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In a 1D elastic collision, it is well-known that the relative velocities of the two objects (before and after the collision) are reversed. When you say reversed do you mean that each object keeps their own velocity just with a change of sign? That would not always be the case for a 1D situation. If a resting block $v_{1,before}=0$ is hit by a moving ...


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Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass: $$ m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1 $$ $$ \frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2} $$ and ...


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Yes , the normal to the surface is the direction of reaction force. And the direction doesnt depend on the material of the object . But note that if friction is considered , direction of net reaction force changes


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It basically means that they just need to cover half the distance. So, you have the distance to be covered, initial velocity(40) and final velocity has to be zero. Finding deceleration won't be an issue.


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The symbol $s_{NN}$ is in OP's context of RHIC the Mandelstam $s$-variable in a Nucleus+Nucleus collision. The $s$-variable is also known as the square of the center-of-mass energy.


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In this cases, momentum is not conserved because of the action of gravity as an external force. When you have a pivoted rod, as in your problem, you can use basically two conservation laws: a) conservation of energy, if the collision is assumed as perfectly elastic; b) conservation of angular momentum about the pivot. As regards b), indeed, if we choose ...


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In Newtonian Mechanics, if a body of mass $\mathtt{m}$ is in free-fall, then gravitational force is responsible for acceleration & hence changing its momentum. Simple, right? The equation of motion is $$\mathtt{m}\cdot a = \mathbf{F_g} = \mathtt{m} \cdot g$$ where $a$ is the net acceleration of the body. Things become intricate when you consider a ...


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When an object falls and hits the ground - which forces are involved to change its momentum? Vectorial sum of all the forces acting on the object will cause the change in momentum of the object. When the object was in free-fall, its momentum was already changing due to gravity(assuming negligible amount of air resistance) and then it hit the ground. ...


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Model the ground as massless critically-damped vertical spring that the particle contacts at zero height. When the particle reaches zero height, it has some KE which is dissipated by the damping mechanism. When in contact with the spring, there are three forces acting on the particle, gravity downward and the damper and spring force upward. The net force ...



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