Question

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly

Answer 1

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:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$, $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$, then the impulse applied by the stick to the park is approximately:

$I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]$

$I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]$

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.