Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
