Question

A 12.4-g marble is dropped from rest onto the floor 1.56 m below. If the marble bounces straight upward to a height of 0.614 m, what is the magnitude of the impulse delivered to the marble by the floor

Answer 1

The impulse is
`I= 0.1116\ kg \cdot m/s`

Step-by-step explanation:

Generally Impulse which the change in momentum is mathematically represented as

`I = m\Delta v`

Where m is the mass with a value 12.4g =
`(12.4)/(1000) = 12.4*10^(-3)kg`

`\Delta v` is the change in velocity which is mathematically represented as

`\Delta v = v_2 -v_1`

Where
`v_1` s the velocity of the marble drooping and
`v_2` is the velocity of the marble bouncing back

setting up in a coordinate y-axis would show that
`v_1` is moving in the negative y-direction so the value would be
`v_1 = -v_1` and
`v_2` is moving in the positive y-direction so the value would be
`v_2 = +v_2`

So the formula for
`\Delta v` would be

`\Delta v = v_2 -( -v_1)`

`\Delta v = v_1 +v_2`

General v i mathematically represented as

`v = √(2gh)`

Substituting this into the formula for
`\Delta v`

`\Delta v = √(2gh_2) + √(2gh_1)`

Now substituting values as given in the question

`\Delta v = √(2 * 9.8 *0.614 ) + √(2 * 9.8 *1.56 )`

`= 8.9986 m/s`

The impulse is

`I = 12.4*10^(-3) * 8.9986`

`I= 0.1116\ kg \cdot m/s`