Conservation of energy
show that when Mass (m1) and kinetic energy (Q1) collide with a stationary particle (m2) it transfers an energy. Using conservation of energy and momentum
Q2 = (4M1M2)/(M1 + M2)^2 x Q1
Cheers
I presume you simply mean find the kinetic energy of particle m2 after the collision.
$\displaystyle m_1v_{10} = m_1v_1 + m2v_2$
where $\displaystyle v_{10}$ is the velocity of m1 before the collision and $\displaystyle v_1, v_2$ are the final velocities.
$\displaystyle \frac{1}{2}m_1v_{10}^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$
Solve the momentum equation for $\displaystyle v_2$ and plug it into the energy equation. After a few steps:
$\displaystyle m_1m_2v_{10}^2 = m_1^2v_{10}^2 - 2m_1^2v_{10}v_1 + (m_1^2 + m_1m_2)v_1^2$
Solve this for $\displaystyle v_1$. Hint: Use the quadratic formula.
$\displaystyle v_1 = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{10}$
So use this to find $\displaystyle v_2$ and find $\displaystyle Q_2$.
-Dan
Hello there,
you need to consider fractional lost in kinetic energy to obtain that equation:
fractional lost in Kinetic Energy
= (lost in Kinetic Energy)/(initial Kinetic Energy) → (1)
given for mass 1:
mass = $\displaystyle m_1$
velocity before collision = $\displaystyle u_1$
velocity after collision = $\displaystyle v_1$
for mass 2:
mass = $\displaystyle m_2$
velocity before collision = $\displaystyle u_2$
velocity after collision = $\displaystyle v_2$
using principle of conservation of momentum:
$\displaystyle
m_1 u_1= m_1 v_1 + m_2 v_2 \longrightarrow (2)
$
from velocity of approach = velocity of separation
$\displaystyle
u_1 - u_2 = v_2 -v_1
$
$\displaystyle
u_1 -0 = v_2 -v_1
$
$\displaystyle
u_1 = v_2 -v_1
$
or
$\displaystyle
v_2 = u_1 +v_1
$
Substituting $\displaystyle
v_2 = u_1 +v_1
$ into (2):
$\displaystyle
m_1 u_1= m_1 v_1 + m_2 (u_1 +v_1)
$
$\displaystyle
m_1 u_1 - m_2 u_1 = m_1 v_1 + m_2 v_1
$
$\displaystyle
(m_1 - m_2) u_1 = (m_1 + m_2) v_1
$
$\displaystyle
v_1=\frac {(m_1 - m_2)} {(m_1 + m_2)}u_1 \longrightarrow(3)
$
From equation (1):
= (lost in Kinetic Energy)/(initial Kinetic Energy)
$\displaystyle
=\frac{ \frac{1}{2}m_1 {u_1}^2 - \frac{1}{2}m_1 {v_1}^2}{\frac{1}{2}m_1 {u_1}^2}
$
$\displaystyle
=\frac{ \frac{1}{2}m_1 ({u_1}^2 - {v_1}^2)}{\frac{1}{2}m_1 {u_1}^2}
$
$\displaystyle
=\frac{ {u_1}^2 - {v_1}^2}{{u_1}^2}
$
$\displaystyle
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
$
$\displaystyle
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
$
$\displaystyle
=\frac{ ({u_1} - (\frac{m_1 - m_2}{m_1 + m_2}){u_1})({u_1} + (\frac{m_1 - m_2}{m_1 + m_2}){u_1})}{{u_1}^2}\longrightarrow using(3)
$
$\displaystyle
=\frac{ ({u_1} (\frac{m_1 + m_2 - m_1 + m_2}{m_1 + m_2})({u_1} (\frac{m_1 + m_2 + m_1 - m_2}{m_1 + m_2}))}{{u_1}^2}
$
$\displaystyle
=\frac{2m_1 \cdot 2m_2}{({m_1 +m_2})^2}
$
$\displaystyle
=\frac{4m_1 m_2}{({m_1 +m_2})^2}
$
lost in kinetic energy is gained by the stationary mass hence we can rewrite the equation as:
kinetic energy of mass m2/ initial kinetic energy
$\displaystyle
=\frac{4m_1 m_2}{({m_1 +m_2})^2}
$
therefore;
kinetic energy of mass m2 = (initial kinetic energy)$\displaystyle (\frac{4m_1 m_2}{({m_1 +m_2})^2})
$
Q2 = $\displaystyle (\frac{4m_1 m_2}{({m_1 +m_2})^2})
$Q1
There is another way to get the answer but i think this should be sufficient unless you really want to know more xD cheers!