1. Momentum and Collisions Motion under the action of time-varying force F(t) Ball hits a wall Newton’s second law Definition of (linear) momentum: Δt=t 2 -t 1 Effect of force on the object’s motion is determined by impulse (Impulse-Momentum Theorem) Momentum of many-particle system

2. The Principle of Conservation of Momentum The total momentum of an isolated system remains constant (is conserved). Proof: Two-particle collision

3. Example: Recoil of a rifle Solution: momentum conservation yields P fx = P 0x = 0 → m B v Bx + m R v Rx = 0 → v Rx = - m B v Bx / m R v Rx = - ( 0.005 kg / 3 kg ) 300 m/s = - 0.5 m/s K B = (1/2)m B v Bx 2 = (1/2)m B (m R v Rx / m B ) 2 = K R (m R /m B ) >> K R

4. Elastic and Inelastic Collisions Elastic collision: total kinetic energy = const (interaction via only conservative forces) Inelastic collision: total kinetic energy ≠ const (nonconservative forces are involved in the interaction: heating, irreversible deformation, explosion, propulsion, etc.) Example of completely inelastic collision

5. Example of the inelastic collision: Ice Skaters Solution: m 1 v f1 + m 2 v f2 = 0 → v f2 = - m 1 v f1 / m 2 v f2 = - 2.5 m/s ( 54 kg / 88 kg ) = - 1.5 m/s Note: K f > K 0 = 0 (Inelastic collision !!!) Important fact: Conservation of the momentum P does not depend on the conservation of the kinetic energy K !

6. Exam Example 18: The Ballistic Pendulum (example 8.8, problem 8.43) A block, with mass M = 1 kg, is suspended by a massless wire of length L=1m and, after completely inelastic collision with a bullet with mass m = 5 g, swings up to a maximum height y = 10 cm. Find: (a) velocity v of the block with the bullet immediately after impact; (b) tension force T immediately after impact; (c) initial velocity v x of the bullet. Solution: V top =0 (a) Conservation of mechanical energy K+U=const (c) Momentum conservation for the collision (b) Newton’s second law yields y L

7. Two-body Elastic Collision

8. The Gravitational Slingshot Effect General property of all elastic collisions: In an elastic collision the relative velocity of two bodies has the same magnitude before and after the collision. In 1-D elastic collision the relative velocity of two bodies changes its sign: v B2x – v A2x = - ( v B1x – v A1x ) Note: This general property is equivalent to conservation of the total kinetic energy. Example of how to give an extra boost to spacecraft -(v B2x – v A2x ) = = v B1x – v A1x = = 20 km/s → K A2 =(29.6/10.4) 2 K A1 = 8·K A1

9. Two-Dimensional Collisions

10. Exam Example 19: Collision of Two Pendulums

11. Center of Mass It is a point that represents the average location for the total mass of a system x y z Center-of-Mass Dynamics from Newton’s 2 nd Law: since (Newton’s third law) Conservation of the momentum in the isolated system x 0 m1m1 m2m2 M x1x1 x2x2 x cm

12. Exam Example 20: Head-on elastic collision (problems 8.48, 8.50) X V 02x V 01x m1m1 m2m2 0 Data: m 1, m 2, v 01x, v 02x Find: (a) v 1x, v 2x after collision; (b) Δp 1x, Δp 2x, ΔK 1, ΔK 2 ; (c) x cm at t = 1 min after collision if at a moment of collision x cm (t=0)=0 Solution: In a frame of reference moving with V 02x, we have V’ 01x = V 01x - V 02x, V’ 02x = 0, and conservations of momentum and energy yield m 1 V’ 1x +m 2 V’ 2x =m 1 V’ 01x → V’ 2x =(m 1 /m 2 )(V’ 01x -V’ 1x ) m 1 V’ 2 1x +m 2 V’ 2 2x =m 1 V’ 2 01x → (m 1 /m 2 )(V’ 2 01x -V’ 2 1x )=V’ 2 2x = (m 1 /m 2 ) 2 (V’ 01x - V’ 1x ) 2 → V’ 01x +V’ 1x =(m 1 /m 2 )(V’ 01x –V’ 1x )→ V’ 1x =V’ 01x (m 1 -m 2 )/(m 1 +m 2 ) and V’ 2x =V’ 01x 2m 1 /(m 1 +m 2 ) (a) returning back to the original laboratory frame, we immediately find: V 1x = V 02x +(V 01x – V 02x ) (m 1 -m 2 )/(m 1 +m 2 ) and V 2x = V 02x +(V 01x – V 02x )2m 1 /(m 1 +m 2 ) y’ X’ (a) Another solution: In 1-D elastic collision a relative velocity switches direction V 2x -V 1x =V 01x -V 02x. Together with momentum conservation it yields the same answer. (b) Δp 1x =m 1 (V 1x -V 01x ), Δp 2x =m 2 (V 2x -V 02x ) → Δp 1x =-Δp 2x (momentum conservation) ΔK 1 =K 1 -K 01 =(V 2 1x -V 2 01x )m 1 /2, ΔK 2 =K 2 -K 02 =(V 2 2x -V 2 02x )m 2 /2→ΔK 1 =-ΔK 2 (E=const) (c)x cm = (m 1 x 1 +m 2 x 2 )/(m 1 +m 2 ) and V cm = const = (m 1 V 01x +m 2 V 02x )/(m 1 +m 2 ) → x cm (t) = x cm (t=0) + V cm t = t (m 1 V 01x +m 2 V 02x )/(m 1 +m 2 )

13. Exam Example 21: Head-on completely inelastic collision (problems 8.86) Data: m 2 =2m 1, v 10 =v 20 =0, R, ignore friction Find: (a) velocity v of stuck masses immediately after collision. (b) How high above the bottom will the masses go after colliding? Solution: (a) Momentum conservation y x h m1m1 m2m2 Conservation of energy: (i) for mass m 1 on the way to the bottom just before the collision (ii) for the stuck together masses on the way from the bottom to the top (b)

14. Rocket Propulsion Momentum conservation in the co-moving with the rocket frame: m dv = - v ex dm → m dv/dt = - v ex dm/dt → thrust F = - v ex dm/dt In the gravity-free outer space for constant v ex :

15. “Missing” Momentum and Energy in a β-Decay of Nuclei