What is the velocity of the ball before the collision?

[UMDstudent]

Homework Statement

A 47.0g ball is fired horizontally with initial speed (Vo) toward a 110g ball that is hanging motionless from a 1.00m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 110g ball swings out to a maximum angle theta = 52.0 degrees.

What was Vo?

Homework Equations

Conservation of Energy
Conservation of momentum
Vfx = ( m1 - m2 / m1 + m2 ) * Vix

The Attempt at a Solution

First off, this is a mastering physics question for my phys161 (Classical Mechanics) class at University of Maryland. I have spent a good 2 hours working through this problem, reading, re-reading the text and have burned through 4 of my 5 available attempts at a correct answer. I know that since this is a Head-On elastic Collision, the velocity of ball 1 will equal zero after the collision and be equivalent to the velocity of ball 2 while on the string.

Should i calculate the velocity of ball 2 (the ball attached to the string) because that would be Vi = Square Root (2 * gravity * Y0) where Y0 is = L(1-cos(theta))?

I need to conceptually understand this problem; although if i figure out the correct answer, that would be nice too :) .

Also, I am not new to these forums as a user; this is just the first time i have gotten so frustrated with a problem that i have brought it to the attention of the forum members :).

Very Respectfully,

Shane

 

[TagutoAza]

Since the mass of the balls are not equal, it would be incorrect to assume that the velocity of ball 1 after the collision is 0. Since ball 1 is less massive than ball 2, it would bounce backward. But conservation of both energy and momentum still applies. Perhaps this will help you move forward with the problem?

 

[UMDstudent]

Heres an idea: Use conservation of energy first, then use conservation of momentum to finish the problem? Using the conservation of energy, we know that Vi = Square root (2*g*final height). This seems incorrect because the height of each ball isn't a factor as their equal until after the collision occurs.


For conservation of momentum, couldn't we calculate the final velocities of the balls and then use that to calculate the initial velocity?

 

[TagutoAza]

The Sqrt[2g*Hf] is correct. The height is a factor after the collision, but this only applies to ball 2. And the Vi that you found would be the velocity of ball 2 immediately after the collision.

Here's another equation to add to your list, in a straight line elastic collision ONLY, the relative velocity of both masses will be:

Vf2-Vf1= -(Vi2-Vi1)

Where f is velocity immediately after the collision and i the velocity immediately before. We can think of it this way because it is essentially a straight line collision, only afterward one of the masses moves up.

Now it's pretty obvious that we want to find Vi1 since that's Vo. Vi2 would just be 0, Vf2 and Vf1 should be solvable now?

 

[UMDstudent]

Based off of that equation : V1i = V2f - V1f

V2f = Square Root (2 * g * y) = Square Root (2 * 9.8m/s) = 4.4 m/s

V2f = 4.4 m/s

V1f = ( 2m1 / m1 + m2 )*(4.4) = 2.63 m/s.

That should be correct :) ?

 

[TagutoAza]

For V2f you forgot to put the y inside the root, y should have been (1-cos 52). Therefore V2f should be Sqrt[2*9.8*(1-cos 52)]=2.74 m/s

I'm not sure if solving for V1f is even necessary, I apologize for my last misleading post. What I was trying to get at is now you have 2 equations,

V1i = V2f - V1f

and using conservation of momentum,

m1*V1i = m1*V1f + m2*V2f

Where the only variables are V1i (or Vo) and V1f, you can then do a simple substitution and find the desired variable.

Edit: The velocity of V1f in your calculation seems to be off though. I didn't calculate it but because m1 is a smaller mass it should have bounced backward after the collision, therefore V1f should be negative.

 

[UMDstudent]

TagutoAza, Thank you for the helpful posts and making me think! Heres my calculations, let me know what you think:

V1f = (m1*V2f - m2*V2f) / 2*m1 = -1.84

V1i = V2f - V1f = 2.74 + 1.84 = 4.58 m/s.

 

[TagutoAza]

No problem! Seems to be the same answers that I got.

On the sidenote, if you're unfamiliar with the relative velocity equation I gave you, you can always throw that equation out. Instead solve the simultaneous equations from the conservation of momentum and kinetic energy to get the same result. The relative velocity equation is so much cleaner to use. Just beware of the special circumstances that need to be met before you can use it (elastic/straight line collision)!