Download Momentum and its conservation and more Lecture notes Classical Mechanics in PDF only on Docsity!
Momentum and its conservation
Newtonian mechanics :
linear or translational (or simply) momentum is the product
mass × velocity = m ~v ≡ ~p
Momentum is more fundamental than velocity – Newton’s 2nd
law actually tells force is rate of change of momentum.
- Momentum gets transferred among bodies but not velocities.
- Momentum depends on reference frame as does velocity but
momentum is conserved in all frames.
- Momentum is conserved in classical & quantum mechanics,
special & general relativity, electrodynamics and QFT.
- Momentum conservation is manifestation of translational
symmetry (or homogeneity) of space.
Momentum : Newtonian mechanics
Single particle momentum ~p = m ~v.
Multi particle momentum ~p =
i mi^ ~vi^.
Defining M =
i mi^ =^
ρ dτ and center of mass coordinate as
~R =
∑^ i^ mi~ri i mi
ρ~r dτ ∫ ρ dτ
, ~vCM = ~R˙ ⇒ ~p = M ~vCM
Center of mass is the most popular reference frame to study
2-body collision, 2/many body decay (explosion perhaps) etc.
CM examples: K&K Pb 3.1 Density of a thin rod of length l varies
with distance x from one end as ρ = ρ 0 x^2 /l^2. Find position of CM?
x dx l
M =
∫ (^) l
0
ρ dτ = (ρ 0 /l^2 )
∫ (^) l
0
x^2 dx = ρ 0 l/ 3
Xcm =
M
∫ (^) l
0
ρ 0 x dx = 3 l 4
CM and mechanics: K&K Pb 3.4 A projectile explodes in two
pieces at the top of its parabolic path at a distance L from the
launch pad. The pieces fly apart horizontally. The mass of one of
the piece is 3 times the other. It is found that the lighter mass
returns to the launching pad. Where does the heavier piece land?
Momentum conservation says two pieces will fly away back to back.
CM continues in parabolic path as long as the pieces are in flight.
−L O L
ms ml
CM
CM
CM
The coordinate of CM is L when it reaches the ground, hence
Xcm = ms xs + ml xl ms + ml ⇒ L = −Lms + 3ms xl ms + 3ms = 3 xl − L 4
So xl = 5L/3 from O or xl = L + 5L/3 = 8L/3 from launch point.
Elastic & inelastic collision
Elastic collision : Momentum and kinetic energy are conserved.
Inelastic collision : Momentum is conserved, not kinetic energy.
Consider an elastic collision, when m 2 is initially at rest.
m1 m
v1i v1f v2f
before collision after
Mom conservation : m 1 v 1 i = m 1 v 1 f + m 2 v 2 f KE conservation : 12 m 1 v (^12) i = 12 m 1 v (^12) f + 12 m 2 v (^22) f Solving above two conservation equations, we get
v 1 i + v 1 f = v 2 f , v 1 f =
m 1 − m 2 m 1 + m 2
v 1 i , v 2 f =
2 m 1 m 1 + m 2
v 1 i
I (^) m 2 moves in direction of v 1 i , for m 1 it depends on sign of m 1 − m 2. I (^) If m 1 = m 2 , m 1 comes to rest and m 2 moves with v 1 i. I (^) If m 2 m 1 , m 1 bounces with v 1 f ≈ v 1 i , m 2 moves with v 2 f ≈ 0. I (^) If m 1 m 2 , m 1 continues as it is with v 1 f ≈ v 1 i and m 2 flies away with v 2 f ≈ 2 v 1 i.
Elastic collision problem : Double ball drop – a light ball is dropped
along with a heavy ball, the small ball rebounds with a high velocity.
Analysis by plain reasoning
v0 2v0 2v0^ 3v
v0 (^) v
v
v
Momentum conservation in Lab frame (since duration of collision
is very small, ignore gravity)
Mv 0 − mv 0 = MvM + mvm Mv 02 + mv 02 = Mv (^) M^2 + mv (^) m^2 ⇒ vm =
3 M − m M + m
v 0
M m gives vm → 3 v 0 and if the balls are dropped from a
height of h, then the small ball rebounds to a height of 9h!!
We will analyze the same problem later in CM frame.
Inelastic collision: No kinetic energy conservation ⇒
Needs a second equation → define coefficient of restitution
e =
speed after collision speed before collision
relative speed of separation after collision relative speed of approach before collision Relative velocity of approach and separation are defined along line of
impact, the common normal for colliding surfaces closest to each other.
m1 m2 m1 m
α1 α2^ β1 β
u1 (^) u2 v v Line of impact
Plane of contact Plane of contact
e =
v 2 cos β 2 − v 1 cos β 1 u 1 cos α 1 − u 2 cos α 2
Value of e ranges between elastic and perfectly inelastic limit
0 ≤ e ≤ 1
Example. Equal mass particles in a 2-dimensional elastic collision
(oblique) emerge at right angles. (Easier when one assumed to be static.)
Since masses are equal, conservation equations lead to
~u = ~v 1 + ~v 2 u^2 = v 12 + v 22
~u · ~u = (~v 1 + ~v 2 ) · (~v 1 + ~v 2 ) u^2 = v 12 + v 22 + 2~v 1 · ~v 2 ⇒ ~v 1 · ~v 2 = 0
Example. A ball strikes a wall with velocity 16 m/s at an angle 30o^ and rebounds. Coefficients of restitution is 0.5. Find the velocity of rebound.
16 m/s (^) v
n
(^30) θ
Use the equation for coefficient of restitution for v 1 n = v 1 sin θ
0 .5 (−16 sin 30o^ − 0) = 0 − v 1 sin θ ⇒ v 1 n = 4m/s Along the surface of contact, i.e. wall, v 1 t = v 1 cos θ v 1 cos θ = 16 cos 30o^ ⇒ v 1 t = 13. 86 m/s Hence,
v 1 =
v (^12) n + v (^12) t = 14. 4 m/s and θ = tan−^1 (v 1 n/v 1 t ) = 16. 1 o
Additional discussion on kinetic energy
Insight can be gained by reorganizing kinetic energy of 2-body
collision into 2 terms –
(i) due to net momentum of the two particles (ii) due to relative velocity of the two particles ~vrel = ~v 1 i − ~v 2 i
~p^2 net = (m 1 ~v 1 i + m 2 ~v 2 i )^2 = m^21 v (^12) i + m^22 v (^22) i + 2m 1 i m 2 i ~v 1 i · ~v 2 i ~v (^) rel^2 = (~v 1 i − ~v 2 i )^2 = v (^12) i + v (^22) i − 2 ~v 1 i · ~v 2 i
Knet =
m 1 v (^12) i +
m 2 v (^22) i
=
2(m 1 + m 2 )
~p net^2 +
m 1 m 2 2(m 1 + m 2 )
~v (^) rel^2
= Kcm + Krel
Kcm is kinetic energy due to motion of CM of 2-body system. This term
is conserved in any 2-body collision, because ~pnet is always conserved.
Additional discussion on CM frame
Recall center of mass coordinate and velocity and velocity ~v (^) i′ of mass mi in center of mass (CM) frame,
~Rcm =
∑^ i^ mi~ri i mi
, ~vcm =
∑^ i^ mi^ ~vi i mi
, ~v (^) i′ = ~vi − ~vcm
Hence, total momentum and its conservation as observed in CM frame,
~pcm =
i
mi ~v (^) i′ =
i
~p i′ = 0 ⇒
i
~p i′, initial =
i
~pi, final = 0
Specializing to 2-body elastic collision : ~p 1 ′i = −~p′ 2 i and ~p′ 1 f = −~p′ 2 f I (^) CM momenta or velocities are always back to back. I (^) Magnitudes of the initial and final CM momenta : p i′ = |~p′ 1 i | = |~p 2 ′i | and p′ f = |~p 1 ′f | = |~p′ 2 f |. I (^) Kinetic energy conservation ⇒ p′ i = p′ f. I (^) Relative speed of approach and separation are equal in all frames.
Momentum conservation & variable mass
Examples : Falling rain drops, rolling snow ball, rocket flight, loading and unloading freight cars, mass accretion etc. A simple example : Two freights of mass m 1 , m 2 moving on same track with speeds u 1 , u 2. When they catch up, they couple together and moves with speed v. Find v and any loss of kinetic energy. Momentum conservation gives v ,
m 1 u 1 + m 2 u 2 = (m 1 + m 2 )v ⇒ v =
m 1 u 1 + m 2 u 2 m 1 + m 2
Difference in kinetic energy before and after collision,
∆K.E. =
m 1 u^21 + m 2 u^22
(m 1 + m 2 )
m 1 u 1 + m 2 u 2 m 1 + m 2
m 1 m 2 (u 1 − u 2 )^2 2(m 1 + m 2 )
Just the same as two bodies in perfect inelastic collision. Where the energy goes?
Variable mass : general discussion
Moving on from discrete mass exchange to continuous process. Newton’s 2nd law in presence of net external force,
d~p dt
= ~F ext^ → ~p(t + δt) − ~p(t) =
∫ (^) t+δt
t
dτ F~ ext(τ )
Mass m(t) moving with velocity ~v (t) either add dm mass having velocity ~u(t) or lose with velocity ~c over time interval t and t + δt. Then while adding mass
~p(t) = m(t) ~v (t) + dm ~u(t) ~p(t + δt) = m(t) ~v (t + δt) + dm ~v (t + δt) = m(t)~v (t) + [m(t) ~v˙ (t) + ˙m(t)~v (t)] δt + O((δt)^2 )
and while losing mass,
~p(t) = m(t) ~v (t) + dm ~v (t) ~p(t + δt) = m(t) ~v (t + δt) + dm (~v (t + δt) − ~c) = m(t)~v (t) + [m(t) ~v˙ (t) + ˙m(t)(~v (t) − ~c)] δt + O((δt)^2 )
Equation of motion of a variable mass system, when mass is added,
m(t) ~v˙ (t) + ˙m(t) (~v (t) − ~u(t)) = ~F ext(t).
Equation of motion of a variable mass system, when mass is ejected,
m(t) ~v˙ (t) − m˙(t) ~c = F~ ext(t)
- Velocity ~u is w.r.t. ground observer while ~c w.r.t. rocket.
- Rate of mass loss m˙ is negative.
- The second equation is used for rocket motion,
Example. An empty freight car of mass M starts from rest under an applied force F. At the same time sand begins to run into the car at steady rate b from a hopper at rest along the track. What will be the speed v of the car when mass m of sand has been transferred?
Sand falls vertically, so imparts no momentum to the car. Initial momentum at t = 0 and final at t = tf are,
~p(0) = 0 ~p(tf ) = (M + m) v = (M + btf ) v
Therefore, using ~p(t + δt) − ~p(t) =
∫ (^) t+δt t ~F extdt, we solve for v along
the track (hence omitting the vector notation),
∫ (^) tf
0
dtF = p(tf ) − p(0) → Ftf = (M + btf ) v
⇒ v = Ftf M + btf
Example. A rope of mass M and length l lies on a frictionless table, with a short portion l 0 hanging out at the edge. Initially the rope is at rest.
Mass per unit length of the rope m = M/l. And x(0) = l 0 , x˙(0) = 0.
At instants t and t + δt, portion of length hanging is x and x + δx. Hence, the external force is due to gravity,
F (t) = (Mx/l) g and F (t + δt) = (M(x + δx)/l) g
Initial and final momenta of the rope are,
p(t) = Mv and p(t + δt) = M(v + δv )
Therefore, from p(t + δt) − p(t) = F (t + δt)(t + δt) − F (t)t, we get
Mδv ≈ (Mg /l)xδt → M v˙ == M x¨ = (Mg /l)x
Hence the general solution and using initial values,
x(t) = Ae
g /lt (^) + Be−
g /lt (^) → x(t) = l 0 cosh(
g /lt)
