1. ## Conservation of energy

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

2. Originally Posted by j-lee00
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.

$m_1v_{10} = m_1v_1 + m2v_2$
where $v_{10}$ is the velocity of m1 before the collision and $v_1, v_2$ are the final velocities.

$\frac{1}{2}m_1v_{10}^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Solve the momentum equation for $v_2$ and plug it into the energy equation. After a few steps:
$m_1m_2v_{10}^2 = m_1^2v_{10}^2 - 2m_1^2v_{10}v_1 + (m_1^2 + m_1m_2)v_1^2$

Solve this for $v_1$. Hint: Use the quadratic formula.
$v_1 = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{10}$

So use this to find $v_2$ and find $Q_2$.

-Dan

3. 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 = $m_1$
velocity before collision = $u_1$
velocity after collision = $v_1$

for mass 2:

mass = $m_2$
velocity before collision = $u_2$
velocity after collision = $v_2$

using principle of conservation of momentum:
$
m_1 u_1= m_1 v_1 + m_2 v_2 \longrightarrow (2)
$

from velocity of approach = velocity of separation
$
u_1 - u_2 = v_2 -v_1
$

$
u_1 -0 = v_2 -v_1
$

$
u_1 = v_2 -v_1
$

or
$
v_2 = u_1 +v_1
$

Substituting $
v_2 = u_1 +v_1
$
into (2):
$
m_1 u_1= m_1 v_1 + m_2 (u_1 +v_1)
$

$
m_1 u_1 - m_2 u_1 = m_1 v_1 + m_2 v_1
$

$
(m_1 - m_2) u_1 = (m_1 + m_2) v_1
$

$
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)
$
=\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}
$

$
=\frac{ \frac{1}{2}m_1 ({u_1}^2 - {v_1}^2)}{\frac{1}{2}m_1 {u_1}^2}
$

$
=\frac{ {u_1}^2 - {v_1}^2}{{u_1}^2}
$

$
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
$

$
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
$

$
=\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)
$

$
=\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}
$

$
=\frac{2m_1 \cdot 2m_2}{({m_1 +m_2})^2}
$

$
=\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
$
=\frac{4m_1 m_2}{({m_1 +m_2})^2}
$

therefore;
kinetic energy of mass m2 = (initial kinetic energy) $(\frac{4m_1 m_2}{({m_1 +m_2})^2})
$

Q2 = $(\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!