free physics video tutorials for all

 

 

MECHANICS

 

Momentum & Impulse

 

Coefficient of Restitution

 

coeffficient of restitution

oblique collisions

 

 

 

The Coefficient of Restitution

 

The Coefficient of Restitution (e) is a variable number with no units, with limits from zero to one.

 

0 ≤ e ≤ 1

 

 

'e' is a consequence of Newton's Experimental Law of Impact, which describes how the speed of separation of two impacting bodies compares with their speed of approach.

note: the speeds are relative speeds

 

 

coefficient of restitution

 

 

If we consider the speed of individual masses before and after collision, we obtain another useful equation:

 

uA = initial speed of mass A

uB = initial speed of mass B

vA = final speed of mass A

vB = final speed of mass B

 

 

relative initial speed of mass A to mass B = uB - uA

 

relative final speed of mass A to mass B = vB - vA

 

 

coefficient of restitution equation

 

 

note: in this equation the absolute of uB - uA and vB - vA are used ( |absolute| so there is no net negative result )

 

 

back to top

 

 

Example

 

A 5 kg mass moving at 6 ms-1 makes a head-on collision with a 4 kg mass travelling at 3 ms-1 .


Assuming that there are no external forces acting on the system, what are the velocities of the two masses after impact?
(assume coefficient of restitution e = 0.5 )

 

 

uA = initial speed of 5 kg mass (mass A)= 6 ms-1

uB = initial speed of 4 kg mass (mass B)= 3 ms-1

mA = 5 kg       mB = 4 kg

vA = final speed of mass A        vB = final speed of mass B

 

 

momentum before the collision equals momentum after

 

hence,       

mA uA + mB uB = mA vA + mB vB

 

also

coefficient of restitution equation

 

 

substituting for e, mA , uA , mB , uB we obtain two simultaneous equations from the conservation of momentum,

 

5 x 0.6 + 4(-3) = 5 vA + 4 vB

 

3 - 12 = 5 vA + 4 vB

 

-9 = 5 vA + 4 vB

 

 

5 vA + 4 vB = -9        (i

 

 

from the coefficient of restitution expression,

 

restitution problem#01

 

0.5(uA - uB) = vB - vA

 

(0.5 x 6) - (0.5(-3)) = vB - vA

 

3 + 1.5 = vB - vA

 

4.5 = vB - vA

 

 

vB - vA = 4.5        (ii

 

 

multiplying (ii by 5 and adding

5 vA + 4 vB = -9

 

- 5 vA + 5 vB = 22.5

 

9 vB = 13.5

 

vB = 1.5 ms-1

from (ii

1.5 - vA = 4.5

 

vA = 1.5 - 4.5 = -3

 

vA = -3 ms-1

 

 

Ans. The velocities of the 5 kg and 4 kg masses are -3 ms-1and 1.5 ms-1, respectively.

 

 

 

back to top

 

 

 

Oblique Collisions

 

For two masses colliding along a line, Newton's Experimental law is true for component speeds. That is, the law is applied twice: to each pair of component speeds acting in a particular direction.

 

Example

 

A particle of mass m impacts a smooth wall at 4u ms-1 at an angle of 30 deg. to the vertical.

 

The particle rebounds with a speed ku at 90 deg. to the original direction and in the same plane as the impact trajectory.

 

What is:


i) the value of the constant 'k' ?
ii) the coefficient of restitution between the wall and the particle?
iii) the magnitude of the impulse of the wall on the particle

 

 

i) There is no momentum change parallel to the wall.

 

coefft. rest prob 01

 

 

 

ii) The coefficient of restitution'e' is the ratio of the speed of separation to the speed of approach:

 

problem part (ii

 

 

 

iii) The impulse is the change of momentum.

Since the vertical unit vectors are unchanged, the momentum change just concerns the horizontal vector components;

 

hence,

 

problem #1 part (iii

 

 

 

 

 

 

 

back to top

 

 

 

this week's promoted video

 

 from Physics Trek

 

 

creative commons license

All downloads are covered by a Creative Commons License.
These are free to download and to share with others provided credit is shown.
Files cannot be altered in any way.
Under no circumstances is content to be used for commercial gain.

 

 

 

 

©copyright a-levelphysicstutor.com 2016 - All Rights Reserved