Momentum, Impulse, and The Law of Conservation of Momentum

1. Momentum, Impulse, and The Law of Conservation of Momentum

2. Definitions: MOMENTUM = a measure of how difficult it is stop a moving object. Its mks units are kg*m/s or N*s. “p” is the symbol for momentum. p = m · v The more mass and velocity an object has, the more difficult it is to stop. More newtons of force are required for a longer time. IMPULSE = a change in an object’s momentum due to a force acting on it over a time interval. F · t = Δp Impulse is force times time, or simply the change in momentum, the units are the same as momentum… N*s or kg*m/s.

3. The equation for impulse is a concise statement of the Impulse-Momentum Theorem which coming straight from Newton’s 2nd Law. F = m · a F = m · Δv/ Δt F · Δt = m · Δv F ·Δt = Δp If no force acts on the object, then it’s momentum won’t change. This is consistent with Newton’s 1st Law which states that an object’s velocity is constant in the absence of an outside net force.

4. According to Newton’s 3rd Law, the law of action-reaction, when two object’s interact over a time interval they exert equal and opposite forces on each other. Since the two object’s exert equal but opposite amounts of force on each other for equal amounts of time, the objects undergo equal and opposite amounts of change in momentum--or impulses. The subscripts 1 and 2 refer to objects 1 and 2 F1 · t1 = - F2 · t2 Δp1 = - Δp2 Δp1 + Δp2 = 0 ΣΔp = 0 The total change in momentum is zero for a system is zero if there are no outside forces. Hey, that sounds like a pretty good law!

5. The Law of Conservation of Momentum The total momentum of a system is constant in the absence of outside forces. NOTE: There can be internal forces like gravity, electric force or a collision and the law of conservation still applies. Expressed mathematically… Σ po = Σ pf NOTE 1: Momentum is a vector quantity because velocity is a vector. Signs must be included to indicate direction based on a coordinate system. NOTE 2: Separate equations must be made for momentum in the x-direction and y-direction. This is often necessary in problems involving glancing collisions.

6. Example: The recoil of a gun after firing a bullet. Before the gun is fired, the bullet and gun are both at rest. Each object has zero momentum, the total momentum of the system is zero. Immediately after the gun is fired, the total momentum of the system must still be zero since there are no outside forces. The bullet has a small mass, but a very large velocity to the right. Its momentum is: mV The gun recoils, or kicks back to left after firing. Its mass is much larger than the bullet, but it’s velocity is much smaller and in the opposite direction. Its momentum is: -Mv. The sum of the bullet’s and gun’s momentum vectors is mV + (-M v) = 0, just as it was before firing.

7. Let’s put some numbers into the gun-bullet scenario... Example: Before the gun is fired both the bullet and gun are at rest. After the gun is fired, a .010 kg bullet is expelled with velocity +400 m/s (to the right). If the gun has mass 1.00 kg, what is its recoil velocity? Knowing the total momentum of the system after the gun is fired (0 N*s), and the final velocity of one object (the bullet), we can calculate the recoil velocity of the gun. Σ pf = pgun + pbullet = mgunvgun + mbulletvbullet 0 = (1.00 kg) vgun + (0.010 kg)(400 m/s) 0 = vgun + 4 so vgun = -4 m/s The gun’s velocity is 4 m/s, backwards compared to the direction bullet’s velocity. This is the recoil we’ve talking about. We know that the total momentum of the system before the gun is fired is 0 N*s. Neither object is moving. Σ po = pgun + pbullet Σ po = mgunvgun + mbulletvbullet Σ po = (1.00 kg)(0) + (0.010 kg)(0) = 0 The total momentum of the gun and bullet must be 0 after firing because momentum is conserved when there are no outside forces.

8. Example: Glancing Collisions, Billiard Balls When the white ball strikes the green ball off-center, both balls are deflected. The white ball which was initially moving in the +y direction is pushed to left. The green ball is pushed up and to the right. After the collision, the combined momentum of the green and white balls in the +y direction is: m v. Momentum is conserved. The white ball has initial momentum: m v in the +y direction. The green ball has no momentum before the collision. The total momentum of the system is m v in the +y direction. After the collision, the combined momentum of the green and white balls in the x direction cancels. One ball moves left, the other right. Momentum is conserved again because there was no momentum in the x direction before the collision.

9. More on Collisions In every collision where there are no outside forces, momentum is always conserved in every direction. This is consistent with Newton’s Laws of Motion. In every collision the total energy of the system is conserved according to the law of conservation of energy. But, it is possible for some of the energy to be transformed into forms other than moving, or kinetic energy.

10. ELASTIC COLLISION = a collision in which the kinetic energy of the objects in the system is conserved. INELASTIC COLLISION = a collision in which the kinetic energy of the objects in the system is not conserved. Collisions between molecules in an ideal gas are elastic. Other collisions one might experience in everyday living are not going to be ideal because kinetic energy from the colliding objects is converted into work done, sound, light or heat.

11. Example: A 2.0-kg ball with initial velocity velocity +6.0 m/s collides elastically with a 4.0-kg ball with initial velocity -2.0 m/s. Find the final velocity of each ball after the collision. +x 6.0 m/s 2.0 m/s 2.0 kg 4.0 kg After the collision we need two equations to find the velocities of the two balls. The orange ball is ball 1. The red ball is ball 2. Σ pf = (2.0 kg)(v1) + (4.0 kg)(v2) = 4 kg*m/s Σ KEf = ½ (2)(v1)2 + ½ (4)(v2)2 = 44 J We’ll solve the momentum equation for v1 and substitute into the KE equation. Since the collision is elastic both momentum and KE must be conserved. Σ po = (2.0 kg)(6.0 m/s) + (4.0 kg)(-2.0 m/s) = 4 kg*m/s Σ KEo = ½ (2)(6)2 + ½ (4)(-2)2 = 44 J After the collision the total momentum of the system must be 4 kg*m/s and the total kinetic energy must be 44 J. 2 v1 + 4 v2 = 4 OR v1 = 2 - 2 v2 (v1)2 + 2 (v2)2 = 44 becomes (2 - 2 v2 )2 + 2 (v2)2 = 44 4 - 8 v2 + 6(v2)2 = 44 solving for v2 yields +10/3 m/s. Plugging into the equation for v1 yields -14/3 m/s. The signs of the balls’ velocities reverse. Ball one, initially moving 6 m/s to the right rebounds at -14/3 m/s to the left. Ball two moving 2 m/s to the left has velocity 10/3 after the collision.