# Explosions,Center of Mass

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1. Jul 3, 2015

### momo1111

1. The problem statement, all variables and given/known data

$2$ masses connected with a solid rod .distace between them is $2L$ ,
total mass is $M$ .

$B$ total mass is $m$ ,length $L$ can freely move between $A$ and $A'$
on flat and frictionless surface. all collisions are elastic.

at the beginning $B$ was stuck to $A$ and all the system was moving to the right ,$v_0$ and kinetic energy $T_0$ .
at some moment ,explosion take place and the system get additional kinetic energy $T$ .

1.what is the velocity of $B$ ,$v_B$ as function of $v_0,T,m,M$ ?
2.what is the velocity of $A+A'$ ,immediately after the explosion ?
3.what is the relation between $T_0$ and $T$ ,if $A$ don't move to the left ?
4.after the explosion $B$ start to move between $A$ and $A'$,what is the time between the
collision of $B$ with $A$ , and the collision between $B$ and $A'$ ?

2. Relevant equations
The total kinetic energy (in the lab frame) is :
$K = K_{cm} +\frac{1}{2} \sum_{n=0}^N m_i (v'_i)^2$

3. The attempt at a solution
before the explosion:
$T = \frac{1}{2}(M+m)v_{cm}^2 + 0 +0$ ,given $v_0 = v_{cm}\hat x$ we get $T = \frac{1}{2}(M+m)v_0^2 + 0 +0$

after :
$T+ T_0 = \frac{1}{2}(M+m)v'^2_{cm} + \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B$
or
$\frac{1}{2}(M+m)v_0^2+ T_0 = \frac{1}{2}(M+m)v'^2_{cm} + \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B$

$v'_{cm}$ = is the velocity of center of mass after the explosion ..so it move the same velocity
as before ,which mean - $v'_{cm} = v_0$ .

so we get :
(1) $0 = Mv'_A +mv'_B$ - momentum
(2) $T_0 = \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B$
i get that if
$m - \frac{m^2}{M} > 0$ then $v'_B = \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}\hat x'$
and $v_B = v_0 \hat x+ ( \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}) \hat x$

Q:
1. is this the solution ?
2.if 1 is o.k then $v_b$ ...easy
3 - what it means "move to the left " ? ...at c.m reference system ,it`s look to me that 2 masses cant move
to the same direction ,do they ? if so , (1) - wrong solution .
any way don't have a clue ..
4.??

2. Jul 3, 2015

### SammyS

Staff Emeritus
What you have given as T above is actually T0.
What you have as T0 on the left side of the above equation is actually T. The $\displaystyle \ \frac{1}{2}(M+m){v_0}^2 \$ is T0.
See what making those correction will do.