## Inelastic Collisions

While collisions in the real world happen in many different and various ways we will restrict our attention to two particular kinds of collisions.

Total momentum is always conserved.SPLAT!-BOING!

SPLAT!

Totally INELASTIC collisions are those in which the two objects collide and REMAIN TOGETHER and go off with a common final velocity.BOING!

Totally ELASTIC collisions are those in which the Kinetic Energy is also conserved.

SPLAT!

(INelastic collisions)

Remember that momentum is a vector.That means that motion to therightcarriespositivemomentum while motion to theleftcarriesnegativemomentum.P _{tot}= p_{1}+ p_{2}P

_{tot,i}= p_{1i}+ p_{2i}p

_{1f}+ p_{2f}= P_{tot,f}

Momentum conservationmeansP _{tot,i}= P_{tot,f}P

_{tot,i}= p_{1i}+ p_{2i}= m_{1}v_{1i}+ m_{2}v_{2i}Remember, velocity and momentum are both

vectors. For straight-line motion, that means a velocity or momentum toward therightispositiveand a velocity or momentum toward theleftisnegative. (Thatshouldsound familiar!)For an

inelastic collision, the two objects stick together and move off with a common velocity. This meansP _{tot,f}= M_{tot}v_{f}= (m_{1}+ m_{2}) v_{f}Now we can use

momentum conservationasP _{tot,i}= P_{tot,f}m

_{1}v_{1i}+ m_{2}v_{2i}= (m_{1}+ m_{2}) v_{f}Often, we might want to know the

final velocity vif we have started with all the initial conditions. Then we can readily solve for v_{f}_{f}:v _{f}= [ m_{1}v_{1i}+ m_{2}v_{2i}] / (m_{1}+ m_{2})

Example:

Two blocks, with masses m_{1}= 1.5 kg and m_{2}= 2.5 kg approach each other with initial velocities v_{1i}= 2.0 m/s and v_{2i}= - 3.0 m/s as shown in this diagram. The two blocks collide in a totally inelastic collision (That means they stick together after they collide). What is their common final velocity after the inelastic collision?We know that

momentum is conserved. So we will find thetotal momentuminitially, before the collision, and set that equal to thetotal momentumfinally, after the collision. Remember,momentum is a vectorso the direction of the motion -- and the sign of the velocity or the momentum -- is very important!P _{tot}= p_{1}+ p_{2}P

_{tot,i}= p_{1i}+ p_{2i}= m_{1}v_{1i}+ m_{2}v_{2i}P

_{tot,i}= (1.5 kg) (2.0 m/s) + (2.5 kg) ( - 3.0 m/s)P

_{tot,i}= 3.0 kg m/s - 7.5 kg m/sP

_{tot,i}= - 4.5 kg m/s

Momentum conservationmeansP _{tot,i}= P_{tot,f}P

_{tot,f}= M_{tot}v_{f}= (4.0 kg) v_{f}P

_{tot,i}= - 4.5 kg m/s = (4.0 kg) v_{f}= P_{tot,f}- 4.5 kg m/s = (4.0 kg) v

_{f}(4.0 kg) v

_{f}= - 4.5 kg m/sv

_{f}= [ - 4.5 kg m/s ] / 4.0 kgv

_{f}= - 1.125m/sThe

negative signmeans that the systemmoves to the left.| More Examples |

Inelastic Collisions in Two Dimensions

Momentum is a vector.This means that when we talk ofmomentum conservationand writeP_{tot,i}=P_{tot,f}we really mean

P _{tot,i,x}= P_{tot,f,x}m

_{1}v_{1ix}+ m_{2}v_{2ix}= ( m_{1}+ m_{2}) v_{fx}P _{tot,i,y}= P_{tot,f,y}m

_{1}v_{1iy}+ m_{2}v_{2iy}= ( m_{1}+ m_{2}) v_{fy}

Example:Two cars collide at an intersection as sketched below. The collide in a totallyinelasc collision. That is, their bumpers and sheet metal become locked together so they will have to be forceably separated after the accident.What is their common velocity -- speed and direction -- after the accident?Car number 1 has a mass of 1200 kg and moves with an initial velocity of 10 m/s

to the Eastwhile car number 2 has a mass of 1500 kg and moves with an initial velocity of 12 m/sto the North.What is the initial total momentum of this two car system?

Remember,

momentum is a vector, so when we writeP_{tot,i}=p_{1i}+ p_{2i}we really mean

P _{tot,i,x}= p_{1ix}+ p_{2ix}P

_{tot,i,x}= m_{1}v_{1ix}+ m_{2}v_{2ix}P

v _{1ix}= 10 m/sv _{2ix}= 0_{tot,i,x}= (1200 kg) (10 m/s) + 0P

_{tot,i,x}= 12 000 kg m/sP _{tot,i,y}= p_{1iy}+ p_{2iy}P

_{tot,i,y}= m_{1}v_{1iy}+ m_{2}v_{2iy}P

v _{2ix}= 0v _{2iy}= 12 m/s_{tot,i,y}= 0 + (1500 kg) (12 m/s)P

_{tot,i,y}= 18 000 kg m/sUsing

momentum conservation, once we know theinitialtotal momentum we also know thefinaltotal momentum.

P _{tot,i,x}= 12 000 kg m/sP

_{tot,f,x}= P_{tot,i,x}P

_{tot,f,x}= 12 000 kg m/sP

_{tot,f,x}= M_{tot}v_{f}P

_{tot,f,x}= (2 700 kg) v_{fx}(2 700 kg) v

_{fx}= 12 000 kg m/sv

_{fx}= [ 12 000 kg m/s ] / 2 700 kgv

_{fx}= 4.44 m/sP _{tot,i,y}= 18 000 kg m/sP

_{tot,f,y}= P_{tot,i,y}P

_{tot,f,y}= 18 000 kg m/sP

_{tot,f,y}= M_{tot}v_{fy}P

_{tot,f,y}= (2 700 kg) v_{fy}(2 700 kg) v

_{fy}= 18 000 kg m/sv

_{fy}= [ 18 000 kg m/s ] / 2 700 kgv

_{fy}= 6.67 m/sNow we know the

componentsof the final velocity,

What is themagnitudeanddirectionof the final velocity?v _{f}= SQRT [ v_{fx}^{2}+ v_{fy}^{2}]v

_{f}= SQRT [ (4.4)^{2}+ (6.67)^{2}] m/sv

_{f}= 8.0 m/stan =

^{opp}/_{adj}= v_{fy}/ v_{fx}tan = 6.67 / 4.44

tan = 1.5

= 56.3

^{o}After the accident, the two cars go off with a final velocity of 8.0 m/s at an angle of 56.3

^{o}as shown

Conservation of MomentumElastic CollisionsReturn to ToC, Momentum(c) 2005, Doug Davis; all rights reserved