fluidistic
Gold Member
- 3,928
- 272
Hi all,
I'm stupefied. In continuation to my last thread (https://www.physicsforums.com/showthread.php?t=277124), I wanted to see if any perfectly plastic collision would make the total mechanical energy of the system go down to a factor of \frac{1}{\sqrt 6}. I got a strange result and I know it cannot be possible so there's at least one error. If you could find it out... I'd be grateful.Say you have initially 2 particles in motion. The first one has a mass m and a speed of v_1. Second one has a mass M and a speed of v_2.
They collide and after the collision they remain attached. They form a new particle of mass M+m and speed v_3. (I assume speed instead of velocity to simplify but I might be wrong by doing this however).
The linear momentum is conserved. Putting away vectors, I have that P_i=P_f \Leftrightarrow mv_1+Mv_2=(M+m)v_3 \Leftrightarrow v_3=\frac{mv_1+Mv_2}{M+m}.
As the mechanical energy is not conserved I still want to find out the energy before and after the collision in order to compare them.
I have that E_i=\frac{mv_1^2+Mv_2^2}{2} while E_f=\frac{(m+M)v_3^2}{2}=\frac{(mv_1+Mv_2)^2}{2(m+M)}.
I want to know how much times the initial energy is greater than the final. So E_i=\alpha E_f \Leftrightarrow \alpha= \frac{mv_1^2+Mv_2^2}{2} \cdot \frac{2(m+M)}{(mv_1+Mv_2)^2} by expanding I finally get that \alpha=\frac{m^2v_1^2+Mmv_2^2+Mmv_1^2+M^2v_2^2}{m^2v_1^2+2mv_1Mv_2+M^2v_2^2}. Note that the numerator and the denominator are almost equal, they only differ by the term "Mm(v_2^2+v_1^2)" in the numerator and "2mv_1Mv_2" at the denominator. But if you set v_1 and v_2 to be \frac {1m}{s} }, there are equal and as a consequence the mechanical energy is conserved...which is obviously wrong. Where did I go wrong?
Thank you very much.
Homework Statement
I'm stupefied. In continuation to my last thread (https://www.physicsforums.com/showthread.php?t=277124), I wanted to see if any perfectly plastic collision would make the total mechanical energy of the system go down to a factor of \frac{1}{\sqrt 6}. I got a strange result and I know it cannot be possible so there's at least one error. If you could find it out... I'd be grateful.Say you have initially 2 particles in motion. The first one has a mass m and a speed of v_1. Second one has a mass M and a speed of v_2.
They collide and after the collision they remain attached. They form a new particle of mass M+m and speed v_3. (I assume speed instead of velocity to simplify but I might be wrong by doing this however).
The linear momentum is conserved. Putting away vectors, I have that P_i=P_f \Leftrightarrow mv_1+Mv_2=(M+m)v_3 \Leftrightarrow v_3=\frac{mv_1+Mv_2}{M+m}.
As the mechanical energy is not conserved I still want to find out the energy before and after the collision in order to compare them.
I have that E_i=\frac{mv_1^2+Mv_2^2}{2} while E_f=\frac{(m+M)v_3^2}{2}=\frac{(mv_1+Mv_2)^2}{2(m+M)}.
I want to know how much times the initial energy is greater than the final. So E_i=\alpha E_f \Leftrightarrow \alpha= \frac{mv_1^2+Mv_2^2}{2} \cdot \frac{2(m+M)}{(mv_1+Mv_2)^2} by expanding I finally get that \alpha=\frac{m^2v_1^2+Mmv_2^2+Mmv_1^2+M^2v_2^2}{m^2v_1^2+2mv_1Mv_2+M^2v_2^2}. Note that the numerator and the denominator are almost equal, they only differ by the term "Mm(v_2^2+v_1^2)" in the numerator and "2mv_1Mv_2" at the denominator. But if you set v_1 and v_2 to be \frac {1m}{s} }, there are equal and as a consequence the mechanical energy is conserved...which is obviously wrong. Where did I go wrong?
Thank you very much.
Last edited: