1027

• Run Multiple Choice Generator at h:\shares\physics
• General College Physics:  Run Core Problems by giving the program qzz as your 'course'--enter this when asked for your 3-digit course number.   Do the Physics Quiz 1025.
• University Physics:  Run Multiple Choice Generator for 201 and choose 'Calculus of Uniformly Accelerated Motion'.
• If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution.

Text Assignments: 

• General College Physics 6th Edition:  Chapter 7 problems 1, 4, 12, 15, 17, 19, 20, 22, 27, 32, 50

• University Physics and General College Physics 5th Edition:  Problems as assigned in Assignments 19-22, along with reading corresponding text.

In a collision, if two objects have masses m1 and m2, what will be the ratio of their changes in velocity?

We know that impulses are equal and opposite.  So changes in momentum are equal and opposite.  Which means that

• m1 `dv1 = - m2 `dv2 so that

`dv2 / `dv1 = - m1 / m2.

If two objects have masses m1 and m2, initial velocities v1 and v2 and final velocities v1' and v2', then what equation expresses how these six variables are related?

We know from above that

• m1 `dv1 = - m2 `dv2

Since `dv1 = v1' - v1 and `dv2 = v2' - v2 so our relationship becomes

• m1 ( v1 ' - v1) = - m2 ( v2 ' - v2).

Expanding by the distributive law we get

• m1 v1' - m2 v1 = - m2 v2' + m2 v2

Rearranging so that the unprimed velocities are on one side and the primed velocities on the other we can easily get

• m1 v1 + m2 v2 = m1 v1' + m2 v2'

This can be read as 'total momentum before collision is equal to total momentum after collision'.

Was kinetic energy conserved in the collision between the steel ball (2.5 cm in diameter and density 7.5 grams / cm^2) and the marble (1/5 the mass of the ball), assuming that the steel ball had initial velocity 65 cm/s, final velocity 45 cm/s, and that the marble (initially stationary) has final velocity 100 cm/s?

The volume of the steel ball is 4/3 pi r^3, with r = 1.25 cm.  We get volume close to 8 cm^3.  With a mass density of 7.5 g / cm^3 we get ball mass 60 grams, approx. 

Assuming marble mass 1/5 of ball mass we get marble mass 1/5 * 60 grams = 12 grams.

KE of the ball before collision:

• .5 m1 v1^2 = .5 * 60 g * (.65 m/s)^2 = 12 g m^2 / s^2

KE of ball after collision:

• .5 m1 v1' ^2 = .5 * 60 g * (.45 m/s)^2 = 6 g m^2 / s^2

KE of marble after collision:

• .5 m2 v2' ^ 2 = .5 * 12 g * (1.00 m/s)^2 = 6 g m^2 / s^2.

Looks like KE is conserved.

With paper between ball and marble, we get initial ball velocity 68 cm/s, final ball velocity 55 cm/s, final marble velocity 76 cm/s.

• This gives us total intial KE of 13 g m^2 / s^2 and a total final KE of about 12.5 g m^2 / s^2.

Tootsie-roll-mediated collision:  Ball goes about 24 cm and marble about 26 cm, implying ball velocity 55 cm/s and marble velocity 62 cm/s after collision.  Is KE conserved?

• Looks like a total KE before is greater than total KE after.

If kinetic energy was not conserved then what might have happened to the part of the kinetic energy of the steel ball that was not transferred to the marble?

The collision causes elastic and inelastic deformations throughout the two objects.  These deformations cause thermal energy changes, and most of the 'lost' KE is dissipated as thermal energy.  Some also goes into sound.

If kinetic energy had been conserved, assuming that the steel ball has 5 times the mass of the marble, then what would have been the after-collision velocities?

We know that

• m1 v1 + m2 v2 = m1 v1' + m2 v2'

and if KE is conserved that

• .5 m1 v1^2 + .5 m2 v2^2 = .5 m1 v1'^2 + .5 m2 v2'^2.

Using algebra we can convert this system of equations into the system

• m1 v1 + m2 v2 = m1 v1' + m2 v2'

• v2 - v1 = - ( v2 ' - v1 ' ).

v2 - v1 is the relative velocity of the objects before collision.

v2` - v1` is the relative velocity of the objects after collision.

What happens if the objects 'stick together' after collision?

Then v2' = v1', so we can just use v ' to represent v2 ' and v1 ' both.

Conservation of momentum becomes

• m1 v1 + m2 v2 = m1 v ' + m2 v '

Factoring out v ' on the right-hand side and switching sides:

• (m1 + m2) v ' = m1 v1 + m2 v2 so that

• v ' = (m1 v1 + m2 v2) / (m1 + m2).

In the case where KE is conserved, we say that the collision is perfectly elastic.

If the objects stick together, the collision is perfectly inelastic. 

A perfectly inelastic collision results in the minimum possible after-collision KE.

Most real-world collisions fall somewhere between these two extremes.

How does the momentum change of the steel ball in the collision compare to that of the marble?