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I am unsure as to why the conservation of kinetic energy can't be used to solve this problem:

An open-box is moving at $1\ \mathrm{m/s}$ on a frictionless surface. If it starts to rain, what happens to the motion of the open box?

If the box weighs $2.0\ \mathrm{kg}$, its initial momentum is: $2 \times 1 = 2$.

Now lets say that at a certain point after the rain starts, the mass of the water that went into the open-box is $3\ \mathrm{kg}$. So the final momentum is: $5 \times V$.

Now I calculated for the final velocity by conservation of momentum: $2 = 5V$, $V = 0.4\ \mathrm{m/s}$.

The initial kinetic energy of the system was: $$\begin{align} E_k &= \frac{1}{2}mv^2 \\ E_k &= \frac{1}{2}(2)(1)^2 \\ E_k &= 1\ \text{joule} \end{align}$$ The final kinetic energy of the system (using the final velocity calculated using the conservation of momentum) is:
$$\begin{align} E_k &= \frac{1}{2}mv^2 \\ E_k &= \frac{1}{2}(5)(0.4)^2 \\ E_k &= 0.4\ \text{joules} \end{align}$$ I am confused as to how the kinetic energy of the system decreased from $1\ \text{joule}$ to $0.4\ \text{joules}$. I thought that the kinetic energy of the system would still be conserved and the velocity would have to decrease since the mass becomes larger, but that is not the case.

If so I could solve it using the conservation of kinetic energy: $$\begin{align} E_k(\text{initial}) &= E_k(\text{final}) \\ 1\ \text{joule} &= \frac{1}{2}(5)(V)^2 \\ V &= 0.63\ \mathrm{m/s} \end{align}$$

I realize I am getting a different velocity from the conservation of momentum method, but I cannot understand why.

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The kinetic energy is not conserved because the rain sticks to the box. That's an inelastic collision.

Also, the momentum of the rain is not constant. First it's vertical, downward and after colliding with the box it is horizontal. The horizontal momentum of the system remains constant because the system is isolated horizontally from outside forces. Vertically, the normal force of the earth on the box/rain system removes vertical momentum from that system.

The speed of the rain also changes from whatever it is before the collision to whatever it is after the collision (you don't tell us a speed for the rain). So you haven't done a full kinetic energy accounting either.

The reason the box slows down is that the rain drops exert a horizontal force backwards on the box when the box makes the rain begin to move forward horizontally (Newton's 3rd Law).

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  • $\begingroup$ I understood everything except for the kinetic energy not being conserved. Could you please further explain how the inelastic collision between the rain and the box works and how it affects the kinetic energy? $\endgroup$ – Duck Nov 7 '17 at 0:18
  • $\begingroup$ An inelastic collision is one in which objects stick together. Considering that outside forces do not act during the collision, the momentum of the system remains constant, but the kinetic energy can change. One can show mathematically that the kinetic energy cannot remain constant if the objects stick together. I'll leave the mathematics for you to work through. It's a good exercise in a physics analysis. $\endgroup$ – Bill N Nov 7 '17 at 0:31
  • $\begingroup$ By the way, there is no general principle for the conservation of kinetic energy or constancy of kinetic energy. It does happen when there is a perfectly elastic collision, but that's a special situation. en.wikipedia.org/wiki/Inelastic_collision $\endgroup$ – Bill N Nov 7 '17 at 0:34
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An experiment confirming this is called the bullet block experiment, where a nail is shot into a stationary small block of wood. One nail is shot at the center of one block and another nail is shot off-center of another.

The experiment shows both blocks reach the same maximum height (momentum) after being shot, however the one shot off-center also has rotational energy. That extra kinetic energy is taken from the different in penetration of the nail in the block.

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