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Mdhiggenz
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Urgent! Pendulum problem elastic collision
An elastic ball (A) with mass m is released from a horizontal position, connected via
a massless string to a rod. When the ball reaches the bottom, it collides with
another elastic ball (B) of twice the mass, aslo connected via a massless string to the rod.
How high can each ball swing, respectively? (The collision is total elastic.)
Assume the +x direction is to the right
a)What is the velocity of ball A after the collision? (5pt)
b)What is the velocity of ball B after the collision? (5pt)
c)How high will ball A swing relative to the bottom position? (5pt)
d)What is magnitude of the tension force (T) in the string connected to the ball A at the
moment right after the collision? (5pt)
A picture of the problem can be found at http://www2.fiu.edu/~leguo/Site/PHY2048_files/ExtraCredit2.pdf
M=Mtotal=3m
Va= initial velocity A
Vb= initial Velocity B
V1= final velocity A
V2= final velocity B
pi= initial momentum
pf= final momentum
ki = initial kinetic energy
kf= final kinetic energy
So what I first did was I wanted to find the initial velocity ( Va) before the collision.
To that I did: 1/2MVa^2=MgH
Va= √2gh
The collision is total elastic which means momentum and energy is conserved
Pi=pf
MaVa=Mav1+Mbv2
Im going to solve for v2:
MaVa-Mav1/Mb=v2
Plugging in known variables
m√2gh-mv1/2m=v2
m's cancel and you get
√2gh-v1/2=v2
Here is where I get confused my main problem lies with h. So I want to make that h cancel
since it is circular motion as the picture shows I try using uniform circular motion
g=Va^2/H
H= Va^2/g
I want to make sure if I am doing good so far. This is a problem that will be on my exam tomorrow so much help would be appreciated.
Homework Statement
An elastic ball (A) with mass m is released from a horizontal position, connected via
a massless string to a rod. When the ball reaches the bottom, it collides with
another elastic ball (B) of twice the mass, aslo connected via a massless string to the rod.
How high can each ball swing, respectively? (The collision is total elastic.)
Assume the +x direction is to the right
a)What is the velocity of ball A after the collision? (5pt)
b)What is the velocity of ball B after the collision? (5pt)
c)How high will ball A swing relative to the bottom position? (5pt)
d)What is magnitude of the tension force (T) in the string connected to the ball A at the
moment right after the collision? (5pt)
A picture of the problem can be found at http://www2.fiu.edu/~leguo/Site/PHY2048_files/ExtraCredit2.pdf
Homework Equations
The Attempt at a Solution
M=Mtotal=3m
Va= initial velocity A
Vb= initial Velocity B
V1= final velocity A
V2= final velocity B
pi= initial momentum
pf= final momentum
ki = initial kinetic energy
kf= final kinetic energy
So what I first did was I wanted to find the initial velocity ( Va) before the collision.
To that I did: 1/2MVa^2=MgH
Va= √2gh
The collision is total elastic which means momentum and energy is conserved
Pi=pf
MaVa=Mav1+Mbv2
Im going to solve for v2:
MaVa-Mav1/Mb=v2
Plugging in known variables
m√2gh-mv1/2m=v2
m's cancel and you get
√2gh-v1/2=v2
Here is where I get confused my main problem lies with h. So I want to make that h cancel
since it is circular motion as the picture shows I try using uniform circular motion
g=Va^2/H
H= Va^2/g
I want to make sure if I am doing good so far. This is a problem that will be on my exam tomorrow so much help would be appreciated.