Momentum Study W/ Springs and Half Balls

grabcad

The Conservation of Linear Momentum is a simple, yet effective method when approaching collision problems. Attached is a YouTube video which demonstrates the collision that takes place with this scenario. In order to understand the variables at play, we must divide the entire collision into smaller intervals:Interval 1: Before 1st CollisionBoth pieces of the first ball are moving with the same initial velocity, so the entire momentum is:p_1 = 2mv_0where p_1 represents the momentum at interval 1, m represents the mass of a half-ball, and v_0 represents the initial velocity of the half-balls.Interval 2: After the 1st CollisionShortly after the 1st collision, the second halves of each of the balls have no velocity. However, the first halves of the balls have some unknown velocities, which can be represented in the momentum equation as:p_2 = mv_1 + mv_2where p_2 is the total momentum at interval 2, v_1 is the velocity of first half-ball on the left, and v_2 is the velocity of the first half-ball on the right.Separately, we can also individually analyze the collision between the to colliding surfaces and model a conservation of momentum equation after it:mv_0 = mv_2From these three equations, we can solve for v_1 and v_2 in terms of v_0:v_1 = v_2 = v_0NOTE: We can also prove that the amount that the spring compresses is the same for both sets of balls by using the conservation of energy approach during interval 1 and 2:mv_0^2 = 1/2mv_1^2 + 1/2mv_2^2 -1/2kx_2^2 + 1/2kx_3^2There is a negative sign because the spring on the left ball is doing work in the negative x-direction.Using our data from previous equations, we see that:x_2 = x_3Interval 3: At the 2nd CollisionAt the 2nd collision, the first halves of the balls have no velocity, and the second halves have some unknown velocity, which can be modeled in this equation:p_3 = mv_3 + mv_4Combining this with the momentum equation from interval 1, we get that:v_0 = 1/2(v_3 + v_4)In order to simplify this, we also have to model the energy at this point in time. We can also assume that there's no spring displacement because there are relatively no oscillations after the 2nd collision.1/2mv_3^2 + 1/2mv_4^2Combining this with the momentum equation, our new equation simplifies down to:1/2(v_3 - v_4)^2 = 0, which implies v_3 = v_4Going back to the momentum equation:v_3 = v_0 = v_4Interval 4: After the 2nd CollisionAfter the 2nd collision, the only notable movement comes from the 2nd ball and its momentum can be modeled by: p_4 = 2mv_5Due to the conservation of momentum:p_4 = p_1, which implies 2mv_0 = 2mv_5, which implies v_5 = v_0So after evaluating the momentum and energy at several specific points of motion, I showed that the velocity stays constant throughout. These calculations can probably be calculated in a numerous amount of ways through other methods however. Physics is very fluid like that.NOTE: These calculations were all under the assumption that m, k, and v_0 were known values.