Name: ______Date: ______Day ______@ ______

Collisions: The Conservation of Momentum and Energy

When momentum and energy are conserved, the physical system is very easy to understand. However, it turns out that all systems in the real world are never conserved – due to frictional and radiation losses. These forces are called dissipative since the energy dissipates away and cannot be recovered. (Note: the total energy is conserved but since the lost energy leaves the system being considered, then energy is not conserved in that system.) When the collision is perfectly elastic, the coefficient of elasticity is identically 1.

Part I: Elastic Collisions

Table 1: Conservation of Momentum and Energy: p = 0; E = 0

Run

/ m1 ( ) / m2 ( ) /

v1( )

/ v1’ ( ) / v2’ ( ) / e / KEBef ( ) / KEAft ( ) / %KE
1
2

Sample Calculations:

e=- (v2’ – v1’)/(v2 – v1)=

KEBef=½ * m1 * v12=

KEAft=½ m1 v12 + ½ m2 v22=

Part II: Inelastic Collisions

Table 2a: Conservation of Momentum: p = 0

Run

/ m1 ( ) /

v1( )

/ m2 ( ) / m ( ) / v’Meas ( ) / v’Thy ( ) / v %
3
4

Sample Calculations:

v’Thy=m1* v1/m=______

v %=(v’Meas – v’Thy) *100/ v’Thy=

Table 2b: Non-conservation of Energy: E ≠ 0

Run

/ m1 ( ) / m ( ) /

v1( )

/ v’ ( ) / KEBef ( ) / KEAft ( ) / %KE
3
4

Sample Calculation:

KE % = 100 * ( KEAft – KEBef ) / KEBef = 100 * (- ) / = ______

Questions: (Answer all questions on this sheet. Do not include scrap paper but show your work.)

1)Show your calculations for KEBef and KEAft for Part II using the data from Run 4.

2)If a rubber ball is dropped from a height of 120 cm, and rebounds to a height of 80 cm, what is the value of ‘e’? (Hint: solve by cutting the problem into two parts: a drop and then a rebound.)

3) What is the value of ‘e’ for a perfectly inelastic collision. (You may use your data from Part II to help you answer this question. Also, show how you got your answer either by reason or arithmetic. You do not need an extra page for this question since the answer, and explanation, is very short.)