a 15.0 g marble is dropped from rest onto the floor 1.44m below.If the marble bounces straight upward to a height of .640m, what is the magnitude of the impluse delivered to the marble by the...

a 15.0 g marble is dropped from rest onto the floor 1.44m below.

If the marble bounces straight upward to a height of .640m, what is the magnitude of the impluse delivered to the marble by the floor? (.133kg*m/s)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember the formula that helps you to evaluate the impulse such that:

`J= m*v`

Since you need to find the magnitude of impulse to the the marble, by the floor, you should calculate the change in momentum `Delta J`  such that:

`J_2 - J_1 = m_2*v_2 - m_1*v_1`

`m_1*v_1`  represents the impulse before the marble touch the ground

`m_2*v_2`  represents the impulse after the marble touch the ground

Evaluating `J_1`  yields:

`J_1 = m_1v_1`

Substituting `sqrt(2gh_1)`  for `v_1`  yields:

`J_1 = m*sqrt(2gh_1) => J_1 = 15*10^(-3)*sqrt(2*9.8*1.44) ` `J_1 = 0.015*sqrt(28.224) => J_1 = 0.079 Kg*m/s`

Since the direction of movement of marble is negative, hence `J_1 = -0.079 Kg*m/s`

Evaluating `J_2`  yields:

`J_2 = m*sqrt(2gh_2) => J_2 =15*10^(-3)*sqrt(2*9.8*0.64)`

`J_2 = 0.053 Kg*m/s`

You need to evaluate the difference `Delta J =J_2 - J_1`  such that:

`Delta J = J_2 - J_1 = 0.053 - (-0.079) = 0.132 Kg*m/s`

Hence, evaluating the magnitude of impulse delivered by the floor to the marble yields `Delta J = 0.132 Kg*m/s.`