- #1
gelfand
- 40
- 3
I would like to check my understanding for this problem :
A puck with mass ##3m## is stationary on a horizontal friction-less surface. It is
being impacted in an elastic head-on collision by another puck with the mass
##2m## traveling with speed ##u## to the right. Find the speed and direction of
motion of each puck after collision in terms of ##u##
**********
So for this I would note that because there's no friction there's going to be no
loss of momentum, and energy is always conserved.
So for the equality of momentum before and after the collision I will have:
$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$
And for the equality of energy I have:
$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$
Here there's no work being done during the movement of the puck, and there's no
energy lost due to friction. The pucks are on the ground, so there's no
potential energy either, which gives$$
KE_0 = KE_f
$$
The kinetic energy for the system is
$$
\frac{1}{2}m_1u_1^2 +
\frac{1}{2}m_2u_2^2 =
\frac{1}{2}m_1v_1^2 +
\frac{1}{2}m_2v_2^2
$$
I can multiply all by ##2##, set the second term of the LHS to zero and input the
given quantities here.
Once this is done I have a set of simultaneous equations that can be solved.
But what I'm not sure about -
What if there's friction? How do I consider this?
It seems that If I have friction then I'll no longer have
$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$
Though I will still have
$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$But this one equation (energy conservation) wouldn't allow me to solve for two
unknowns?
I see there are other questions appearing that have similar meanings - I have tried to be explicit in my understanding here though, so if there's anything that "doesn't look right" please say.
Thanks
A puck with mass ##3m## is stationary on a horizontal friction-less surface. It is
being impacted in an elastic head-on collision by another puck with the mass
##2m## traveling with speed ##u## to the right. Find the speed and direction of
motion of each puck after collision in terms of ##u##
**********
So for this I would note that because there's no friction there's going to be no
loss of momentum, and energy is always conserved.
So for the equality of momentum before and after the collision I will have:
$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$
And for the equality of energy I have:
$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$
Here there's no work being done during the movement of the puck, and there's no
energy lost due to friction. The pucks are on the ground, so there's no
potential energy either, which gives$$
KE_0 = KE_f
$$
The kinetic energy for the system is
$$
\frac{1}{2}m_1u_1^2 +
\frac{1}{2}m_2u_2^2 =
\frac{1}{2}m_1v_1^2 +
\frac{1}{2}m_2v_2^2
$$
I can multiply all by ##2##, set the second term of the LHS to zero and input the
given quantities here.
Once this is done I have a set of simultaneous equations that can be solved.
But what I'm not sure about -
What if there's friction? How do I consider this?
It seems that If I have friction then I'll no longer have
$$
m_1 v_1 + m_2 v_2 =
m_1 u_1 + m_2 u_2
$$
Though I will still have
$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$But this one equation (energy conservation) wouldn't allow me to solve for two
unknowns?
I see there are other questions appearing that have similar meanings - I have tried to be explicit in my understanding here though, so if there's anything that "doesn't look right" please say.
Thanks