# Collision in a free fall

• Fede Aguilera
The collision is elastic3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for ## \vec{v_a}=2\vec{v_b} ## since they are in opposite directions.Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if ## \lvert v_a \rvert = \lvert 2v_b \rvert ## after collision then ## \lvert v_b \rvert = \lvert 2v_a \rvert ## before collision. So we are left with the relatively simple task of solving:$$v - gt = 2gt #### Fede Aguilera ## Homework Statement A ball A is dropped from the top of a building of height h and simultaneously a ball B is thrown upwards and both balls collide. After the collision the ball A has the double the velocity of ball B Determine the fraction of the building where the balls collide. ## Homework Equations mA.VA+mB.VB=mA.2V'+mB.V' VA=-g.t VB=Vo-g.t ## The Attempt at a Solution I tried with conservation of momentum, conservation of energy, but I can't get rid of the masses, they are always there and can't get them out of the equations, so I can't finish the problem. Is there a special case when an object gets the double of the other's object velocity after a collision? I looked for it but I didn't find anything PS: The answer is a fraction (obviously) but there are only numbers in it, so that's the reason I put it here and not in the Advanced Physics Homework. Thanks in advance. I think you have to assume the masses are the same. You are already assuming work is conserved, which seems the greater assumption, but I think you need that too. haruspex said: I think you have to assume the masses are the same. You are already assuming work is conserved, which seems the greater assumption, but I think you need that too. Are you sure about that? Then the problem will state that "An equal ball B is thrown simultaneously" but I'll try that out and see where it takes me. Thanks. Fede Aguilera said: Are you sure about that? Then the problem will state that "An equal ball B is thrown simultaneously" but I'll try that out and see where it takes me. Thanks. I analysed the number of unknowns and number of equations, and it seemed to you needed to assume both conservation of work and equal masses. On that basis i obtained a solution, and it appears entirely consistent. Can you tell me which equations you used? I can't conclude the problem yet. haruspex said: I analysed the number of unknowns and number of equations, and it seemed to you needed to assume both conservation of work and equal masses. On that basis i obtained a solution, and it appears entirely consistent. How about initial speed of B? Will the result be independent of it, even assuming equal mass and elastic collision? nasu said: How about initial speed of B? Will the result be independent of it, even assuming equal mass and elastic collision? In this problem, no specific values emerge for anything - it's all a matter of ratios. There are specific ratios between all the speeds. If the speeds before collision are vA down and vB up, and the speeds after collision wA up and wB down, you are given wA = 2*wB. It follows from conservation laws and equal masses that 2*vA = vB. you will also find that the initial upward speed of B is 3*wB. • nasu and Fede Aguilera Are you saying that you determined the fraction of the building's height with just the given data? What is this fraction? I got it. Thank you haruspex! @Nasu: If you care the answer is 1/3 from the top or 2/3 from the bottom So there is only one initial speed that satisfies the conditions. Nice. Fede Aguilera said: I got it. Thank you haruspex! @Nasu: If you care the answer is 1/3 from the top or 2/3 from the bottom Hmm... I got 1/6 from the top. How do I know that a ball falling in free fall, which collides with another object in free fall, is experiencing 0 gravity? Lunippa said: How do I know that a ball falling in free fall, which collides with another object in free fall, is experiencing 0 gravity? Presumably because both are in free fall and therefore not accelerating relative to each other. I think there are a number of assumptions one must make with this problem: 1) Masses are equal 2) The collision is elastic 3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for ## \vec{v_a}=2\vec{v_b} ## since they are in opposite directions. Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if ## \lvert v_a \rvert = \lvert 2v_b \rvert ## after collision then ## \lvert v_b \rvert = \lvert 2v_a \rvert ## before collision. So we are left with the relatively simple task of solving:$$ v - gt = 2gt $$leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):$$ 2(v - gt) = gt $$neilparker62 said: I think there are a number of assumptions one must make with this problem: 1) Masses are equal neilparker62 said: I think there are a number of assumptions one must make with this problem: 1) Masses are equal 2) The collision is elastic 3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for ## \vec{v_a}=2\vec{v_b} ## since they are in opposite directions. Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if ## \lvert v_a \rvert = \lvert 2v_b \rvert ## after collision then ## \lvert v_b \rvert = \lvert 2v_a \rvert ## before collision. So we are left with the relatively simple task of solving:$$ v - gt = 2gt $$leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):$$ 2(v - gt) = gt $$2) The collision is elastic 3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for ## \vec{v_a}=2\vec{v_b} ## since they are in opposite directions. Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if ## \lvert v_a \rvert = \lvert 2v_b \rvert ## after collision then ## \lvert v_b \rvert = \lvert 2v_a \rvert ## before collision. So we are left with the relatively simple task of solving:$$ v - gt = 2gt $$leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):$$ 2(v - gt) = gt $$Thank you! But what if the mass of the ball is a lot smaller and the 2 objects collide falling in the same direction, towards earth? But what if the mass of the ball is a lot smaller and the 2 objects collide falling in the same direction, towards earth? Are you still referring to the situation in which one object is dropped and the other thrown or fired upwards? And the dropped object is the lighter one ? neilparker62 said: 3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for ## \vec{v_a}=2\vec{v_b} ## since they are in opposite directions. Whoah there. What guarantees that they are in opposite directions? jbriggs444 said: Whoah there. What guarantees that they are in opposite directions? Assuming elastic collision between two equal masses. Before collision:$$p_1=mv_1 p_2=-mv_2$$where the - sign indicates the two velocities are in opposite directions - this will be the case per OP's post. Collision impulse (elastic collision):$$Δp = 2μΔv=m(v_1+v_2)$$where reduced mass μ=m/2 and relative velocity Δv=v1+v2 Immediately after collision:$$p_1=mv_1-Δp=mv_1-m(v_1+v_2)=-mv_2p_2=-mv_2+Δp=-mv_2+m(v_1+v_2)=mv_1
The respective momenta simply exchange during collision and so if the velocities were opposite before collision, they remain so after collision. Since both objects are in free fall, I am assuming that g does not affect the normal collision dynamics at all.