Chapter 8 Momentum, Impulse and Collisions 8.1

Momentum and Impulse

In the previous two chapters we have reformulated the Newton’s second law in terms of conservation of energies (kinetic, potential, internal). We can also express it as ! ⃗ i = d⃗p (8.1) F dt i where ⃗p is a new physical quantity known as momentum. In this course we define it as ⃗p ≡ m⃗v. (8.2)

Note, however, that in context of the so-called Hamiltonian mechanics (which is the starting point in both quantum mechanics and statistical mechanics) it is defined as " # ∂E x, dx dt ⃗p = . (8.3) ) ∂( dx dt For example if

then

% $ %2 $ 1 d⃗x 1 d⃗x = m + kx2 E ⃗x, dt 2 dt 2 ⃗p =

" x# ∂E ⃗x, d⃗ dt x ∂( d⃗ ) dt

which is in agreement with (8.2).

96

=m

d⃗x dt

(8.4)

(8.5)

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The units of momentum are given by [Momentum] = [Mass] × [V elocity]

(8.6)

and for example in SI units 1 N · 1 s = 1 kg · m/s2 .

(8.7)

Alternatively one can write units of momentum as [Momentum] = [Mass] × [V elocity] = [Mass] × [Acceleration] × [T ime] = [F orce] × [T ime]

(8.8)

where the latter form suggests that when some force is applied to a system for some time then, it may effect the momentum. In fact if we define impulse (of the time-independent net force) as ! ⃗J = ⃗ i ∆t F (8.9) i

then from (8.1) we get ⃗J = ⃗pf − ⃗pi . When the net force is time dependent the impulse is defined as & tf ! ⃗J = ⃗ i dt F ti

(8.10)

(8.11)

i

and from (8.1) we get ⃗J =

&

tf

ti

! i

⃗ i dt = F

&

tf

ti

d⃗p dt = ⃗pf − ⃗pi dt

(8.12)

which is the same as equation (8.10). Note also that if the average force is defined as ' tf ( ⃗ i Fi dt ⃗ avg ≡ ti (8.13) F tf − ti

then the impulse is just similarly to (8.9).

⃗J = F ⃗ avg (tf − ti )

(8.14)

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Similarly to the energy conservation which is fundamentally due to timeshift symmetry of physics laws, the momentum conservation is due to spaceshift symmetry. For this reason the conservation of energy expresses changes caused by force in time ! ⃗J = ⃗ i dt = ⃗pf − ⃗pi F (8.15) i

and the conservation of momentum expresses the changes caused by forces in space ! ⃗ i · d⃗l = Kf − Ki . W = F (8.16) i

The fact that the two expressions look so much alike might be surprising at first but this is what lead people to eventually discover a more fundamental and unified conserved quantity known as energy-momentum tensor as well as other conserved quantities such as electric charge. In fact discovery of new symmetries and developing theories based on these symmetries is what physicists did for the most part of the 20th century. Example 8.1 . We can now go back to the example 6.5 where we considered a race of two iceboats on a frictionless frozen lake. The boats have ⃗ masses m and 2m, and the wind exerts the same constant horizontal force F on each boat. The boats start from rest and cross the finish like a distance s away. Which boat crosses the final line with greater momentum.

Although it is true that the final kinetic energy of both boats will be the

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same, K1 = K2 1 1 (m) v12 = (2m) v22 2 2 and the final velocities where not the same v1 √ = 2. v2

(8.17)

(8.18)

and thus momenta are related by v1 √ p1 = = 2. p2 v2

(8.19)

This is due to the fact that the same forces were acting for different periods of time. Using the impulse-momentum theorem we can conclude that F ∆t1 = mv1 F ∆t2 = mv2

(8.20)

and thus

∆t1 v1 √ = = 2. (8.21) ∆t2 v2 Example 8.2. You throw a ball with a mass of 0.40 kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. (a) Find the impulse of the net force on the ball during its collision with the wall. (b) If the ball is in contact with the wall for 0.010 s, find the average horizontal force that the wall exerts on the ball during impact.

The momentum-energy theorem implies ⃗J = ⃗pf − ⃗pi .

(8.22)

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Assuming that the x-axis is chosen in the direction of motion of the ball before collision we get ⃗J = m (⃗vf − ⃗vi ) ) * ) * = (0.40 kg) −30 m/sˆi − 20 m/sˆi = (−20 N·s) ˆi.

(8.23)

Now if the ball is in contact with the wall for ∆t = 0.010 s then ˆ ⃗ ⃗ avg = J = (−20 N·s) i = 2000 Nˆi F ∆t 0.010 s

8.2

(8.24)

Conservation of Momentum

When a given system is analyzed it often useful to distinguish two types of forces: • internal forces (forces exerted on each other by objects inside the system) • external forces (forces exerted on the system by objects outside of the system) When there are no external forces the systems is said to be closed or isolated. For isolated systems one can write down the impulse-momentum theorem for each object separately ⃗ A on B = d⃗pB F dt d⃗ ⃗ B on A = pA F dt

(8.25)

but because of the Newton’s third law ⃗ A on B = −F ⃗ B on A F

(8.26)

we get d⃗pA d⃗pB =− . dt dt If we now define the total momentum of all particles as ⃗ = ⃗pA + ⃗pB P

(8.27)

(8.28)

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then we get the law of conservation of total momentum ⃗ dP = 0. dt This is true for an arbitrary collection of particles, i.e. ⃗ = ⃗pA + ⃗pB + ⃗pC + ... = mA ⃗vA + mB ⃗vB + mC ⃗vC + ... P

(8.29)

(8.30)

given that there are no external (but only internal) forces acting on these particles. Note that since (8.29) is a vector equation it must be satisfied along all three components dPx = d(mA VAx +mB VdtBx +mC VCx +...) = 0 dt dPy d(mA VAy +mB VBy +mC VCy +...) = =0 dt dt dPz = d(mA VAz +mB VdtBz +mC VCz +...) = 0 (8.31) dt and for arbitrary time interval Pix = Pf x Piy = Pf y Piz = Pf z

⇒ ⇒ ⇒

mA VAix + mB VBix + mC VCix + ... = mA VAf x + mB VBf x + mC VCf x + ... mA VAiy + mB VBiy + mC VCiy + ... = mA VAf y + mB VBf y + mC VCf y + ... mA VAiz + mB VBiz + mC VCiz + ... = mA VAf z + mB VBf z + mC VCf z(8.32) + ...

Example 8.4. A marksman holds a rifle of mass mR = 3.00 kg loosely, so it can recoil freely. He fires a bullet of mass mB = 5.00 g horizontally with a velocity relative to the ground of vBx = 300 m/s. What is the recoil velocity vRx of the rifle? What are the final momentum and kinetic energy of the bullet and rifle?

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Conservation of momentum implies mB ⃗vBi + mR ⃗vRi = mB ⃗vBf + mR ⃗vRf 0 = mB vBx + mR vRx mB vRx = − vBx mR 5.00 × 10−3 kg (300 m/s) vRx = − 3.00 kg vRx = −0.500 m/s.

(8.33)

So the final momenta and kinetic energies are ⃗pR = (3.00 kg) (−0.500 m/s) ˆi = −1.50 kg · m/s ˆi " # ⃗pB = 5.00 × 10−3 kg (300 m/s) ˆi = 1.50 kg · m/s 1 KR = (3.00 kg) (−0.500 m/s)2 = 0.375 J 2 # 1" 5.00 × 10−3 kg (300 m/s)2 = 225 J. KB = 2

(8.34)

Example 8.6. Consider two batting robots on a frictionless surface. Robot A, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. It collides with robot B, which has mass 12 kg and is initially at rest. After the collision, robot A moves at 1.0 m/s in a direction that makes an angle α = 30◦ with its initial direction. What is the final velocity of robot B.

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Conservation of momentum implies mA ⃗vA1 = mA ⃗vA2 + mB ⃗vB2

(8.35)

or along x-axis mA vA1 = mA vA2 cos(α) + mB vB2x mA (vA1 − vA2 cos(α)) vB2x = mB (20 kg) vB2x = (2.0 m/s − 1.0 m/s cos(30◦ )) = 1.89 m/s (12 kg)

(8.36)

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and along y-axis 0 = −mA vA2 sin(α) + mB vB2y mA vB2y = − vA2 sin(α) mB (20 kg) (1.0 m/s) sin(30◦) = −0.83 m/s vB2y = − (12 kg) and so

8.3

⃗vB2 = 1.89 m/s ˆi − 0.83 m/s ˆj.

(8.37)

(8.38)

Inelastic Collisions

In this section we will discuss instantaneous events - we call collisions which suddenly change the kinetic energies of objects. • If forces between colliding objects are conservative, then the total kinetic energy right before and right after collision is the same and the the collision is called elastic. ∆K = 0

(8.39)

• If the forces are non-conservative, then the total kinetic energy is not conserved and the collision is called inelastic (or completely inelastic is objects stick together). ∆K ̸= 0 (8.40) The key point is that although the (kinetic) energy might not be conserved, the momentum is still conserved for both elastic and inelastic collisions: ⃗ =0 ∆P

(8.41)

Consider a completely inelastic two-bodies collision, i.e. ⃗vA2 = ⃗vB2 = ⃗v2

(8.42)

then conservation of momentum implies mA ⃗vA1 + mB ⃗vB1 = (mA + mB ) ⃗v2 or ⃗v2 =

mA ⃗vA1 + mB ⃗vB1 . (mA + mB )

(8.43)

(8.44)

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For example if object B was originally at rest, then v2 =

mA vA1 (mA + mB )

(8.45)

and then it is easy to show that after the (completely inelastic) collision the total kinetic energies are 1 2 mA vA1 2 $ %2 1 1 mA 1 m2A 1 2 2 mA v2 + mB v2 = (mA + mB ) vA1 = (8.46) v2 = 2 2 2 (mA + mB ) 2 (mA + mB ) A1

K1 = K2

and so

K2 mA =