# How to solve this question?

## Apr 15, 2018

#### Answer:

a) Not answer
b) not answered
c) L/Msqrt(k(M+m)
d)$2 L$
Read disclaimer

#### Explanation:

Disclaimer: I'm fairly sure my method is not that expected from the question, but I found it interesting and wanted to share my attempt anyway. I think its wrong because I effectively skipped parts a and b. Any feedback on this would be great!

Assumptions I've made which might be wrong:

• all the energy transferred in the spring is transferred into kinetic energy for the marble and piston when the piston returns to equilibrium
• the marble and piston travel together as a single particle until the piston returns to equilibrium piston, then the marble is projected off.

Consider the problem in stages. $\text{Energy in stretching the string} = \frac{1}{2} k {L}^{2}$

Due to the conservation of energy, this will all be transformed to kinetic energy when the piston returns to ${x}_{0}$, its starting position.

This is the kinetic energy possessed by the whole system, this being the piston and marble.

Let $x$ be the final speed of the marble and piston combined.

${E}_{k} = \frac{1}{2} \left(M + m\right) {x}^{2}$

$\frac{1}{2} k {L}^{2} = \frac{1}{2} \left(M + m\right) {x}^{2}$

${x}^{2} = \frac{k {L}^{2}}{M + m}$

color(blue)(x=sqrt((kL^2)/(M+m)

Now, consider when the marble slips from the piston. The piston will be stationary at the end, so has a final velocity of 0.
We can find the final speed of the marble using the conservation of momentum.

From conservation of momentum
Total momentum before = total momentum after

$\left(M + m\right) \sqrt{\frac{k {L}^{2}}{M + m}} = M v$

$M v = \sqrt{{\left(M + m\right)}^{2}} \sqrt{\frac{k {L}^{2}}{M + m}}$

$= \sqrt{\frac{k {L}^{2} {\left(M + m\right)}^{\cancel{2}}}{\cancel{M + m}}}$

${M}^{2} {v}^{2} = k {L}^{2} \left(M + m\right)$

${v}^{2} = {L}^{2} / {M}^{2} k \left(M + m\right)$

color(blue)(v=L/Msqrt(k(M+m))

for part d), we will do the inverse.

Let the marble be projected off at speed $2 v$ From conservation of momentum
Momentum before=momentum after

$2 M v = x \left(M + m\right)$

$2 L \sqrt{k \left(M + m\right)} = x \left(M + m\right)$

color(blue)(x=(2Lsqrt(k(M+m)))/(M+m)

Now, we'll use our energy idea from earlier to find our new length back for the spring, ${L}_{1}$ $K E = \frac{1}{2} \left(M + m\right) {\left(\frac{2 L \sqrt{k \left(M + m\right)}}{M + m}\right)}^{2}$

$= \frac{\cancel{M + m}}{2} \cdot \frac{4 {L}^{2} k \cancel{\left(M + m\right)}}{\cancel{{\left(M + m\right)}^{2}}}$

$= 2 k {L}^{2}$

${E}_{\text{pot" " in spring}} = \frac{1}{2} k {L}_{1}^{2}$

By conservation of energy;

$\frac{1}{2} \cancel{k} {L}_{1}^{2} = 2 \cancel{k} {L}^{2}$

${L}_{1}^{2} = 4 {L}^{2}$

color(blue)(L_1=2L

So to get double the speed out, double the distance you pull the spring. (rather boring final answer, if I do say so).

Apr 16, 2018

See below.

#### Explanation:

Movement equation

$k \left(x - {x}_{0}\right) + \left(m + M\right) {d}^{2} / \left({\mathrm{dt}}^{2}\right) \left(x - {x}_{0}\right) = 0$

Multiplying by $\frac{d}{\mathrm{dt}} \left(x - {x}_{0}\right)$

$\frac{1}{2} k \frac{d}{\mathrm{dt}} {\left(x - {x}_{0}\right)}^{2} + \left(m + M\right) \frac{1}{2} \frac{d}{\mathrm{dt}} {\left(\frac{d}{\mathrm{dt}} \left(x - {x}_{0}\right)\right)}^{2} = 0$

Calling $x - {x}_{0} = y$ and $\frac{d}{\mathrm{dt}} \left(x - {x}_{0}\right) = v$ we have after integration

$\frac{1}{2} {y}^{2} + \frac{1}{2} \left(m + M\right) {v}^{2} = C$

this relationship works just until the marble loses contact with the piston.

Also integrating directly the movement equation

$k y + \left(m + M\right) \ddot{y} = 0$

which is a harmonic movement equation results

$y = {y}_{0} \sin \left(\sqrt{\frac{k}{m + M}} t + {\phi}_{0}\right)$

and

$v = \dot{y} = \sqrt{\frac{k}{m + M}} {y}_{0} \cos \left(\sqrt{\frac{k}{m + M}} t + {\phi}_{0}\right)$

Apr 16, 2018

#### Answer:

See parts (c) and (d) as a separate answer.

#### Explanation:

(a) Keeping in view the given figure, and considering $x = {x}_{0}$ as the origin, equation of motion for the system can be obtained from Newton's Second Law of Motion and Hook's Law as

${F}_{\text{net}} = \left(M + m\right) \frac{{d}^{2} x}{\mathrm{dt}} ^ 2 = - k x$

Rewriting it as a second order differential equation of SHM.

$\frac{{d}^{2} x}{\mathrm{dt}} ^ 2 + \frac{k}{M + m} x = 0$ ......(1)

(b) Above equation has a general sinusoidal solution as

$x = A \sin \setminus \omega t + B \cos \setminus \omega t$ ......(2)
where $\omega = \sqrt{\frac{k}{M + m}}$

Now velocity is given by

$\dot{x} = \omega A \cos \omega t - \omega B \sin \omega t$ ......(3)

Displacement at $t = 0$ is $= - L$. From (2) we get

$- L = A \sin \setminus \left(\omega \times 0\right) + B \cos \setminus \left(\omega \times 0\right)$
$\implies B = - L$

Velocity at $t = 0$ is $= 0$. From (3) we get

$0 = \omega A \cos \left(\omega \times 0\right) - \omega B \sin \left(\omega \times 0\right)$
$\implies A = 0$

Solution (2) becomes

$x = - L \cos \setminus \omega t$

Velocity becomes

$\dot{x} = \omega L \sin \omega t$

It is clear that equation of motion and its solution are valid so long as piston and marble move together as single entity.

Apr 16, 2018

#### Answer:

From other posted solution we got

Position

$x = - L \cos \setminus \omega t$ ......$\left(i\right)$

Velocity

$\dot{x} = \omega L \sin \omega t$ ........$\left(i i\right)$

#### Explanation:

(c) Let velocity of the piston marble system be maximum at $t = T$. After this point, piston starts to slow down, the contact between the two is lost and marble carries on its forward motion with this constant velocity. Due to presence of $\sin \omega t$ factor in the expression for velocity, maximum velocity is reached when

$\sin \omega T = 1$
$\implies \omega T = \frac{\pi}{2}$

$\therefore$ from $\left(i i\right)$

${\dot{x}}_{\max} = \omega L$
$\implies {\dot{x}}_{\max} = \sqrt{\frac{k}{m + M}} L$ ......$\left(i i i\right)$

Inserting value of $\omega T$ in expression for position we get

$x \left(T\right) = - L \cos \setminus \left(\frac{\pi}{2}\right)$
$x \left(T\right) = 0 = {x}_{0}$

(d) From $\left(i i i\right)$ we see that

${\dot{x}}_{\max} \propto L$

Therefore, in order to achieve twice the speed of marble on separation the piston must be pushed/pulled back twice the initial value.

$= 2 L$