Why can't I use a try block around my super() call?

So, in Java, the first line of your constructor HAS to be a call to super... be it implicitly calling super(), or explicitly calling another constructor. What I want to know is, why can't I put a try block around that?

My specific case is that I have a mock class for a test. There is no default constructor, but I want one to make the tests simpler to read. I also want to wrap the exceptions thrown from the constructor into a RuntimeException.

So, what I want to do is effectively this:

public class MyClassMock extends MyClass {
     public MyClassMock() {
         try {
             super(0);
         } catch (Exception e) {
             throw new RuntimeException(e);
         }
     }
 
     // Mocked methods
 }
 

But Java complains that super isn't the first statement.

My workaround:

public class MyClassMock extends MyClass {
     public static MyClassMock construct() {
         try {
             return new MyClassMock();
         } catch (Exception e) {
             throw new RuntimeException(e);
         }
     }
 
     public MyClassMock() throws Exception {
         super(0);
     }
 
     // Mocked methods
 }
 

Is this the best workaround? Why doesn't Java let me do the former?

---

My best guess as to the "why" is that Java doesn't want to let me have a constructed object in a potentially inconsistent state... however, in doing a mock, I don't care about that. It seems I should be able to do the above... or at least I know that the above is safe for my case... or seems as though it should be anyways.

I am overriding any methods I use from the tested class, so there is no risk that I am using uninitialized variables.

---