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

Question

4 years ago

9

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.

Physics

1 answer:

anastassius [24] · Answer 1 · 2022-01-13T04:21:36+03:00

Using the initial momentum vector as a basis, the change in momentum vector Δp for the cart is drawn as shown in the attachment.

<h3>Further explanation</h3>

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.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

Initial speed of cart = v_i = v

Final speed of cart = v_f = v

<u>Unknown:</u>

The change in momentum of cart = I = ?

<u>Solution:</u>

$I = \Delta p$

$I = p_f - p_i$

$I = mv_f - mv_i$

$I = m ( v_f - v_i )$

$I = m ( -v - v )$

$I = m ( -2v )$

$I = -2mv$

$I = -2p_i$

$\texttt{ }$

<em>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.</em>

The vector <em>Δp could be drawn as shown </em><em>in the attachment.</em>

$\texttt{ }$

<h3>Learn more</h3>