Respuesta :
Answer:
[tex]r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Explanation:
Given,
- mass of the first particle = [tex]m_1[/tex]
- velocity of the first particle = [tex]v_o[/tex]
- mass of the second particle = [tex]m_2[/tex]
- velocity of the second particle = [tex]v_2 = 0[/tex]
- Time interval = [tex](t_1\ -\ t_o)[/tex]
Let [tex]v_{cm}[/tex] be the initial velocity of the center of mass of the system of particle at time [tex]t_o[/tex]
[tex]\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}[/tex]
Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'
- [tex]x_1\ =\ 0[/tex]
- [tex]x_2\ =\ d[/tex]
Center of mass of the system of particles
[tex]x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\[/tex]
Hence, at time [tex]t_0[/tex], the center of mass of the system is at [tex]x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}[/tex] at an initial speed of [tex]v_{cm}[/tex]
Both the particles are assumed to be the point masses, therefore at the time [tex]t_1[/tex] the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.
Let [tex]r_{cm}[/tex] be the distance traveled by the center of mass of the system of particles in the time interval [tex](t_1\ +\ t_0)[/tex]
From the kinematics,
[tex]s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.
