Can we use equation of motion for a mass that has changed?

Question

There is a mass A attached to a horizontal spring fixed at an end. An equal mass B comes along and hits the mass with velocity $v$. Can I use the equation of motion $v^2=u^2+2as$, to calculate the acceleration of the masses after the moving mass hits the one on the spring and the spring gets compressed?

I got the final velocity as $v/2$ by using equation of conservation of linear momentum of a system. I took initial velocity as $0$, since the mass A was at rest initially. I was wondering whether this formula would work because initially there was the mass A and now the mass is doubled? The u=0 was for mass A but now final velocity was for the system of two masses.

Answer 1

There are two equations of motion: You have one prior to collision and after collision. It's somewhat unclear what you are trying to do, but if you are trying to obtain the equations of motion after collision, you would want to follow the steps:

1. calculate the new momentum using ${p_{obj1}} + {p_{obj2}} = {p_{obj1\& 2}}$.

2. calculate the velocity from the momentum using $p = mv$ where $m$ is the total mass

3. calculate the kinetic energy using $T = 0.5mv^2$

4. you know the spring potential energy is $V = 0.5kx^2$. Assume all kinetic energy is is applied to the spring and converted to potential energy. This means that you have a situation before compression and after compression:

Before

$T_{before} = something$

$V_{before} = 0$ (assuming the spring is at equilibrium at the time of impact)

After

$T_{after} = 0$ (it is no longer moving)

$V_{after} = something$

Given that you now know $V_{after}$, you can simply use $V_{after}=0.5kx^2$ and solve for x being the distance from the spring equilibrium (i.e. the compression).

If you have expressions for both energies T and V, you can express fully the equations of motion, although I'm not sure that's what you are supposed to do in this assignment.

The above method is the energy approach, which you would usually use.

Kinematics (approach using the equations of motion):

Edit: No, you can't use $ma = kx$ and simply solve for x. You should consider both the acceleration and x as functions of time.

By stating $ma = kx$ you obtain the equations of motion

\begin{array}{l} \ddot x\left( t \right) = - \frac{k}{m}x\\ x\left( t \right) = A\sin \left( {t\sqrt {\frac{k}{m}} } \right) \end{array}

$x(t)$ can be found to be the solution where A is the oscillation amplitude (how compressed the spring becomes)

Edit2 (more help):

Obtain the derivative of the solution, and have it be the initial velocity

$$\dot x\left( 0 \right) = v_0$$

You should be able to obtain the result from this equation.

I should probably stress that you would usually use the energy or "work" approach for solving this kind of tasks. But it can be done using the equations of motion in this case because $x(t)$ is analytical.

Answer 2

In your example it is relatively to choose a system with constant mass and so Newton’s laws can be used.

You start off with a pure collision type problem involving two separate masses, one moving and one not and those two masses are the system. À collision between the masses occurs over a relatively short period of time so any external impulse (force) due to the spring on the system of two masses is negligible compared to the internal impulses (forces) the masses suffer during their collision.
It becomes a collision with linear momentum being conserved and the final velocity of the two masses being half the original velocity of the mass which was moving before the collision.

Once the velocity of the two masses has been found the next phase of the analysis can be done he system now being the 2 masses and the spring with the kinetic energy of the 2 masses plus the spring potential energy being constant.
The maximum compression of the spring occurs when the velocity of the two masses is zero.