Answers


1. Let m1 =0.143 Kg , m2 = 0.309 Kg , v1 = + 0,800 m/s ,
v2 = - 2.30 m/s.
m1v1 + m2v2 = m1V1 + m2V2 conservation of momentum
m1v1^2+m2v2^2=m1V1^2+m2V2^2 conservation of energy 
Solve
V1 = (m1-m2) v1/(m1+m2) + 2 m2 v2/(m1+m2)
V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)
V1 = - 3.44 m/s
V2 = - 0.338 m/s

2. \begin{displaymath} m_1 v_{i1} = m_1 v_{f1} + m_2 v_{f2}, \end{displaymath}


where $v_{f1}$ and $v_{f2}$ are the final velocities of the first and second objects, respectively. Since the collision is elastic, the total kinetic energy must be the same before and after the collision. Hence, 
\begin{displaymath} \frac{1}{2} m_1 v_{i1}^2 = \frac{1}{2} m_1 v_{f1}^{ 2} + \frac{1}{2} m_2 v_{f2}^{ 2}. \end{displaymath}


Let $x=v_{f1}/v_{i1}$ and $y=v_{f2}/v_{i1}$. Noting that $m_2/m_1=3$, the above two equations reduce to 

\begin{displaymath} 1 = x + 3 y, \end{displaymath}


and 
\begin{displaymath} 1 = x^2 + 3 y^2. \end{displaymath}


Eliminating $x$ between the previous two expressions, we obtain 
\begin{displaymath} 1 = (1-3 y)^2 + 3 y^2, \end{displaymath}


or 
\begin{displaymath} 6 y (2 y-1)=0, \end{displaymath}


which has the non-trivial solution $y=1/2$. The corresponding solution for $x$ is $x= (1-3 y)=-1/2$.It follows that the final velocity of the first object is 

\begin{displaymath} v_{f1} = x v_{i1} = -0.5\times12 = -6 {\rm m/s}. \end{displaymath}


The minus sign indicates that this object reverses direction as a result of the collision. Likewise, the final velocity of the second object is 
\begin{displaymath} v_{f2} = y v_{i1} = 0.5 \times 12 = 6 {\rm m/s}. \end{displaymath}

Walang komento:

Mag-post ng isang Komento

Mag-subscribe sa: Mga Post (Atom)