How to solve this question?

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4 Answers
Apr 15, 2018

Answer:

a) Not answer
b) not answered
c) #L/Msqrt(k(M+m)#
d)#2L#
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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.

enter image source here

#"Energy in stretching the string"=1/2kL^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=1/2(M+m)x^2#

#1/2kL^2=1/2(M+m)x^2#

#x^2=(kL^2)/(M+m)#

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

Now, consider when the marble slips from the piston.

enter image source here

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

#(M+m)sqrt((kL^2)/(M+m))=Mv#

#Mv=sqrt((M+m)^2)sqrt((kL^2)/(M+m))#

#=sqrt((kL^2(M+m)^cancel2)/cancel(M+m))#

#M^2v^2=kL^2(M+m)#

#v^2=L^2/M^2k(M+m)#

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


for part d), we will do the inverse.

Let the marble be projected off at speed #2v#

enter image source here

From conservation of momentum
Momentum before=momentum after

#2Mv=x(M+m)#

#2Lsqrt(k(M+m))=x(M+m)#

#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#

enter image source here

#KE=1/2(M+m)((2Lsqrt(k(M+m)))/(M+m))^2#

#=cancel(M+m)/2*(4L^2kcancel((M+m)))/cancel((M+m)^2)#

#=2kL^2#

#E_"pot" " in spring"=1/2kL_1^2#

By conservation of energy;

#1/2cancelkL_1^2=2cancelkL^2#

#L_1^2=4L^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

Answer:

See below.

Explanation:

Movement equation

#k(x-x_0) + (m+M)d^2/(dt^2)(x-x_0) = 0#

Multiplying by #d/(dt)(x-x_0)#

#1/2 k d/(dt)(x-x_0)^2+(m+M)1/2 d/(dt)(d/dt(x-x_0))^2 = 0#

Calling #x-x_0 = y# and #d/(dt)(x-x_0) = v# we have after integration

#1/2 y^2+1/2(m+M)v^2 = C#

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

Also integrating directly the movement equation

#k y +(m+M) ddot y = 0#

which is a harmonic movement equation results

#y = y_0 sin(sqrt(k/(m+M))t+phi_0)#

and

#v = dot y = sqrt(k/(m+M))y_0 cos(sqrt(k/(m+M))t+phi_0)#

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_"net"=(M+m)(d^2x)/dt^2=-kx#

Rewriting it as a second order differential equation of SHM.

#(d^2x)/dt^2+(k)/(M+m)x=0# ......(1)

(b) Above equation has a general sinusoidal solution as

#x=Asin\ omegat+B cos\ omega t# ......(2)
where #omega=sqrt((k)/(M+m))#

Now velocity is given by

#dot x=omegaAcosomegat-omegaBsinomegat# ......(3)

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

#-L=Asin\ (omegaxx0)+B cos\ (omega xx0)#
#=>B=-L #

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

#0=omegaAcos(omegaxx0)-omegaBsin(omegaxx0)#
#=>A=0#

Solution (2) becomes

#x=-L cos\ omega t#

Velocity becomes

#dot x=omegaLsinomegat#

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\ omega t# ......#(i)#

Velocity

#dot x=omegaLsinomegat# ........#(ii)#

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 omegat# factor in the expression for velocity, maximum velocity is reached when

#sinomegaT=1#
#=>omegaT=pi/2#

#:.# from #(ii)#

#dotx_max=omegaL#
#=>dotx_max=sqrt(k/(m+M))L# ......#(iii)#

Inserting value of #omegaT# in expression for position we get

#x(T)=-L cos\ (pi/2)#
#x(T)=0=x_0#

(d) From #(iii)# we see that

#dot x_maxpropL#

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

#=2L#