# 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.

# (INelastic collisions)

Remember that momentum is a vector. That means that motion to the right carries positive momentum while motion to the left carries negative momentum.

Ptot = p1 + p2

Ptot,i = p1i + p2i

p1f + p2f = Ptot,f

Momentum conservation means

Ptot,i = Ptot,f

Ptot,i = p1i + p2i = m1 v1i + m2 v2i

Remember, velocity and momentum are both vectors. For straight-line motion, that means a velocity or momentum toward the right is positive and a velocity or momentum toward the left is negative. (That should sound familiar!)

For an inelastic collision, the two objects stick together and move off with a common velocity. This means

Ptot,f = Mtot vf = (m1 + m2) vf

Now we can use momentum conservation as

Ptot,i = Ptot,f

m1 v1i + m2 v2i = (m1 + m2) vf

Often, we might want to know the final velocity vf if we have started with all the initial conditions. Then we can readily solve for vf:

vf = [ m1 v1i + m2 v2i ] / (m1 + m2)

Example:

Two blocks, with masses m1 = 1.5 kg and m2 = 2.5 kg approach each other with initial velocities v1i = 2.0 m/s and v2i = - 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 the total momentum initially, before the collision, and set that equal to the total momentum finally, after the collision. Remember, momentum is a vector so the direction of the motion -- and the sign of the velocity or the momentum -- is very important!

Ptot = p1 + p2

Ptot,i = p1i + p2i = m1 v1i + m2 v2i

Ptot,i = (1.5 kg) (2.0 m/s) + (2.5 kg) ( - 3.0 m/s)

Ptot,i = 3.0 kg m/s - 7.5 kg m/s

Ptot,i = - 4.5 kg m/s

Momentum conservation means

Ptot,i = Ptot,f

Ptot,f = Mtot vf = (4.0 kg) vf

Ptot,i = - 4.5 kg m/s = (4.0 kg) vf = Ptot,f

- 4.5 kg m/s = (4.0 kg) vf

(4.0 kg) vf = - 4.5 kg m/s

vf = [ - 4.5 kg m/s ] / 4.0 kg

vf = - 1.125m/s

The negative sign means that the system moves to the left.

Inelastic Collisions in Two Dimensions

Momentum is a vector. This means that when we talk of momentum conservation and write

Ptot,i = Ptot,f

we really mean

 Ptot,i,x = Ptot,f,x m1 v1ix + m2 v2ix = ( m1+ m2 ) vfx Ptot,i,y = Ptot,f,y m1 v1iy + m2 v2iy = ( m1+ m2 ) vfy

Example: Two cars collide at an intersection as sketched below. The collide in a totally inelasc 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 East while car number 2 has a mass of 1500 kg and moves with an initial velocity of 12 m/s to the North.

What is the initial total momentum of this two car system?

Remember, momentum is a vector, so when we write

Ptot,i = p1i + p2i

we really mean

Ptot,i,x = p1ix + p2ix

Ptot,i,x = m1 v1ix + m2 v2ix

 v1ix = 10 m/s v2ix = 0
Ptot,i,x = (1200 kg) (10 m/s) + 0

Ptot,i,x = 12 000 kg m/s

Ptot,i,y = p1iy + p2iy

Ptot,i,y = m1 v1iy + m2 v2iy

 v2ix = 0 v2iy = 12 m/s
Ptot,i,y = 0 + (1500 kg) (12 m/s)

Ptot,i,y = 18 000 kg m/s

Using momentum conservation, once we know the initial total momentum we also know the final total momentum.

 Ptot,i,x = 12 000 kg m/s Ptot,f,x = Ptot,i,x Ptot,f,x = 12 000 kg m/s Ptot,f,x = Mtot vf Ptot,f,x = (2 700 kg) vfx (2 700 kg) vfx = 12 000 kg m/s vfx = [ 12 000 kg m/s ] / 2 700 kg vfx = 4.44 m/s Ptot,i,y = 18 000 kg m/s Ptot,f,y = Ptot,i,y Ptot,f,y = 18 000 kg m/s Ptot,f,y = Mtot vfy Ptot,f,y = (2 700 kg) vfy (2 700 kg) vfy = 18 000 kg m/s vfy = [ 18 000 kg m/s ] / 2 700 kg vfy = 6.67 m/s

Now we know the components of the final velocity,

vfx = 4.44 m/s and vfy = 6.67 m/s
What is the magnitude and direction of the final velocity?
vf = SQRT [ vfx2 + vfy2 ]

vf = SQRT [ (4.4) 2 + (6.67) 2] m/s

vf = 8.0 m/s

tan = opp/adj = vfy / vfx

tan = 6.67 / 4.44

tan = 1.5

= 56.3o

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