isn't the energy (not the relative speed) involved in the collision higher than that in a batch hitting a static object?
Yes it is. It's double. Conservation of energy applies. You give some kinetic energy x to one object to make it move, and you give the same kinetic energy x to another object. When they collide head on, the collision energy is 2x. See the Wikipedia LHC article which says this:
"The protons will each have an energy of 7 TeV, giving a total collision energy of 14 TeV".
The relative speed is arguably double too, though IMHO it's better to refer to the closing speed. Again from the Wikipedia article:
"At this energy the protons have a Lorentz factor of about 7,500 and move at about 0.999999991 c, or about 3 metres per second slower than the speed of light".
One proton comes in this way → at 299,792,455 m/s, the other comes in this way ← at 299,792,455 m/s. Their closing speed is 599,584,910 m/s or just under 2c. Check out the closing speed section of the Wikipedia faster-than-light article. Note that the velocity addition formula:
$$s = {v+u \over 1+(vu/c^2)}$$
...tells you how fast you would measure another object approaching you, and this is always going to be less than c. If you were one of the protons, you might claim that the other proton was coming at you at 0.999999999999999959c, and you might call this the relative speed. But make no mistake, closing speeds can approach 2c. Just think about watching two spaceships setting out towards one another on a collision course from stars 4 light years apart. You would see them crash into one another after 2 years.