Change in velocity of an object as it enters water

Question

So if I drop a block which is denser than water, it will sink.

Now if I have 2 identical blocks. I drop them at different heights above the water. Block $A$ is dropped from $20\mathrm{m}$. Block B is dropped from $5\mathrm{m}$.

How do I calculate the respective velocity just after their impact with water? Logically, Block $A$ will have a higher velocity at $1\mathrm{m}$ depth compared to Block $B$ since Block $A$ has a higher initial velocity.

But this is disproved by some simple calculations which shows that both block will have the same velocity at $1\mathrm{m}$ depth since their velocity just after impact with water is $0\mathrm{m}/\mathrm{s}$.

Let mass of both blocks be $2\mathrm{kg}$
Let $g=10\mathrm{N}/\mathrm{kg}$
Let time of impact be $0.2\mathrm{s}$

Final Velocity $$A=rt(2gh)=rt(2\cdot10\mathrm{N}/\mathrm{kg}\cdot20\mathrm{m})=20\mathrm{m}/\mathrm{s}$$

Final velocity $$B=rt(2gh)=rt(2\cdot10\mathrm{N}/\mathrm{kg}\cdot5\mathrm{m})=10\mathrm{m}/\mathrm{s}$$

Force on water by

$$A = \frac P t=\frac{mv}{t}=\frac{2\mathrm{kg}\cdot20\mathrm{m}/\mathrm{s}}{0.2\mathrm{s}}=200\mathrm{N}$$

Decceleration due to impact

$$a=F/m=-200\mathrm{N}/2\mathrm{kg}=-100\mathrm{m}/\mathrm{s}^2$$

Final Velocity just after impact

$$v=u+at=20\mathrm{m}/\mathrm{s}+(-100\mathrm{m}/\mathrm{s}^2)(0.2\mathrm{s})=0\mathrm{m}/\mathrm{s}$$

If you do the same for block $B$, you will get the same results. Block $B$ will be at zero metre per second when it just enters the water.

But of course we know that is not true since experimentally, object dropped at a greater height will have a higher velocity in water at the same depth compared to the same object dropped from a lower height.

So what is wrong with the calculation or theory?

Answer 1

You said "Force on water by A = P/t =mv/t =(2kg × 20m/s)÷0.2s =200N". To get this formula you basically assumed,

1, The end speed is zero;

2, The impact force is constant during the impact time.

So there is no surprise that by doing circular calculation, you find out that the speed of the block after impact is zero.

Answer 2

When you drop a block into water, a few things happen:

1. there is a collision, in which momentum is transferred. The block loses momentum, the water gains some. We often treat this as an instantaneous effect in simple physics, such as colliding two blocks on wheels.
2. the block was travelling through air but is now travelling through water. Essentially both are fluids, and both resist its motion, the water resists it more. The way that air resistance works on a free-falling block is, if you travel at a low speed, air resistance is very low (less than the force of gravity) so the block speeds up, if you travel at a high speed air resistance is very much higher (more than the force of gravity) so the block slows down. So whatever speed the block was travelling at, it speeds up or slows down to a fixed "terminal velocity". Ignoring turbulence, in water the block will do the same, but the resistance is higher so the terminal velocity is much smaller. It will speed up or slow down to that speed.
3. the momentum/energy transferred to the water in the collision and afterwards manifests as energy of water waves, bodily movement of water, and extremely tiny levels of heat.
4. Because water is dense, it has quite a lot of resistance to sudden bulk impact. In simple terms the air moves aside when the block "hits" it, but water finds it much harder to move aside. For example "belly flopping" from a height the water will basically not move apart fast enough to stop you being killed, whereas the air above it parted easily to let you travel downwards. The importance of this is that the blocks will probably be slowed down a lot at impact, whatever height dropped from, if its more than a trivial height. The block dropped from higher will lose more speed than the lower block, but then again, it was moving faster in the first place and will still be moving faster after impacting the surface.
5. the block does not become momentarily stationary. It does not stop, it does not have 0 m/s velocity, any more than you stop and then start moving again if your bike hits a tree. It slows down and distorts, slightly, and the water begins to move and distorts its shape, more, as momentum and energy are transferred during the fraction of a second the collision takes to complete.

At the heights you are discussing, 5m or 20m, the blocks will not be travelling at their terminal velocity (in air) when they hit the water. So the block dropped from higher, will be travelling faster, because of its different starting point.

To determine exactly what happens immediately upon impact, you'd need to know more about how water behaves, and I'm not confident of suggesting mathematics for how a dense ideal fluid acts in collision. Its well beyond basic physics.

The bullets above suggest that the block dropped from higher will still be moving faster - but because its moving faster the water resistance slows it down faster. Remember that they will both slow down to a much slower terminal velocity in water. So by the time the blocks are even a short distance below the water surface, they are both probably slowed to a very similar speed, and this explains what you observe.

Update

I found some other questions that tackle the question you ask - better understanding of point 5: Try these:

• Why do steel Newton's balls work?
• After elastic collision, where position of center of stationary ball? Does ball stop instantly?
• How does a (near) elastic collision work at a microscopic level?
• Why do the surfaces in contact of two colliding balls deform?
• Is there such a thing as instantly stopping?
• Classical mechanics and the speed of a train-mosquito collision, when perfectly rigid bodies