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The question is: if
So I think I can think of this as: $$F = \frac{dp}{dt} = m \cdot \frac{v_i - 0}{\Delta t} = \frac{p_i}{\Delta t}$$ Since both have same momentum, given same force and momentum, time will be the same? Is this right? Then how do I do the stopping distance one? |
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You're right about the stopping time, if you continuously apply a constant force, this will indeed be true. The stopping distance will probably not be the same, as the pingpong-ball is moving much faster initially (why?). Can you determine the velocity of the ball as a function of time? How will you use this velocity for determining the stopping distance? |
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Another way of looking at the distance is to look at the kinetic energies $\tfrac12 m v^2$, which will be greater for the less massive ping-pong ball if the momentum is the same. The change associated with reducing the kinetic energy to zero will be equal to the force times the stopping distance, so if the forces are the same then the distance will be greater for the ping-pong ball. |
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Let the initial momentum be $p_0$, corresponding to a kinetic energy $E_0 = \frac{p_0{}^2}{2m}$. After a time $\Delta t$, the momentum will decrease by $\Delta p = F\cdot\Delta t$, and after a travelling distance $\Delta s$, the energy will decrease by $\Delta E = F\cdot\Delta s$. Stopping the ball is characterized by $\Delta p = p_0$, yielding $$ \Delta t = \frac {p_0}F $$ independent of the ball's mass. It can also be characterized by $\Delta E = E_0$, yielding $$ \Delta s = \frac {E_0}F = \frac {p_0{}^2}{2mF} $$ in inverse proportion to the ball's mass. |
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