How High Will Each Ball Swing After Elastic Collision?

AI Thread Summary
The discussion revolves around a pendulum problem involving two elastic balls in a collision scenario. Ball A, with mass m, is released and collides with stationary ball B, which has twice the mass. The conservation of momentum and energy principles are applied to derive the velocities of both balls post-collision. Ball A's velocity after the collision is calculated to be 1/3 of its initial velocity in the opposite direction, while ball B's velocity is 2/3 of ball A's initial velocity in the positive direction. Participants seek verification of their calculations and approaches to ensure accuracy before an upcoming exam.
Mdhiggenz
Messages
324
Reaction score
1
Urgent! Pendulum problem elastic collision

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.
 
Physics news on Phys.org


I'm attempting the same problem right now.

I took a different approach after final the initial velocity.

Since one object is stationary, I used the equations:

vaf = (ma - mb) / (ma + mb) * vai

vbf = (2ma) / (ma + mb) * vai

plugging in that gave me

(m-2m) / (m+2m) * √2gh
= -m / 3m * √2gh = -1/3 * √2gh

and

2m / m+2m * √2gh = 2m/3m *√2gh = 2/3 * √2gh

so the velocity of ball A after the collision is 1/3 the initial velocity in the opposite direction and the velocity of ball B is 2/3 the initial velocity of ball A in the +x direction

Can someone verify if this is correct?
 


churchmeng said:
I'm attempting the same problem right now.

I took a different approach after final the initial velocity.

Since one object is stationary, I used the equations:

vaf = (ma - mb) / (ma + mb) * vai

vbf = (2ma) / (ma + mb) * vai

plugging in that gave me

(m-2m) / (m+2m) * √2gh
= -m / 3m * √2gh = -1/3 * √2gh

and

2m / m+2m * √2gh = 2m/3m *√2gh = 2/3 * √2gh

so the velocity of ball A after the collision is 1/3 the initial velocity in the opposite direction and the velocity of ball B is 2/3 the initial velocity of ball A in the +x direction

Can someone verify if this is correct?


That looks okay :approve:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top