Here is the information that I have. Vinny is wearing a suit made of Velcro (the soft fuzzy side). Starting from rest, “velcro vinny” runs along level ground accelerating at 7.6m/s2 for 3 seconds. At exactly 3 seconds, he collides with a stationary crate full of ninjas (m= 209.10 kg). This crate is conveniently covered in the hard prickly side on Vinny’s suit. Vinny and the CFON stick together and slide along the same level ground (mk=0.5) coming to rest 35 meters from Vinny’s original starting point. What is Velcro Vinny’s mass? PLEASE HELP! I AM TERRIBLE AT PHYSICS. email: mykayliegirl@yahoo.com
I NEED TO FIND THE MASS OF AN OBJECT IN AN INELASTIC COLLISION.
Here is the information that I have. Vinny is wearing a suit made of Velcro (the soft fuzzy side). Starting from rest, “velcro vinny” runs along level ground accelerating at 7.6m/s2 for 3 seconds. At exactly 3 seconds, he collides with a stationary crate full of ninjas (m= 209.10 kg). This crate is conveniently covered in the hard prickly side on Vinny’s suit. Vinny and the CFON stick together and slide along the same level ground (mk=0.5) coming to rest 35 meters from Vinny’s original starting point. What is Velcro Vinny’s mass? PLEASE HELP! I AM TERRIBLE AT PHYSICS. email: mykayliegirl@yahoo.com
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October 27th, 2009 at 10:26 pm
m1v1 + m2v2 = (m1+m2) v
m1 = vinny’s mass
v1 = solve for velocity at t = 3 with a = 7.6
m2 = 209.10
v2 = 0
v = since it stops 35 meters away, balance an equation to solve for v, where your deacceleration is calculated by the frictional force 0.5
October 31st, 2009 at 6:19 am
Momentum is conserved
Find his speed
V = at = (7.6)(3) = 22.8 m/s
p = mV = 22.8m
The momentum after he sticks to the ninjas is P =(m+M)v
p= P
22.8m = (m + 208.1)v
m =(208.1v)/(22.8-v)
Where v = the speed of the combined masses just after they stick together.
KE=work done = uNd
Distance vinny accelerated = (1/2)at^2= .5 * 7.6 * 9 = 34.2 m
Distance combined masses slid = 35-34.2 = 0.8 m
v = sqrt(2ugd)= sqrt(2*.5*9.8*0.8)= 2.8 m/s
Go back and find m
m =(208.1v)/(22.8-v)
m = (208.1*2.8)/22.8-2.8 = 582.68/20
m =29.13400 Kg
November 3rd, 2009 at 6:41 am
Just remember: Physics (especially mechanics) is a totally empirical science, meaning it’s based off experiments, observations, etc.
In inelastic collisions, momentum is conserved but kinetic energy is NOT. You CAN’T solve this problem using any form of conservation of energy method because in inelastic collisions, energy is lost into vibrational motion, caused by the two inertias trying to conform together as one.
so, the momentum of the system before will be equal to the momentum of the system afterward. Below:
mv = mv
(mass of vinny)(V before collision) = (mass of vinny and box)(velocity of combined system after collision)
IMPORTANT
In order to find the mass of vinny, you need to find the velocity of before the collision, velocity of vinny and the box after collision and the combined mass.
Below are the steps to take to find these unknowns.
Hopefully, I’m not too vague or too thorough.
____
FINDING THE VELOCITY BEFORE COLLISION
so, firstly, the velocity of vinny right before collision can be easily found using the UAM formula: vf = vo + at. where
vf = final velocity (or in this case, the velocity right before collision)
vo = initial velocity (in this case, from rest, so zero)
a = acceleration (7.6 m/s/s)
t = time (3seconds)
Plugging in everything gives you:
Vf = 0 + 7.6(3) = 22.8 m/s
_____
FINDING THE VELOCITY AFTER COLLISION
Ok, so, we know at the coefficient of kinetic friction is 0.5, and we also know that it took 35 meters for it to stop.
We’re gonna focus on friction, and use that to find the rate of deceleration, and then use a UAM equation to find the speed of the combined mass right after the collision.
Ff = Fn(mk) = ma
Substitute Fn = (combined mass)g and mk = 0.5
Ff = (combined mass)(-9.8m/s/s)(0.5) = (combined mass)( a)
NOTE: combined masses cancel out!
you get:
-9.8(0.5)= a
a = - 4.9m/s/s (this makes sense b/c a negative acceleration means the object is slowing down)
Now, to find the velocity right after the collision
We have to use another UAM equation!
We know these already:
Vo = ?
Vf = 0
a = -4.9 m/s/s
x = 35 meters
Use the UAM equation: Vf^2 = Vo^2 + 2ax
0 = Vo^2 + 2(-4.9)(35)
Vo = 18.52 m/s
____________________
Alright! Now we’re ready!
Mass of vinny = M
Momentum before = momentum afterward
(M)(22.8) = (M+ 209.10)(18.52)
Solve for M.
22.8M = 18.52M + 3872.59
4.28M = 3872.59
M = 904.81 kg
the mass of vinny is 904.81 kg! Haha.. that’s like.. about 400 some pounds.. lol
Yay!!
OMGG i still remember physics!! haha.
November 5th, 2009 at 9:03 am
I can´complete the answer as I don´t understand the meaning of (mk=0.5) but there might be some help
1) v= at so after 3 s v (Vinny) = 22.8 m/s
By conservation of moment m*22.8 =(m+209.1)V
The collision is at d = 1/2*7.6*(3)^2= 34.2m
so the distance after collision is 0.8m
supposing that 0.5 is the friction factor ,the friction force is
0.5*g(m+209.1) and its work W = 0.5*g(m+209.1)*0.8 which must be equal to the kinetic energy after collision = 0.5(m+209.1)V^2
As V^2 = [(22.8^2*m^2/(m+209.1)^2]
0.5g(m+209.1)*0.8= =.5(m+209.1) * 22.8^2*m^2 /(m+209.1)^2
so g* 0.8(m+209.1)^2=22.8^2m^2
which gives you x= 29.27 kg