Tuesday

December 6, 2016
Posted by **Jared** on Wednesday, November 29, 2006 at 12:02am.

There are two balls hanging vertically from a horizontal plane. Ball A is pulled back and is set up to strike ball B.

Given for ball A -

mass = 1.5 kg.

Height = .3 m (measured from a horizontal plane from Ball B)

initial speed = 5.00 m/s (as it leaves your hand)

Given for Ball B -

mass = 4.60 kg

speed = 0 m/s (implied)

Question a) is -

Using the principle of conservation of mechanical energy, find the speed of Ball A just before impact.

Question b) is -

Assuming an elastic collision, find the velocities (magnitude and direction) of both balls just after the collision.

Thanks!!

a) This part is easy. Set the initial potential energy of ball A, M g h, equal to its kinetic energy just before it hits Ball B

M g H = (1/2) M V^2

H = 0.3 m and you know what g is. M cancels out, solve for V. M is the mass of ball A in this case.

b) You need to write equations of conservation of both momentum and kinetic energy. There will be two unknowns, the velocities of the two balls after collision. You may get a negative velocity for ball A after collision if it bounces backwards, since ball A is more massive.

The algebra is a bit messy for this one, but give it a try. Two equations are enough to solve for two unknowns.

Let V be the initial velocity of ball A that you get in part (a). Let va be the final velocity of ball A and vb be the fnal velocity of ball B. Let ma and mb be the two masses.

ma V = ma va + mb vb (momentum)

ma V^2 = ma va^2 + ma vb^2 (energy)

va = V - (mb/ma) vb

Make the substitution for va into the energy equation and solve for vb

You can save some work in b) by working in the center of mass frame. In the center of mass frame the total momentum is zero. Conservation of energy then implies that the velocities get reversed after the collision.

To see this, note that yo can write the kinetic energy as:

pa^2/(2ma) + pb^2/(2mb)

where pa and pb are the momenta of particles a and b, repectively.

But in the center of mass frame pb = - pa, so the energy is just:

pa^2[1/(2ma)+ 1/(2mb)]

After the collision the energy must be the same. But after the collision the energy is given by the same expression where you now have to replace pa by the momentum of ball A after the collision.

This means that the energy can only stay the same if pa^2 stays the same. this means that pa changes sign and that pb also changes sign (the other possiblity wherethey don't change sign corresponds to the balls not colliding wit each other).

So, all you have to do is to change the signs of the velocities in the center of mass frame.

Suppose you are in a frame that is moving at velocity V'. Then in your frame the total momentum is:

P = ma (V-V') - mb V'.

P is zero if you choose your velocity V' as:

ma (V-V') - mb V' = 0 -->

V' = ma V/(ma + mb)

The velocity of ball A before collision in the center of mass frame is thus:

va' = V - V'

and ball B has a velocity of:

vb' = -V'

Therefore after the collision the velocities (indicated by double primes) are:

va'' = -va' = V' - V

vb'' = V'

To transform back to the original frame you add back V':

va = va'' + V' = 2V' - V

vb = vb'' + V' = 2V'

va and vb are the velocities after the collision in the original frame.