Solved Problem on Linear Momentum
ES
FR
PT-BR
advertisement
In the figure, mA = 1 kg and mB = 2 kg, the plane is frictionless,
and the spring has a negligible mass. Initially, the spring is compressed between the blocks then the
system is released at rest. The blocks are pushed apart, and the spring falls for not being attached to
any of them. Body B acquires a speed of 0.5 m/s. Determine the elastic potential energy of the
spring at the moment the system is released.
Problem data:
- Mass of block A: mA = 1 kg;
- Mass of block B: mB = 2 kg;
- Speed of block B: vB = 0.5 m/s.
Initially, the total energy of the system is in the form of elastic potential energy of the spring (Us), and the initial speeds of the blocks are zero (v0A = v0B = 0) since the system is initially at rest. When the system is released, the spring begins to push the blocks, the elastic potential energy of the spring begins to convert into kinetic energy of blocks A and B, \( K_{\small A} \) and \( K_{\small B} \), due to the speed that the blocks acquire. Finally, when the spring is completely distended the total energy of the system will be in the form of kinetic energy of the blocks moving with speeds vA and vB = 0.5 m/s (Figure 1).
Solution
The total energy of the system will be given by the Principle of Conservation of Mechanical Energy
\[
\begin{gather}
E_i=E_f\\[5pt]
U_s=K_{\small A}+K_{\small B}
\end{gather}
\]
the kinetic energy is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{K=\frac{mv^2}{2}}
\end{gather}
\]
\[
\begin{gather}
U_{s}=\frac{m_{\small A}v_{\small A}^2}{2}+\frac{m_{\small B}v_{\small B}^2}{2} \tag{I}
\end{gather}
\]
The final speed of block A is obtained from the Conservation of Momentum
\[
\begin{gather}
p_i=p_f
\end{gather}
\]
The momentum is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{p=mv}
\end{gather}
\]
\[
\begin{gather}
p_{\small A i}+p_{\small B i}=p_{A f}+p_{B f}\\[5pt]
m_{\small A}v_{0\small A}+m_{\small B}v_{0\small B}=m_{\small A}v_{\small A}+m_{\small B}v_{\small B}
\end{gather}
\]
substituting the problem data
\[
\begin{gather}
1\;\mathrm{kg}\times 0+2\;\mathrm{kg}\times 0=(1\;\mathrm{kg})v_{A}+(2\;\mathrm{kg})\times\left(0.5\;\mathrm{\frac{m}{s}}\right)\\[5pt]
-(1\;\mathrm{\cancel{kg}})v_{A}=1\;\mathrm{\frac{\cancel{kg}.m}{s}}\\[5pt]
v_{A}=-1\;\mathrm{m/s}
\end{gather}
\]
Substituting the problem data and the speed of block A, found above, in the equation (I)
\[
\begin{gather}
U_s=\frac{(1\;\mathrm{kg})\times\left(-1\;\mathrm{\frac{m}{s}}\right)^2}{2}+\frac{(2\;\mathrm{kg})\times\left(0.5\;\mathrm{\frac{m}{s}}\right)^2}{2}\\[5pt]
U_s=\frac{1\;\mathrm J}{2}+\frac{(2\;\mathrm{kg})\times(0.25\;\mathrm{\frac{m^2}{s^2}})}{2}\\[5pt]
U_s=\frac{1\;\mathrm J}{2}+\frac{0.5\;\mathrm J}{2}\\[5pt]
U_s=\frac{1.5\;\mathrm J}{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{U_s=0.75\;\mathrm J}
\end{gather}
\]
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .