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A small object is sliding along a level surface with negligible friction, and has mass m, and constant speed v_0, when it collides with a brick resting on the surface. If the object recoils backward at (v_0)/3, and is in contact with the brick for time delta t, what average force did the brick exert on the object? Assume the object is moving along x-direction.

A. F_net = 0
B. F_net= -((mv_0)/delta t)
C. F_net = -((4mv)/(3delta t))
D. F_net = -((mv_0)/ delta t)
E. F_net = -((3mv_0)/(2delta t))
F. F_net = -((2mv_0)/(3delta t))

Respuesta :

cjmejiab

To solve this problem we will apply the concepts related to the Impulse-Momentum theorem. On this theorem it is specified that the Momentum is defined as the product between the mass and the velocity of the object, and the Impulse as the product between the Force and time. Both concepts fully comparable to each other. Mathematically they can be described as,

[tex]\Delta p = m \Delta v[/tex]

And

[tex]\Delta P = F_{net} \Delta t[/tex]

Here,

[tex]m = mass\\\Delta v = \text{Change in velocity}\\F_{net} = \text{ Net Force}\\\Delta t = \text{Change in time}\\[/tex]

According to the value given we have that the change in velocity in the momentum is

[tex]\Delta p = m(-v_0 - \frac{v_0}{3})[/tex]

[tex]\Delta p = -4 m \frac{(v_0)}{3}[/tex]

Using this value to find the Net force we have that

[tex]F_{net} = \frac{\Delta P}{\Delta t}[/tex]

[tex]F_{net} =\frac{-4 m \frac{v_0}{3}}{\Delta t}[/tex]

[tex]F_{net}  = \frac{-4 mv_0}{3 \Delta t}[/tex]

Therefore the correct answer is C.

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