Impulse and average force of ball

[polymerase]

Homework Statement

A ball of mass m moving with velocity v_{i} strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is \theta_{i} as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is \Deltat, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.

http://session.masteringphysics.com/problemAsset/1010992/25/MLM_e2.jpg

Homework Equations

What is the magnitude F of the average force exerted on the ball by the wall?

How do you even start this question...? i don't know what its talking about. Any comments would be great

 

[learningphysics]

What is the definition of impulse?

or rather...

the average force = mass*average acceleration

so what is the average acceleration...

 

[polymerase]

What is the definition of impulse?

or rather...

the average force = mass*average acceleration

so what is the average acceleration...

ok well impulse is the change in linear momentum so, F\Deltat = m\Delta\stackrel{\rightarrow}{v} but in this case it isn't linear...and I am not sure what you mean by average acceleration. but the thing is i was told by my teacher, the answer does not depend on the variable \Delta\stackrel{\rightarrow}{p} or mv_{i}

 

[learningphysics]

Hmmm... I'm not sure how the answer can't depend on mvi...

Momentum is linear... think of the momentum along the x-axis (perpendicular to the wall), and the momentum along the y-axis (parallel to the wall).

F_x*(\Delta t) = mv_{xfinal} - mv_{xinitial}

F_y*(\Delta t) = mv_{yfinal} - mv_{yinitial}, but Fy is just 0 according to the question... so

0 = mv_{yfinal} - mv_{yinitial}

You are also given that the collision is perfectly elastic... therefore the final kinetic energy = initial kinetic energy.

so try to use these 3 equations (impulse in the x-direciton, impulse in the y-direction, conservation of kinetic energy)

to solve for Fx.

 

[polymerase]

Hmmm... I'm not sure how the answer can't depend on mvi...

Momentum is linear... think of the momentum along the x-axis (perpendicular to the wall), and the momentum along the y-axis (parallel to the wall).

F_x*(\Delta t) = mv_{xfinal} - mv_{xinitial}

F_y*(\Delta t) = mv_{yfinal} - mv_{yinitial}, but Fy is just 0 according to the question... so

0 = mv_{yfinal} - mv_{yinitial}

You are also given that the collision is perfectly elastic... therefore the final kinetic energy = initial kinetic energy.

so try to use these 3 equations (impulse in the x-direciton, impulse in the y-direction, conservation of kinetic energy)

to solve for Fx.

How come you need the three equations? why can't you just rearrange the Fx to get you

F_{x}= \frac{mv_{xfinal} - mv_{xinitial}}{(\Delta t)}

note: my mistake, the answer does depend on mv_{xi} but it does not depend on mv_{xf}

so in light of this information, i guess i have to ask you how these 3 equations actually relate to one another.

 

[learningphysics]

To get rid of the extra variables... if you just use that equation, you have vf, vinitial, thetafinal, thetainitial...

you should be able to write the final answer just in terms of vinitial and thetainitial...

 

[polymerase]

To get rid of the extra variables... if you just use that equation, you have vf, vinitial, thetafinal, thetainitial...

you should be able to write the final answer just in terms of vinitial and thetainitial...

so is the final answer just F = mv_{ix}/t

 

[learningphysics]

so is the final answer just F = mv_{ix}/t

not quite... how did you get that?

 

[polymerase]

not quite... how did you get that?

i don't know...i don't know how to set the three equations equals to one another...

 

[learningphysics]

i don't know...i don't know how to set the three equations equals to one another...

we'll start with this one:

F_x = \frac{mv_{xf} - mv_{xi}}{(\Delta t)}

write vxf in terms vf and thetaf... write vxi in terms of vi and thetai... vf is the magnitude of the final velocity... vi is the magnitude of the initial velocity. vxf = -vfsin(thetaf) (because it is going towards the left)

F_x = \frac{-mv_{f}sin(\theta_{f}) - mv_{i}sin(\theta_{i})}{(\Delta t)} (1)

KEfinal = KEinitial

(1/2)mvf^2 = (1/2)mvi^2

gives

vf = vi.

next equation:

0 = mv_{yf} - mv_{yi}

so

0 = v_{yf} - v_{yi}

v_{yf} = v_{yi}

v_fcos(\theta_f) = v_icos(\theta_i)

using vf = vi from the kinetic energy conservation equation, we get:

v_icos(\theta_i) = v_icos(\theta_i)

cos(\theta_f) = cos(\theta_i)

hence \theta_f = \theta_i

So the point of all this was to show that vf = vi and \theta_f = \theta_i

what do you get when you plug these 2 into (1)

 

[polymerase]

we'll start with this one:

F_x = \frac{mv_{xf} - mv_{xi}}{(\Delta t)}

write vxf in terms vf and thetaf... write vxi in terms of vi and thetai... vf is the magnitude of the final velocity... vi is the magnitude of the initial velocity. vxf = -vfsin(thetaf) (because it is going towards the left)

F_x = \frac{-mv_{f}sin(\theta_{f}) - mv_{i}sin(\theta_{i})}{(\Delta t)} (1)

KEfinal = KEinitial

(1/2)mvf^2 = (1/2)mvi^2

gives

vf = vi.

next equation:

0 = mv_{yf} - mv_{yi}

so

0 = v_{yf} - v_{yi}

v_{yf} = v_{yi}

v_fcos(\theta_f) = v_icos(\theta_i)

using vf = vi from the kinetic energy conservation equation, we get:

v_icos(\theta_i) = v_icos(\theta_i)

cos(\theta_f) = cos(\theta_i)

hence \theta_f = \theta_i

So the point of all this was to show that vf = vi and \theta_f = \theta_i

what do you get when you plug these 2 into (1)

So...final answer is F = \frac{-2msin\theta_{i}v_{i}}{(\Delta t)} ?

 

[learningphysics]

Yes, but since they only want the magnitude you'd leave off the minus in the final answer.

 

[polymerase]

Yes, but since they only want the magnitude you'd leave off the minus in the final answer.

thanks you thank you thank you