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Momentum and Collisions. F(t). Motion under the action of time-varying force. Newton’s second law. Δ t=t 2 -t 1. Definition of (linear) momentum:. Momentum of many-particle system. Effect of force on the object’s motion is determined by impulse. Ball hits a wall.
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Momentum and Collisions F(t) Motion under the action of time-varying force Newton’s second law Δt=t2-t1 Definition of (linear) momentum: Momentum of many-particle system Effect of force on the object’s motion is determined by impulse Ball hits a wall (Impulse-Momentum Theorem)
The Principle of Conservation of Momentum The total momentum of an isolated system remains constant (is conserved). Proof: Two-particle collision
Example: Recoil of a rifle Solution: momentum conservation yields Pfx = P0x = 0 → mBvBx + mRvRx = 0 → vRx = - mBvBx / mR vRx = - ( 0.005 kg / 3 kg ) 300 m/s = - 0.5 m/s KB = (1/2)mB vBx2= (1/2)mB (mRvRx / mB)2 = KR (mR/mB)>> KR
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
Important fact: Conservation of the momentum P does not depend on the conservation of the kinetic energy K ! Example of the inelastic collision: Ice Skaters Note: Kf> K0 = 0 (Inelastic collision !!!) Solution: m1vf1 + m2vf2 = 0 → vf2 = - m1 vf1 / m2 vf2 = - 2.5 m/s ( 54 kg / 88 kg ) = - 1.5 m/s
Exam Example 18: The Ballistic Pendulum(example 8.8, problem 8.43) y 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. L Vtop=0 Find: (a) velocity v of the block with the bullet immediately after impact; (b) tension force T immediately after impact; (c) initial velocity vx of the bullet. Solution: (a) Conservation of mechanical energy K+U=const (b) Newton’s second law yields (c) Momentum conservation for the collision
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: vB2x – vA2x = - ( vB1x – vA1x ) Note: This general property is equivalent to conservation of the total kinetic energy. The Gravitational Slingshot Effect Example of how to give an extra boost to spacecraft -(vB2x – vA2x) = = vB1x – vA1x = = 20 km/s → KA2=(29.6/10.4)2 KA1 = 8·KA1
Center of Mass It is a point that represents the average location for the total mass of a system z y Center-of-Mass Dynamics from Newton’s 2nd Law: x since (Newton’s third law) Conservation of the momentum in the isolated system m2 M m1 0 x1 x xcm x2
Exam Example 20: Head-on elastic collision (problems 8.48, 8.50) Data: m1, m2, v01x, v02x Find: (a) v1x, v2x after collision; (b) Δp1x, Δp2x , ΔK1, ΔK2 ; (c) xcm at t = 1 min after collision if at a moment of collision xcm(t=0)=0 V01x y’ V02x m1 X’ m2 Solution: In a frame of reference moving with V02x, we have V’01x= V01x- V02x, V’02x = 0, and conservations of momentum and energy yield m1V’1x+m2V’2x=m1V’01x→ V’2x=(m1/m2)(V’01x-V’1x) m1V’21x+m2V’22x=m1V’201x→ (m1/m2)(V’201x-V’21x)=V’22x= (m1/m2)2(V’01x- V’1x)2→ V’01x+V’1x=(m1/m2)(V’01x–V’1x)→ V’1x=V’01x(m1-m2)/(m1+m2) and V’2x=V’01x2m1/(m1+m2) (a) returning back to the original laboratory frame, we immediately find: V1x= V02x+(V01x– V02x)(m1-m2)/(m1+m2) and V2x = V02x +(V01x– V02x)2m1/(m1+m2) 0 X (a)Another solution: In 1-D elastic collision a relative velocity switches direction V2x-V1x=V01x-V02x. Together with momentum conservation it yields the same answer. (b) Δp1x=m1(V1x-V01x), Δp2x=m2(V2x-V02x) → Δp1x=-Δp2x(momentum conservation) ΔK1=K1-K01=(V21x-V201x)m1/2, ΔK2=K2-K02=(V22x-V202x)m2/2→ΔK1=-ΔK2(E=const) • xcm = (m1x1+m2x2)/(m1+m2) and Vcm = const = (m1V01x+m2V02x)/(m1+m2) • → xcm(t) = xcm(t=0) + Vcm t = t (m1V01x+m2V02x)/(m1+m2)
Exam Example 21: Head-on completely inelastic collision (problems 8.86) Data: m2=2m1, v10=v20=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? y Solution: (a) Momentum conservation m1 Conservation of energy: (i) for mass m1 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) h x m2
Rocket Propulsion Momentum conservation in the co-moving with the rocket frame: m dv = - vex dm → m dv/dt = - vex dm/dt → thrust F = - vex dm/dt In the gravity-free outer space for constant vex :
