Question

A ball is kicked from a location 7, 0, −8 (on the ground) with initial velocity −11, 19, −5 m/s. The ball's speed is low enough that air resistance is negligible. (a) What is the velocity of the ball 0.4 seconds after being kicked? (Use the Momentum Principle!) v = Incorrect: Your answer is incorrect. m/s

Answer 1

Answer:

$\vec{v} = <-11,15.08,-5>$

Explanation:

given,

location of the ball ⟨7,0,−8⟩

initial velocity of the ball ⟨-11,19,−5⟩

time = 0.4 s

speed of the ball = ?

using Momentum Principle

change in momentum = Force x time

$m \vec{v} - m \vec{u}= \vec{F}\times \Delta t$

$\vec{v} =\vec{u} + \dfrac{\vec{F}}{m}\times \Delta t$

Net force acting in this case will be equal to force due to gravity because air resistance is negligible.

F_net = F_g = ⟨0 ,-9.8 m , 0⟩

now,

$\vec{v} = <-11,19,-5>+ \dfrac{<0 ,-9.8 m , 0>}{m}\times (0.4-0)$

$\vec{v} = <-11,19,-5>+ <0 ,-3.92 , 0>$

$\vec{v} = <-11,15.08,-5>$

hence, the velocity of the ball 0.4 s after being kicked is equal to $\vec{v} = <-11,15.08,-5>$

Answer 2

The velocity of the ball 0.4 seconds after being kicked is; v = (-11, 15.08, -5) m/s

Velocity Vector

We are given;

• Location of the ball = (7,0,−8)

• Initial velocity of the ball = (-11, 19, −5)

• Time = 0.4 s

From the Momentum Principle, we know that;

Change in momentum = Impulse

Thus;

m(v - u) = F * Δt

Divide through by m and make v the subject to get;

v = u + [(F/m)Δt]

In this question, due to the fact that air resistance is negligible, the net force acting will be equal to force due to gravity. Thus;

F_net = F_g = mg

Since acceleration due to gravity is 9.8 m/s², then;

F = (0, -9.8m, 0)

Thus;

v = (-11, 19, −5) + [((0, -9.8m, 0)/m) * (0.4 - 0)]

v = (-11, 19, −5) + (0, -3.92m, 0)

v = (-11, 15.08, -5) m/s

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