}\end{aligned}

Impulse-momentum for balls A and B.

x-direction ←

\begin{aligned}m_{B} \nu_{0}+0 & =m_{A} \nu_{A}+m_{B}\left(\nu_{B}\right)_{x}+m_{B}\left(\nu_{B}\right)_{t} \\(2)(10)+0 & =2 \nu_{A}+2\left[\left(\nu_{B}\right)_{n} \cos 40^{\circ}+6.4279 \sin 40^{\circ}\right] \\2 \nu_{A}+ & 2\left(\nu_{B}\right)_{n} \cos 40^{\circ}=11.7365 \quad \quad \text{(1)}\end{aligned}

Coefficient of restitution. 

\begin{gathered}(e=0.8) \\\left(\nu_{B}\right)_{n}=\left(\nu_{A}\right)_{n}=e\left[0-\left(\nu_{0}\right)_{n}\right] \\\left(\nu_{B}\right)_{n}-\nu_{A} \cos 40^{\circ}=-(0.8)(10) \cos 40^{\circ} \quad \quad \text{(2)}\end{gathered}

Solving Eqs. (1) and (2) simultaneously,

\nu_{A}=6.6566 \mathrm{~m} / \mathrm{s} \quad\left(\nu_{B}\right)_{n}=-1.0291 \mathrm{~m} / \mathrm{s}

As ball A moves from the impact location to the lowest point on the path, the spring compresses and the elevation decreases. Since friction is negligible, energy is conserved.

\begin{aligned}T_{1}+V_{1} & =T_{2}+V_{2} \\\frac{1}{2} m_{A} \nu_{A}^{2}+\left(V_{e}\right)_{1}+\left(V_{g}\right)_{1} & =\frac{1}{2} m_{A} \nu_{2}^{2}+\left(V_{e}\right)_{2}+\left(V_{g}\right)_{2}\end{aligned}

Position 1: (Just after impact.)

\begin{aligned}T_{1} & =\frac{1}{2} m_{A} \nu_{A}^{2}=\frac{1}{2}(2)(6.6566)^{2}=44.3101 \mathrm{~J} \\\left(V_{e}\right)_{1} & =0 \quad \text { (The spring is unstretched. }) \\\left(V_{g}\right)_{1} & =0 \quad \text { (Datum) }\end{aligned}

Position 2: (Lowest point on path.)

T_{2}=\frac{1}{2} m_{A} \nu_{2}^{2}=\frac{1}{2}(2) \nu_{2}^{2}=\nu_{2}^{2}

For the spring,

\begin{gathered}x_{2}=l_{2}-l_{0}=0.4 \mathrm{~m}-1.2 \mathrm{~m}=0.8 \mathrm{~m} \\F_{e}=k x_{2}=(100)(0.8)=80 \mathrm{~N} \\\left(V_{2}\right)_{e}=\frac{1}{2} k x_{2}^{2}=\frac{1}{2}(100)(0.8)^{2}=32 \mathrm{~J}\end{gathered}

Elevation above datum: \quad \quad \quad h_{2}=-0.4 \mathrm{~m}

\left(V_{2}\right)_{g}=m_{A} g h_{2}=(2)(9.81)(-0.4)=-7.848

Conservation of energy:

\begin{gathered}44.310+0+0=\nu_{2}^{2}+32-7.848 \\\nu_{2}^{2}=20.158 \mathrm{~m}^{2} / \mathrm{s}^{2} \quad \nu_{2}=4.489 \mathrm{~m} / \mathrm{s}\end{gathered}

Normal acceleration at lowest point on path:

a_{n}=\frac{\nu_{2}^{2}}{\rho}=\frac{20.158}{0.7}=28.798 \mathrm{~m} / \mathrm{s}^{2} \quad \mathbf{a}_{n}=28.8 \mathrm{~m} / \mathrm{s}^{2} \uparrow

Apply Newton’s second law to the ball.

\begin{aligned}& +\uparrow  \Sigma F =m a_n:N-mg-F_e-ma_n\\& N=m g+F_{e}+m a_{n}  \\& =(2)(9.81)+80+(2)(28.798) \quad \quad \quad \quad  N= 157.2  N \blacktriangleleft\end{aligned}