Questions about Explosions and Center of Mass in a Homework Problem

[momo1111]

Homework Statement

[/B]
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' ?

Homework 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

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
<br /> \frac{1}{2}(M+m)v_0^2+ T_0 = \frac{1}{2}(M+m)v&#039;^2_{cm} + \frac{1}{2}Mv&#039;^2_A +\frac{1}{2}mv&#039;^{2}_B

v&#039;_{cm} = is the velocity of center of mass after the explosion ..so it move the same velocity
as before ,which mean - v&#039;_{cm} = v_0 .

so we get :
(1) 0 = Mv&#039;_A +mv&#039;_B - momentum
(2) T_0 = \frac{1}{2}Mv&#039;^2_A +\frac{1}{2}mv&#039;^{2}_B
i get that if
m - \frac{m^2}{M} &gt; 0 then v&#039;_B = \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}\hat x&#039;
and v_B = v_0 \hat x+ ( \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}) \hat xQ:
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 can't move
to the same direction ,do they ? if so , (1) - wrong solution .
any way don't have a clue ..
4.??

 

[SammyS]

Homework Statement

[/B]
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&#039;
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&#039; ,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&#039;,what is the time between the
collision of B with A , and the collision between B and A&#039; ?

Homework Equations

The total kinetic energy (in the lab frame) is :
K = K_{cm} +\frac{1}{2} \sum_{n=0}^N m_i (v&#039;_i)^2

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

What you have given as T above is actually T₀.

after :
T+ T_0 = \frac{1}{2}(M+m)v&#039;^2_{cm} + \frac{1}{2}Mv&#039;^2_A +\frac{1}{2}mv&#039;^{2}_B
or
<br /> \frac{1}{2}(M+m)v_0^2+ T_0 = \frac{1}{2}(M+m)v&#039;^2_{cm} + \frac{1}{2}Mv&#039;^2_A +\frac{1}{2}mv&#039;^{2}_B

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.

v&#039;_{cm} = is the velocity of center of mass after the explosion ..so it move the same velocity
as before ,which mean - v&#039;_{cm} = v_0 .

so we get :
(1) 0 = Mv&#039;_A +mv&#039;_B - momentum
(2) T_0 = \frac{1}{2}Mv&#039;^2_A +\frac{1}{2}mv&#039;^{2}_B
i get that if
m - \frac{m^2}{M} &gt; 0 then v&#039;_B = \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}\hat x&#039;
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 can't move
to the same direction ,do they ? if so , (1) - wrong solution .
any way don't have a clue ..
4.??

See what making those correction will do.