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

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

Step-by-step 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[(m)/(s)\right]` and
`\vec {v_(2)} = 25\,\hat{i}\,\left[(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[(m)/(s) \right]`

`I = 7\,\hat{i}\,\left[(kg\cdot m)/(s) \right]`

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