Lot of collisions, lot of compressions!

Classical Mechanics Level pending

Two bodies $A$ and $B$ of masses $2m$ and $m$ are respectively placed on a smooth floor. They are connected by a light spring. A third body of mass $m$ moves with a velocity $v$ along the line joining $A$ and $B$ collides elastically with $A$ , as shown in the figure. At a certain time $t$ after the collision, it is found that instantaneous velocities of $A$ and $B$ are the same. Further at that instant, the compression of the spring is found to be $x$ . Determine the spring constant, $k$ .

\dfrac{mv^2}{x^2}

\dfrac{2mv^2}{3x^2}

\dfrac{4mv^2}{9x^2}

\dfrac{8mv^2}{27x^2}

2 solutions

Skanda Prasad
Feb 2, 2017

We have to use the Work Energy Theorem.

$U_i+K_i+W_{ncf}=U_f+K_f$ $0+\dfrac{1}{2}(2m)v^2+0=\dfrac{1}{2}kx^2+\dfrac{1}{2}(m+2m)v_c^2$ $2mv^2=kx^2+3mv_c^2$

By conservation of linear momentum, we get $v_c^2=\dfrac{4}{9}v^2$ $2mv^2=kx^2+3m(\dfrac{4}{9}v^2)$

$\implies$ $6mv^2-4mv^2=3kx^2$ $\implies$ $k=\dfrac{2mv^2}{3x^2}$

I am not able to understand how you have written the first equation.
Initially, the block $C$ of mass $m$ is moving and it will not stop after the collision is over. We will first have to make the equations of collisions between the block $C$ and $A$ and then conserve the energy to get the answer as $\dfrac{8mv^2}{3kx^2}.$
I think what you have done will be correct if the mass of block $C$ is also $2m$ .

Rohit Gupta - 4 years, 4 months ago

I'm really sorry, I didn't get you...Can you please post your solution so that I can correct my mistake? I came across this question in a test...There might have been an error in the question...That I can't say...But this was the answer I got and I had marks for it too..It was an integer type question. Answer was supposed to be in the form $\dfrac{nmv^2}{3x^2}$ I had given answer as $2$ for $n$ value...The answer was correct...There's a possibility of error in my solution too...my approach might have been wrong...

So please post your solution here itself. Later I can edit the question, solution and options accordingly or @Calvin Lin can delete the problem if he wishes to do so.

Skanda Prasad - 4 years, 4 months ago

Ok, I will post my solution. Then we can discuss further what can be done.

Rohit Gupta - 4 years, 4 months ago

@R G Just forget this problem, I'm not able to decide who is correct... Your solution seems correct to me too...Sorry for the late reply, I was not at home for a few days... Once you reply to this comment, I'll be deleting this question. Waiting for your reply. @Calvin Lin , if you and other experts feel that this question can be edited in some way, I'll do it or you let the concerned person do it, please reply to me, else I'll delete this problem. If it is kind of misguiding people, let this problem not be there.

Skanda Prasad - 4 years, 3 months ago

Rohit Gupta
Feb 16, 2017

First, we will calculate the velocity of $A$ after an elastic collision by $C$ . For this, we will make two equations one of momentum conservation and the other of the coefficient of restitution.

Conserving momentum. $mv = mv_C + 2m v_A$ $v = v_C + 2v_A$

Equation of coefficient of restitution, $e = 1 = \frac{v_A - v_C}{v}$ $v = v_A - v_C$ Adding the two equations we will get $v_A = \frac{2v}{3}$

Now, the system of $A$ and $B$ is also isolated we can conserve their momentum as well. Let at the instant the when the compression in the spring is $x$ their speeds are equal to $v'$ .

Conserving the momentum, $2m v_A = 3m v'$ $v' = \frac{4v}{9}$

As there is no dissipative force, we can conserve the mechanical energy of the system $AB$ after the collision, $\begin{gathered} \frac{1}{2}2mv_A^2 = \frac{1}{2}3mv{'^2} + \frac{1}{2}k{x^2} \\ 2m\frac{{4{v^2}}}{9} = 3m\frac{{16{v^2}}}{{81}} + k{x^2} \\ k = \left( {\frac{8}{9} - \frac{{16}}{{27}}} \right)\frac{{m{v^2}}}{{{x^2}}} = \frac{8}{{27}}\frac{{m{v^2}}}{{{x^2}}} \\ \end{gathered}$

Lot of collisions, lot of compressions!

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