The problem is as follows:
The figure from below shows a person whose mass is $60\,kg$ is standing over a chunk of ice whose mass is $125\,kg$ which was initially at rest. The person runs across the ice chunk with a speed of $3.7\,\frac{m}{s}$ with respect of the ice chunk. What will be the speed of the ice chunk with respect of the ice rink?.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&0.6\,\frac{m}{s}\\ 2.&3.6\,\frac{m}{s}\\ 3.&24\,\frac{m}{s}\\ 4.&1.2\,\frac{m}{s}\\ 5.&24\,\frac{m}{s}\\ \end{array}$
I'm stuck with this problem as I'm not sure how to proceed with the appropiate calculation for the speed of the ice chunk. Should I consider that there is a conservation of momentum?. How exactly should I put the equation for the momentum?. Can someone help me with this?. I would really like to show something but in this problem I'm stuck at the very beginning.
I believe that for this problem I have to consider the momentum at the beginning is the same at the end but how can I put these into math?.
Conservation of linear momentum.
Initial momentum is zero.
Let the velocity of the ice chunk relative to the ground be $-v$ (it has to go backwards).
The velocity of the person is $3.7$ m/s relative to the ice chunk, but $3.7-v$ m/s relative to the ground. That is very important.
Write down an expression for the total momentum of person and ice chunk. Set it equal to zero. Solve to find $v$.
In response to a comment... If you want to assume that the velocity of the ice chunk is $v$ m/s forwards, then you would have to make the person's velocity be $3.7+v$ m/s. Say total momentum is zero and solve. You will discover that $v$ is a negative value (going in the opposite direction to what you thought).