A small toy cart equipped with a spring bumper rolls toward a wall with a speed of v . The cart rebounds from the wall, with the same speed v . The sketch below shows the initial velocity vector v⃗i , final velocity vector v⃗f , and the initial momentum vector p⃗i . Using the initial momentum vector as a basis, draw the change in momentum vector Δp⃗ for the cart.

The orientation and length of your vector will be graded. The location will not be graded.

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

Initial speed of cart = v_i = v

Final speed of cart = v_f = v

Unknown:

The change in momentum of cart = I = ?

Solution:

[tex]I = \Delta p[/tex]

[tex]I = p_f - p_i[/tex]

[tex]I = mv_f - mv_i[/tex]

[tex]I = m ( v_f - v_i )[/tex]

[tex]I = m ( -v - v )[/tex]

[tex]I = m ( -2v )[/tex]

[tex]I = -2mv[/tex]

[tex]I = -2p_i[/tex]

[tex]\texttt{ }[/tex]

From the results above, we can conclude that the change in momentum vector Δp is twice the initial momentum vector p_i but in opposite direction.

The vector Δp could be drawn as shown in the attachment.

[tex]\texttt{ }[/tex]

Learn more

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Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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