What equation can I use?

Question

Which equation would most benefit me to determine the time for an air track glider to pass through a photogate. I will have the distance the glider will travel, the width of the glider, mass of the glider, also mass added to glider, and the accelerating weight. This was given to me as a extra credit challenge problem in my Physics class. So can ANYONE help me? I REALLY need the extra credit.

Answer 1

The air track glider is effective to demonstrate the equations of motion with almost no friction.

I am assuming the weight will be suspended on a pulley and the glider will accelerate down the track and the photo gate will detect the entrance and exit of the glider.

use
s(t)=s0+v0*t+.5*a*t^2
to describe the motion of the glider.
First, compute a.

Using fbd of the weight
m1*g-T=m1*a
where T is the tension in the wire.
fbd of the glider
m2*a=T
so
m1*g-m2*a=m1*a
a=m1*g/(m1+m2)

Sources of error here are the friction and mass of the pulley as it gains rotational kinetic energy, air resistance of the mass as it drops, and the combination of friction and air resistance of the glider as it moves down the track. In addition, the mass of the wire will retard acceleration since it adds mass that must gain kinetic energy, and add to the mass descending, which will increase the kinetic energy of the system greater than predicted.

Now that you know a, set s0 = 0 and s1 as the distance from the photogate to the starting point of the front of the glider.
What you want to do is calculate the velocity of the glider as it enters the photo gate
s1=.5*a*t1^2
where t1 is the time it will take for the front edge of the glider to reach the photo gate.
t1=sqrt(2*s1/a)
and now the velocity can be calculated
v1=a*t1
ok, now the width of the glider is w, so
w=v1*t+.5*a*t^2
where t= the time it takes the glider to traverse the gate
this is a quadratic and all variables are known except t. You must choose the correct root.

I checked my work by conceiving an approximate set-up so you can see the equations play out.
I assumed that m1=100 g, m2= 200 g, g=9.81, s1= 3 m, and w=15 cm
Using these conditions:
a=m1*g/(m1+m2)

a=3.27 m/s^2

t1=sqrt(2*s1/a)
t1=1.3546 s

v1=a*t1
v1=4.4294 m/s

w=v1*t+.5*a*t^2
or
-w+v1*t+0.5*a*t^2=0

the positive root is 0.033451 s

The reason that there is a negative root is that the equation of motion could also describe what would happen if the glider started on the other side of the photo gate and had a velocity that pulls the weight upward. Since the a is constant, under these conditions it would travel to the right under a constant decelleration and then travel according to the conditions above. Keep in mind that the equations of motion are linear, but the action of releasing a weight to pull the glider is non-linear.

On review of the question, you stated that you have the mass that will be added to the glider. How this gets added in the experiment will change the solution I provided.
I will assume that the teacher wants you to understande conservation of momentum. Let's say the glider moves a distance of sp and the additional mass is added to the glider in a collision in such a manner that there was no momentum of the mass in the direction of the motion of the glider. Under these conditions, find tp, which is the time the glider moves until it collides with the additional mass. Also find vp, which is the glider speed at the moment of collision. Since the collision is inelastic, energy will not be conserved, but momentum will be. So
m2*vp=(m2+mp)vc
where mp is the additional mass, and the vc is the velocity after collision.
vc=m2*vp/(m2+mp)

After collision, the a will change since the glider is now more massive. So after collision,
a2=m1*g/(m1+m2+mp)

and the time from the collision point to the photo gate is governed by the equation
s(t)=vp*t+.5*a2*t^2
if now s1 is the distance from the front of the glider at the moment the collision ends to the edge of the photo gate, then
s1=vp*t1+.5*a2*t1^2
The source of error here is the impulse time of the collision since the acceleration of the glider will not be constant during that time.

Solve for t1 using the positive root, and then find the new v1 and use
w=v1*t+.5*a2*t^2 to compute the transit time through the gate.

Best of luck

j

Answer 2

because you've x AND x² contained in the equation someplace, it really is a quadratic. you want to get it contained in the kind: ax² + bx + c = 0 So our equation is: x² - 7x + 38 = 5x + 3 We actually merely favor each little thing on the left side of the equation, so allow's subtract 5x on both side: x² - 12x + 38 = 3 Now we subtract 3 on both side: x² - 12x + 35 = 0 we've a quadratic expression in which a = a million, b = -12, and c = 35. There are 2 common thoughts of adjusting. a million) Factoring: (x - 7)(x - 5) = 0 x = 7 or 5 2) Quadratic formula x = (-b ±?(b² - 4ac)) / 2a = (12±?((-12)² + (-4)(a million)(35)) / (2*a million) =(12±?(one hundred forty four - 100 and forty))/2 =(12±?(4))/2 = (12+2)/2 or (12-2)/2 = 14/2 or 10/2 = 7 or 5