# Explosion of 2 Carts on a Platform (Momentum)

• uSee2
In summary: In a closed system with no escape, momentum can only increase. The more massive the object, the less it changes velocity after each collision.
kuruman said:
For whatever it's worth, here is a plot of the positions of the two cart and platform as a function of time. The parameters of the masses are 1, 2 and 80 units (1 unit = 10 kg). The corresponding initial velocities were -4 m/s, +2 m/s and 0 m/s. When the two carts stick together their common velocity is -9.6×10-2 m/s while the platform's velocity is +3.6×10-3 m/s.

View attachment 319607

And here is part B. I took the mass of the platform to be equal to the mass of the Earth, 6×1023 units.

View attachment 319609
I don't understand the green lines. Why do both suddenly jump from 0 to 12 when the light cart hits its buffer? And what does the mass matter in case A, with the brakes on?

uSee2 said:
I'm not sure if I'm intepreting this right, but why didn't the platform move much when the red cart (by that I mean the red line) collided with its buffer? Was it just because it was so massive?
haruspex said:
I don't understand the green lines. Why do both suddenly jump from 0 to 12 when the light cart hits its buffer? And what does the mass matter in case A, with the brakes on?
I understand the green lines. They are the result of my misguided effort at multitasking: simultaneously watch football on television, celebrate the impending arrival of the new year and post here. After the first three seconds the data points for the platform shift from the position of the center to the position of the right end. I have edited the post to fix that blunder.

The first plot addresses the situation "Suppose the platform's wheels are free to roll, but the platform has much more mass than the two carts." I considered a platform mass of 80 times the mass of the lighter cart to be "much more." The second plot addresses the engaged brake situation in which case the mass of the platform is equivalent to the mass of the Earth. I added plot titles to distinguish the two.

Finally, I should note my appreciation at the clarity of thought resulting from sobriety the morning after.

erobz
haruspex said:
Using the approach in post #1, you don’t need to analyse each bounce that carefully. You just need to note that
- when the light mass hits its buffer the heavier is only half way
- the light mass bounces back with only slightly diminished speed, and the platform picks up even less
- the light mass will be nearly back to the start point when the heavy mass hits its buffer
- ergo, the masses reunite well to the right
But that OP asks for the position of the platform relative to ##x=0## when the small masses unite, not where the position of the small impacting masses are relative to the center of the cart when they reunite. Maybe I'm under thinking it, but that seems less obvious to me.

will the center of the platform be to the left, to the right, or at x = 0 when the carts collide again and stick?

Is it the case that since both small masses are on the RHS of ##x = 0 ## that the center of mass of platform must be on the LHS of ##x= 0##, such that the combined COM of all three masses remains at ##x= 0## for the entire duration of the collisions?

Its dawning on me that should be the case given no external forces are acting on the system.

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My inclination is to analyze the problem in detail. If m is either of the masses and ##m_p## is the mass of the much higher mass platform, the equations of conservation of momentum and conservation of energy for an elastic collision (multiplied by a factor of two) are: $$mv_0+m_pv_{p0}=mv_1+m_pv_{p1}\tag{1}$$and $$mv_0^2+m_pv_{p0}^2=mv_1^2+m_pv_{p1}^2\tag{2}$$Rearranging these slightly yields:$$m_1(v_1-v_0)=-m_p(v_{p1}-v_{p_0})\tag{3}$$and$$m_1(v_1^2-v_0^2)=-m_p(v_{p1}^2-v_{p_0}^2)\tag{4}$$If we divide Eqn. 3 by Eqn. 4, we obtain: $$v_1+v_0=v_{p1}+v_{p_0}\tag{5}$$Solving Eqns. 1 and 5 for the velocities after the collision yields: $$v_{1}=\frac{(m-m_p)}{(m+m_p)}v_0+\frac{2m_p}{m_p+m}v_{p0}\tag{6}$$ and$$v_{p1}=\frac{(m_p-m)}{(m_p+m)}v_{p0}+\frac{2m}{m_p+m}v_{0}\tag{7}$$Linearizing these equations with respect to (small) ##\epsilon=m/m_p## then gives: $$v_1=-(1-2\epsilon)v_0+2(1-\epsilon)v_{p0}\tag{8}$$and$$v_p=(1-2\epsilon)v_{p0}+2\epsilon v_0\tag{9}$$

The first collision occurs at t = 3 s between the left mass and the left bumper. Prior to this collision, the velocity of the left mass is -4 m/s and the velocity of the platform is zero. From the above equations, after this collision, the velocity of the platform is ##v_p=-8\epsilon\ m/s## is ##v_L=(4-8\epsilon_L)\ m/s##, where ##\epsilon_L=m_L/m_p##. The location of the left mass after this collision is at $$x_L=-12+(4-8\epsilon_L)(t-3)$$The location of the right bumper after this first collision is at $$x_{RB}=12-8\epsilon_L (t-3)$$The location of the right mass after this first collision is at $$x_R=6+2(t-3)$$So the collision between the right mass and the right bumper occurs when $$x_R=x_{RB}=6+2(t-3)=12-8\epsilon_L (t-3)$$at time $$t=3+\frac{3}{1+4\epsilon_L}=6-12\epsilon_L$$Prior to this collision with the right bumper, the velocity of the right mass is ##v_R=2\ m/s## and the velocity of the platform is ##v_p=-8\epsilon_L\ m/s##. After this collision with the right bumper, the velocity of the right mass is $$v_R=-2(1-2\epsilon_R)-16(1-\epsilon_R)\epsilon_L$$ and the velocity of the platform is $$v_p=-8\epsilon_L(1-2\epsilon_R)+4\epsilon_R$$Linearizing with respect to the ##\epsilon##s and taking into account that ##\epsilon_R=2\epsilon_L##, these relationships reduce to $$v_R=-2-8\epsilon_L$$ and $$v_p=0$$
According to this, the platform stops moving after the collision of the right mass with the right bumper. According to the previous equations, the right bumper us at $$x_R=12-24\epsilon_L$$ when this collision occurs. So the center of the platform is displaced to the left ##24\epsilon_L##.

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nasu and SammyS
erobz said:
since both small masses are on the RHS of x=0 that the center of mass of platform must be on the LHS of x=0, such that the combined COM of all three masses remains at x=0 for the entire duration of the collisions?
Yes. From post #1:
"After their collision, the 2 carts are to the right of x = 0. Center of mass originally was at x = 0, so the platform had to move to the left to balance it out to keep the center of mass in the same position. As such, the platform is to the left from x = 0."

erobz
To get a better idea how this all plays out in a "practical" case, I have used Eqns. 6 and 7 of post #39 to establish the details of what transpires for the specific case where the mass of the platform is 100 kg. Here are the results of the calculations:

The collision between the left mass and the left bumper occurs at 3 seconds at x = -12 m. After this collision, the velocities of the masses and the platform are $$v_L=3.619\ m/s$$$$v_R=2.0\ m/s$$and $$v_P=-0.381\ m/s$$The total momentum of the two masses and the platform remains zero.

The collision between the right mass and the right bumper occurs at 5.52 seconds at x = 11.04 m. After this collision, the velocities of the masses and the platform are $$v_L=3.619\ m/s$$$$v_R=-2.327\ m/s$$and $$v_P=0.0519\ m/s$$The total momentum of the two masses and the platform remains zero.

The final collision between the two masses occurs at 7.86 seconds at x = 5.59 m. The center of the platform at this time is at x = -0.84 m

nasu, SammyS and uSee2

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