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Determine the correct graph(s) that show the energy-momentum 4-vectors for a particle with the given worldline. Consider a completely inelastic collision between two identical masses A and B, where B is initially stationary. The final "rest mass" of the AB lump and the increase in rest mass of an object containing internal parts are also discussed.
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Which graph(s) could correctly show the energy-momentum 4-vectors for a particle with the worldline shown?Or choose (e) more than one could be correct
In a completely inelastic collision, the two objects stick together afterwards. Consider a completely inelastic collision between two identical masses A and B, where B is initially stationary.The initial velocity of A cannot exceed c, so the final velocity cannot exceed c/2.A] trueB] false
In a completely inelastic collision, the two objects stick together afterwards. Consider a completely inelastic collision between two identical masses A and B, where B is initially stationary.The initial velocity of A cannot exceed c, so the final velocity cannot exceed c/2.FALSE.
In a completely inelastic collision, the two objects stick together afterwards. Consider a completely inelastic collision between two identical masses A and B, where B is initially stationary.The final “rest mass” of the AB lump is:A] 2mB] smaller than 2mC] larger than 2m
Consider a completely inelastic collision between two identical masses A and B, where B is initially stationary.The final “rest mass” of the AB lump is larger than 2m. The “rest mass” of an object containing internal parts will increase if the internal parts move faster in the rest frame of the object. The rest mass will also increase if there is any stored internal energy.(Consider two masses connected by a spring… it’s clear from P conservation that the mass increases when the masses are moving. But when the spring is stretched or compressed, they aren’t moving, but the mass can’t be different from when they are moving.)
We will call m the “effective mass” or even sometimes just the mass. Px = m v cPt = m c2 = mc2 + K At low speeds, these expressions turn into Newtonian formulas for P and K.