Optimizing velocity in inelastic collision?

It is given that the stretch of the spring is $0.1$ m. and the spring force is $10$ N. So the force constant of the spring is $100$ N/m. Energy stored in the spring is $\dfrac{1}{2}\times 100\times (0.1)^2=0.5$ Joules. So velocity acquired by the block $A$ is $u=\dfrac{1}{\sqrt M}$ . Since the collision is perfectly inelastic, the two blocks will move together with velocity $v$ obtained from the momentum conservation equation as $v=\dfrac{\sqrt M}{M+4}=\dfrac{1}{\sqrt M+\dfrac{4}{\sqrt M}}$ . Using A.M.-G.M. inequality, we get the maximum value of $v$ as $v_{max}=V=\dfrac{1}{4}$ when $\sqrt M=\dfrac{4}{\sqrt M}$ or $M=4$ . Hence $V\times M=\dfrac{1}{4}\times 4=\boxed 1$ .

Potential energy in the compressed spring. ${ E }_{ s }=\frac { 1 }{ 2 } k{ x }^{ 2 }$

Hooke’s law $F=kx$ or $k=F/x$

${ E }_{ s }=\frac { 1 }{ 2 } \frac { F }{ x } { x }^{ 2 }\\ { E }_{ s }=\frac { 1 }{ 2 } Fx\\ { E }_{ s }=\frac { 1 }{ 2 } 10*0.1\\ { E }_{ s }=0.5J$

Kinetic energy ${ E }_{ k }=\frac { 1 }{ 2 } m{ v }^{ 2 }$

Initially both blocks have 0 velocity. No kinetic energy.

All the springs potential energy is transferred to Kinetic energy in the movement of Block A.

${ E }_{ s }={ E }_{ k\\ }\\ 0.5j=\frac { 1 }{ 2 } { m }_{ A }{ { v }_{ A } }^{ 2 }\\ { v }_{ A-before }=\sqrt [ 2 ]{ \frac { 1 }{ { m }_{ A } } }$

An inelastic collision conservation of momentum is observed. Energy will be lost to deformation of the blocks, vibrations, noise, heat etc. after the collision both blocks will be moving at the same velocity with no separation between them.

The total mass moving towards the right will be the sum of the two blocks.

${ v }_{ A-after }={ v }_{ B-after }\\ { v }_{ B-before }=0$

Momentum = mass time velocity.

$(p=mv\\ { p }_{ A-before }+{ p }_{ B-before }={ p }_{ A-after }+{ p }_{ B-after }\\ { m }_{ A }{ v }_{ A-before }+{ m }_{ B }{ v }_{ B-before }={ m }_{ A }{ v }_{ A-after }+{ m }_{ B }{ v }_{ B-after }\\ { m }_{ A }\sqrt [ 2 ]{ \frac { 1 }{ { m }_{ A } } } +0={ m }_{ A }{ v }_{ A-after }+4{ v }_{ B-after }\\ { m }_{ A }\sqrt [ 2 ]{ \frac { 1 }{ { m }_{ A } } } =({ m }_{ A }+4){ v }_{ B-after }\\ { v }_{ B-after }={ { m }_{ A }\sqrt [ 2 ]{ \frac { 1 }{ { m }_{ A } } } }/{ ({ m }_{ A }+4) })\\ { v }_{ B-after }=\frac { \sqrt [ 2 ]{ { m }_{ A } } }{ { (m }_{ A }+4) }$

Equation relates the velocity of block B after the collision as it relates to the mass of block A.

The problem has become; determine the value of mass for block A that returns the largest value for velocity of block B in this expression. It has been more that 40 years since I have done any calculus, I don’t recall much. I believe the solution would be to find the first derivative of this relatively simple expression and then equate that first derivative (slop of the line) to = 0. I decided to leave that to someone that is more comfortable with calculus.

I took the practical route (for me). I took a couple of minutes, inserted the expression in a excel spreadsheet and had the excel solver determine the highest possible value for the velocity.

It seems to me that there may be some question as to whether anyone will ever read this post but if someone has a different approach I would very much like to see it.

Maximum value for Velocity of block B is 0.25m/s

The mass of block A that is required to achieve that maximum velocity = 4kg

Not unexpected, the mass of block A must equal the mass of block B in order to achieve the maximum velocity after the collision.

Answer

${ m }_{ A }{ v }_{ B }=4*0.25=1$