1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time of collision of two lead spheres

  1. Aug 23, 2016 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Two uniform lead spheres each have mass 5000kg and radius 47cm. They are released from rest with their centres 1m apart and move under their mutual gravitational attraction. Show that they will collide in less than 425s.

    2. Relevant equations
    By Gauss' Law, force on a sphere exerted by the other is ##F = - m^2 G/R^2##.

    3. The attempt at a solution
    The acceleration of the spheres as they approach each other increases as an inverse power of R. To get an upper bound of the collision time, assume constant acceleration throughout and use standard kinematical equations. This gave me the solution. I'm interested in solving for the time more analytically by solving a DE.

    The spheres approach each other in a straight line so force on one from other is $$F = - \frac{m^2 G}{R^2} = m \frac{dv}{dt} \Rightarrow - \frac{m G}{R^2} = \frac{dv}{dt} = \frac{dv}{dR}v$$ Solving for v gives $$ \frac{1}{2} v^2 = mG \left(\frac{1}{R} - \frac{1}{R_0}\right)$$ where at ##R_0## the configuration is at rest. Then $$\frac{dR}{dt} = \sqrt{2mG} \sqrt{ \frac{R_0-R}{RR_0}}$$ so need to then solve $$\sqrt{R_0} \int_{R_0}^R \frac{\sqrt{R'}}{\sqrt{R_0-R'}} dR' = \sqrt{2mG} \int_0^T dt$$ where T is the collision time. The integral on lhs can be solved by ##R' = R_0 \sin^2 \theta## where ##R_0 = 0.06 m## and ##R=0.03m##. (The system will collide at the CoM) . For these values of R and R_0, and choosing the integration range to be ##[-\pi/2, -0.253], | \sin \theta| = - \sin \theta## and ##|\cos \theta| = \cos \theta##. This gives $$T \approx -1.12 \frac{1}{\sqrt{mG}}$$ while in the solution assuming constant acceleration it was $$T \approx 0.245 \frac{1}{\sqrt{mG}}$$ so the numbers are off in my analysis.

    Thanks for any pointers!
     
  2. jcsd
  3. Aug 23, 2016 #2
    i was wondering about limits of R as initially the centres are one meter apart?
    one can take the grav. pull keeping the mass at the centre as equivalent positions!
    no doubt the collision will take place as the spheres will touch each other i.e. when the centers are at 2 times the radius...
     
  4. Aug 23, 2016 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right, but you have not shown your steps after that, so we cannot see where you go wrong.
     
  5. Aug 23, 2016 #4

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Are you sure about the 1/2 on the left side of the equation where you solve for v? If you view this as essentially an energy conservation equation with the change in kinetic energy on the left and the change in potential energy on the right, the 1/2 assumes that only one sphere moves instead of both.
     
  6. Aug 23, 2016 #5
    What do you think is wrong with this equation?
     
  7. Aug 23, 2016 #6

    CAF123

    User Avatar
    Gold Member

    $$ \sqrt{R_0} \int_{R_0}^R \frac{\sqrt{R'}}{\sqrt{R_0-R'}}dR' \rightarrow \int_{\theta_1}^{\theta_2} \frac{\sqrt{R_0 \sin^2 \theta}}{\sqrt{\cos^2 \theta}} 2 R_0 \sin \theta \cos \theta d \theta.$$ When ##R'=R_0, \sin^2 \theta = 1## so that ## \theta = \pm \pi/2##. When R'=0.03, R_0=0.06, ##\sin^2 \theta = 1/2## so ##\theta \approx \pm 0.253##. Between ##\theta \in [-\pi/2, -0.253]##, we get ##\sqrt{\sin^2 \theta} = -\sin \theta## and ##\sqrt{\cos^2 \theta} = \cos \theta## Therefore have $$-2 R_0^{3/2} \int_{-\pi/2}^{-0.253} \sin^2 \theta d \theta \approx -0.0229.$$ This gives $$T \approx -0.016 \frac{1}{\sqrt{mG}}$$ With a negative in front on the rhs (as I think Chestermiller was hinting at, see below) this evaluates to about 28s.

    I think it implies that R grows with t rather than gets smaller so I need a -ve in front on the rhs.
     
  8. Aug 23, 2016 #7

    CAF123

    User Avatar
    Gold Member

    I think I am just stating that the kinetic energy gain of one of the spheres is through the change in its potential from ##R_0## to ##R##.
     
  9. Aug 23, 2016 #8
    Yes. That is correct. The negative root is the one to use.
     
  10. Aug 23, 2016 #9

    CAF123

    User Avatar
    Gold Member

    @Chestermiller, thanks :) is there any physics behind the positive root and do you agree with the answer I derived of 28s?
    At the instant of collision, the speeds of the spheres will be zero and then so as to conserve momentum the spheres will start going in opposite directions...so the positive root entails this part of the motion?
     
  11. Aug 23, 2016 #10
    The force balance, rather than the energy balance, should tell you which sign to use for dR/dt. I think your conclusion regarding the separation of the bodies following the collision is correct.
    EDIT: Actually, after further consideration , the negative sign still applies. But there is an initial condition that needs to be included in the equation.
     
    Last edited: Aug 24, 2016
  12. Aug 24, 2016 #11

    CAF123

    User Avatar
    Gold Member

    The force on sphere a) from sphere b) is the same force as that from a point mass at the centre of b). For this reason, I think my limits on the R' integration should be R_0=1 and R=0.94, which is the distance of the CoM's at the initial and collision times.

    Yes, disregard my previous post in reply to you -> R is the relative separation of the masses so dR/dt=2v, thanks.
     
    Last edited: Aug 24, 2016
  13. Aug 24, 2016 #12

    CAF123

    User Avatar
    Gold Member

    Yup I see that now too. My limits for the R' integration are ##R_o = 1## and ##R=0.94##. With the negative in place and using dR/dt=2v and proceeding through with the analysis I obtain T=421s. The maximum acceleration of a sphere is right before it collides, when R=0.94. Assuming this largest value of acceleration throughout the whole motion from release from rest to collision will provide a lower bound on the total time to collision, T. This comes out at about 398s. So I want to find T such that 398 < T < 425....My value of 421 satisfies this so I think my working is correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Time of collision of two lead spheres
  1. Dart strikes lead sphere (Replies: 10)

Loading...