Question

One end of an ideal spring is fixed to a wall at origin O and the axis of the spring is parallel to x-axis. A block of mass $m = 1\;{\rm{kg}}$ is attached to the free end of the spring and it is performing SHM. Equation of position of the block in the coordinate system shown is $x = 10 + \sin 10t$, $t$ is in second and $x$ in cm. Another block of mass, \[M = 3\;{\rm{kg}}\] moving towards the origin with velocity $30\;{\rm{cm/s}}$ collides with the block performing SHM at $t = 0$ and gets stuck to it. Calculate:(i) new amplitude of oscillations (ii) new equation for position of the combined body (iii) loss of energy during collision. Neglect friction

Answer 1

Hint: According to the conservation of momentum, the momentum after collision will be the same as that of the total of the momentum possessed.Conservation of momentum is a major law of physics which states that the momentum of a system is constant if no external forces are acting on the system. It is embodied in Newton’s First Law or The Law of Inertia.

Complete step by step answer:
(i)
It is given in the question that the position coordinate of the block with mass,
$m = 1\;{\rm{kg}}$, is $x = 10 + \sin 10t$.
Now we know that the velocity is calculated as the derivative of position. So here, we can calculate the derivative of the position coordinate to find the velocity. Thus,
$v = \dfrac{{dx}}{{dt}}$
So we can substitute for $x$,
$v = \dfrac{{d\left( {10 + \sin 10t} \right)}}{{dt}}\\
\Rightarrow v = 10\cos t$
As the given condition is that the body is performing SHM at $t = 0$, substituting for $t$, the above equation takes the form,
$v = 10 \times \cos 0\\
\Rightarrow v = 10$
Thus we get the velocity as $10\;{\rm{cm/s}}$.

Now let us consider the other body with mass \[M = 3\;{\rm{kg}}\] moving towards the origin with velocity $30\;{\rm{cm/s}}$ collides with this body in SHM. And it get stuck to it. We know that after the collision both the bodies move toward the origin with the same speed. Now we can apply the conservation of momentum.As the momentum is calculated as $p = mv$, where$m$ is the mass and $v$ is the velocity, the momentum for both the cases can be calculated as below.Before collision, the total momentum is equal to the sum of the momentum possessed by the two bodies.For the body of mass, $m = 1\;{\rm{kg}}$, the velocity is $10\;{\rm{cm/s}}$ and hence the momentum is,
${P_1} = 1\;{\rm{kg}} \times 10\;{\rm{cm/s}}\\
\Rightarrow {P_1} = {\rm{ 10}}\;{\rm{kg}}\;{\rm{cm/s}}$

For the second body, with mass \[M = 3\;{\rm{kg}}\]and velocity $30\;{\rm{cm/s}}$, the momentum is
${P_2} = 3\;{\rm{kg}} \times 30\;{\rm{cm/s}}\\
\Rightarrow {\rm{ 90}}\;{\rm{kg}}\;{\rm{cm/s}}$
Let us suppose that after collision, the velocity of the bodies is $V$. Therefore the conservation of momentum in this case can be written as,
$\left( {m + M} \right)V = \left( {90 + 10} \right)\;{\rm{kg}}\;{\rm{cm/s}}\\
\Rightarrow \left( {1 + 3} \right)\;{\rm{kg}}\left( V \right) = 100\;{\rm{kg}}\;{\rm{cm/s}}\\
\Rightarrow V = \dfrac{{100\;{\rm{kg}}\;{\rm{cm/s}}}}{{\left( {1 + 3} \right)\;{\rm{kg}}}}\\
\Rightarrow V = 25\;{\rm{cm/s}}$

As the energy for the spring is calculated as, $\dfrac{1}{2}kA$ where $A$ is the amplitude and $k = m{\omega ^2}$.Now from the equation $x = 10 + \sin 10t$, which is of the form, $y = A\sin \omega t$, we have $\omega = 10$. And thus, the value of $k$ is, $k = 1 \times {10^2} = 100$. So we can now apply the conservation of energy. Thus we have,
$\dfrac{1}{2}\left( {m + M} \right){V^2} = \dfrac{1}{2}k{A^2}\\
\Rightarrow \dfrac{1}{2} \times 4 \times 25 = \dfrac{1}{2} \times 100 \times {A^2}\\
\Rightarrow {A^2} = 1\\
\therefore A = 1\;{\rm{cm}}$
Thus the new amplitude is $1\;{\rm{cm}}$.

(ii) After collision, the total mass of the body is $\left( {m + M} \right) = 1 + 3 = 4\;{\rm{kg}}$. And the new $\omega $ is,
$\omega ' = \sqrt {\dfrac{k}{{m + M}}} \\
\Rightarrow \omega ' = \sqrt {\dfrac{{100}}{4}} \\
\Rightarrow \omega ' = \sqrt {25} \\
\therefore \omega ' = 5\;{\rm{rad/s}}$
Therefore from the position coordinate we have the new position as,
$x = 10 - \sin 5t$
Here as the collision results in the inversion, minus sign is taken.

(iii) The loss of energy can be calculated as the difference between the initial energy and the final energy, thus we have,
$E' = \left( {\dfrac{1}{2} \times 1 \times {{\left( {.1} \right)}^2} + \dfrac{1}{2} \times 3 \times {{.3}^2}} \right) - \dfrac{1}{2} \times 4 \times {.25^2}\\
 \Rightarrow E' = 0.005 + 0.135 - 0.125\\
 \therefore E' = 0.015\;{\rm{J}}$

Note:When a position coordinate is given, the velocity can be calculated as the derivative of the point. When two bodies collide, they move after collision with a different velocity from their initial velocities.Basically, in the case of collision, the kinetic energy before the collision and after the collision remains the same and is not converted to any other form of energy.It can be either one-dimensional or two-dimensional. In the real world, perfectly elastic collision is not possible because there is bound to be some conversion of energy, however small.