Yes. If you are able to construct P1 such that SHA1(P1) = IV1 and P2 s.t. SHA1(P2) = IV2, then<p><pre><code>    SHA1(P1 + padding(P1) + M1) = SHA1(P2 + padding(P2) + M2) = f020486f071bf11053547a86f4a7153b3c950f4b
</code></pre>
where padding(M) is the padding appended to message M as the first step of computing SHA1.<p>(edit: clarified padding)