In the field of consumer behavior, an impulse purchase or impulse buying is an unplanned decision by a consumer to buy a product or service, made just before a purchase. One who tends to make such purchases is referred to as an impulse purchaser, impulse buyer, or compulsive buyer. Research findings suggest that emotions, feelings, and attitudes play a decisive role in purchasing, triggered by seeing the product or upon exposure to a well crafted promotional message.
So I have: ##H(\omega)=(\exp(-i\omega)-\exp(i\omega))\exp(i\omega)##, I denote by ##Z(\omega)=\exp(i\omega)##, to get: ##H(\omega)=Z(-\omega)Z(\omega)-Z(\omega)^2##, now, I want to find ##h[n]##, I think it should be: ##h[n]=z[-n]*z[n]+z[n]*z[n]##.
But I am not sure how to calculate the...
TL;DR Summary: An impulse is given to the pendulum so that it moves in 3 dimensions. What equations apply throughout its motion?
A particle of mass ##m## is suspended from a string of length ##\ell##. The string is then deflected at an angle ## \theta ##, where the particle and string are in...
(bottom graph relates only to c)
(a)
i. The students can calculate the area under the graph to find the impulse exerted on the block. This is because the area under a force vs. time graph is the change in momentum or the impulse.
ii. Knowing that the graph is linear and begins at around 3 N...
Just going over some basic equations to help someone out and i have noticed that i get 2 different answers when using 2 different but very similar equations for impulse.
first eq. I=F*t
lets say that i have a small rocket motor that has a force of 5.95 newtons and a impulse of 2.5 n-s...
I have done the first two parts of this question.
✅(i) Find the clubhead's change of momentum. ( dp = p_f - p_i = (0.225*18 - 0.225*30) = -2.7 kg ms^-1 )
✅(ii) Find the impulse given to the golf ball. (F dt = dp = 2.7 Ns from part (i) with just a change in the units. Because of the...
For question 1.
I am stuck. I know that the equation involves time and possibly rate, should solve for distance. But not sure how to set it up with information given.
2. Ft= m 🔺️ v
F(3)= (100kg)(30m/s)
3 s= 3000 kg m/s
Same applied to question 3.
3. F(2)= (100kg)(-30m/s)
F(2) = -3000 kg m/s...
i tried to calculate the the kinetic energy by using the impulse law and got that the speed is 150Ns/65kg then I calculated the Energy and later the power. My answer is was 433Watts. How would you solve this question?
For this,
Can someone please tell me why they integrate the impulse over from ##t_i## to ##t_f##? Why not from ##j_i## to ##j_f##? It seems strange integrating impulse with respect to time.
Many thanks!
So i was able to solve the angular velocity part but i don't know how to find the velocity of centre of mass . For the first part i simply conserved momentum about COM because if i consider the particles as a part of the same system as rod the collision are internal forces . I am mainly...
When the upper block collides with the wall, the impulse to it will be:
$$ \int (-N+ f)dt = -2mu$$
Where N is the contact force from the wall.
Since N is really huge, the impulse from friction in the first equation above can be ignored, hence the equation become:
$$ \int -Ndt = -2mu$$...
I have tried this same approach three times and I got the same answer. I can't figure out what's wrong. Btw answer is 12mu/(3+cos2α)
And yes, sorry for my shitty handwriting. If you can't understand the reasoning behind any step then please let me know.
hello i would like to get some help with this problem.
At first it try to calculate the impulse by the area but i found it too difficult
Then i try to solve it by the forumla J= F(t-t0), but the problem is that i don't know what F is so i try to solve it like this
F10) = 8
F(33) = -13
so
EF =...
i would like to get some help and to understand why my answer is incorrect , here is how i did the first and second part.
about the first part i did it right but i don't understand what I am doing wrong in the second part i tried -k and i get -23.997 and i also try +k and i get 23.997 but they...
Summary:: I'd like to check my understanding of standard problems where a billiard ball resting on a plane is hit horizontally at some height above its center
So the situation is that a ball of mass ##m## and radius ##r## is at rest on a horizontal surface. There is friction between the ball...
Hi
I've tried solving this question but it seems that I flipped the direction of the impulse, what did I interpret wrong? the question didn't give any clue on their direction before so I couldn't infer the direction of the impulse. It also just gave me the magnitude without the direction. I...
a) Assuming it is inelastic as it is accelerating and therefore kinetic energy is not conserved? Intuitively this doesn't feel right...
b) change in momentum = p initial - p final = force x time
p initial = 0 as at rest
p final = 0.25 x v2
force x time = area under graph = 0.1
therefore 0.25 x...
So far for the word problem I have: A 100 g particle, traveling at 40 m/s, collides inelastically with another 100g particle traveling towards it at 30 m/s.
Now from the equation provided we need the question to ask us to find delta t, and that's simple enough but I'm not sure what that 1/2 is...
The misconception came up from the following problem: "A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds. Which cart (#1 or #2) has the greatest acceleration? "
I know the answer is the following (I looked it...
Sometimes there are forces which act for very small time known as impulsive forces. We cannot measure such large forces acting in a very short time but we can ##\Delta p##. ##\Delta p=F\Delta t##.This quantity is defined as Impulse.
Why do we keep introducing new quantities? When is a new...
a) From impulse-momentum theorem I have: ##J=mv## so ##v=\frac{J}{m}## and since the ball doesn't slip ##v=\Omega b## so ##\Omega=\frac{J}{mb}## and ##\dot{\theta}=\frac{v}{a+b}=\frac{\Omega b}{a+b}##.
b) I considered the angular impulse: ##-J(a+b)=I_0 \Omega_0 \Rightarrow...
I attempted to do mvf-mvi to find the impulse, but had trouble figuring out what to use as v (where does the angle of 3degrees come in?), and thought that there had to be more to the problem considering the other details I was given. I then attempted to maybe calculate the kinetic energy lost...
I'm trying to model the linear collision of a bat and a ball using the conservation of angular momentum. The ball is a point particle with at rest wrt the axis of rotation, and the bat is being treated as a rod of negligible radius. I have had to work through several problems involving a ball...
Answer is (a)
I thought it would be (b) due to conservation of momentum - so final momentum of the hockey stick is equal to the initial momentum of the ball. I assume this isn't correct because there are other external forces acting (air resistance?) Is that sound?
What we know:
The ball is dropped at the tip A with some speed ##v_0## and rebounds with speed ##v##. This collision produces an angular impulse, changing the angular momentum of the bar with the flywheels.
Solution inspired by an answer provided by @TSny in the similar question.
Angular...
We know that impulse is
$$\vec J = \vec F \Delta t = \Delta \vec p$$
Let ##l, m## be the length of single rod and its mass respectively.
Analyzing torques and forces on each rod separately we have:
Rod ##AC##:
$$F\Delta t +N_x\Delta t = mV_{ac,x} \space\space\text{ eq. }(1)$$
$$F\Delta t\cdot...
1. When an object attached to a fixed point with a string, is given a velocity and the string goes taut.
So it says in this book (Applied Mathematics 1 by L. Bostock and S. Chandler) that when the string goes taut, the component of the velocity of the particle becomes zero in the direction...
I wrote:
J = m(2/3)v0 - mv0 for part a but I'm not sure if that's correct or if I need something else?
I have no idea how to do part b. I wrote F * t = m(2/3)v0 - v0 but I honestly don't know what I'm doing.
Hello friends,
I have a problem with a exercise sheet. Given is the impulse response of a discret element. The task is to draw the block diagram. But I think that the solution in the sheet is wrong. Because based on the difference equation (Exercise.pdf) there should be 3 delay elements. I have...
Summary:: Griffiths problem 8.5
Problem 8.5 of Griffiths (in attachment)
I already solved part (a), and found the momentum in the fields to be $$\textbf{p}=Ad\mu_0 \sigma^2 v \hat{\textbf{y}}$$
In part (b), I am asked to find the total impulse imparted on the plates if the top plate starts...
The rubber hammer is more elastic so the time of contact will increase. Ft=m(v-u)
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less...
Here is my calculation:
F = ma
50N = 1050kg * a
a = 0.0476m/s²
S = ut + ½at ²
1000m = 0t + ½(0.0476)t²
t = 204.980s
y = 204.980s (time to travel 1000m)
since impulse = momentum,
F * t = mv
F * x = m * distance covered/y
50N * x = 1050kg * 1000m/204.980s
50N * x = 5122.450N⋅s
x = 102.440s...
In a volleyball match a player hits a 280g mass ball with a force given by . A force acts for 4 meters and the ball acquires a final velocity given by . (a) What is the momentum the player has applied to ball? (b) What is the final speed of the ball?
Thank you.
a)
F = J/t
F = 4000 N*s / 0.35 s
F = 11429 N
b) I was going to equate impulse to the change in momentum and solve for v' (final velocity). Then use v' to solve for ΔEk. set ΔEk = Fd and solve for d. (The question never mentioned an angle of inclination, so I thought it would be ok to use W =...
Hints given:
-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time t.
-Use a kinematic equation to relate the final velocity and time to the distance traveled.
-What is his initial velocity?
My...
Black powder has specific impulse of around 80s, while rocket candy has up to 130s of specific impulse. Does that mean I could replace the propellant in a BP cartridge with 80/130 of the weight in rocket candy and obtain the same performance in an idealized gun? (as in without considering...
As an analogue, if 5J of work is done on an object then the linear KE might increase by 2J and the angular by 3J, so the work is divided between the linear and rotational forms.
Now suppose there is a sphere sliding on a frictionless surface. If an impulse of magnitude 1Ns is applied to the...
Hi there,
Just asking a logistics question since I want to be sure I am approaching this problem correctly.
My professor showed me an example of a bullet being fired from a barrel, given its initial velocity was 0. The change in time was 0.1 seconds. The mass of the bullet is 0.02 kg. The...
Homework Statement: Joe Varsity kicks a football of mass 0.9 kg. As his foot makes contact with the ball, it exerts a force which gradually increases to a maximum value over 5 milliseconds, then falls immediately to zero, as shown in the graph above. The force is given by the equation...
After impact VCos(alpha) will be normal to the inclined, now i calculated for theta by using [ Tan(theta) = vCos(alpha)/vSin(alpha),. Tan(theta)= Cot(alpha) ]. I don't know how to solve further to get value of theta, according to book thta = alpha.
Summary: I want to crush a single layer of granules from a single material that are 400-600um in size using a flat, circular probe, which will generate a force v time or force v distance curve. The quantity I am trying to measure is the granule (tensile) strength of the material.
I believe...
As the title says, I am studying this topic for my control systems fundamentals course. I think I intuitively understand the meaning of the convolution integral that relates input, output and the impulse response, but I am failing to prove it graphically.
For example, the intuitive explanation...
Now, the net vertical impulse on the wedge should be zero. It's quite obvious from the figure that the ground will also exert an impulse of ##J cos 30°## on the wedge. But they've given the answer as ##J sin 30°##.
They're wrong, right?
The fig. 1.1(a) is a mass m attached to a spring that is fixed to a wall. I don't understand what does "a sudden momentum impulse" means. Is it an external force o what?
I imagined that the new equation of motion would be
md^2x/dt^2+dp1/dt-kx=0
md^2x/dt^2+mdv1/dt-kx=0
is this the equation i...