Impulse-Momentum

Impulse In some situations there is no way to directly measure a force: The normal force on my feet when I land after jumping The force caused by a ball hitting a bat Instead of trying to describe the force, we describe how the system reacts to the force.

𝐽 𝑎𝑣𝑔 = 𝐹 𝑎𝑣𝑔 Δ𝑡 Impulse Towards this end we define the impulse: 𝐽 𝑎𝑣𝑔 = 𝐹 𝑎𝑣𝑔 Δ𝑡 Just like with work, we want to describe how the system will react to a given impulse. Note: the Textbook uses 𝐼 to denote impulse, while the AP Exam uses 𝐽.

Suppose an average impulse 𝐽 𝑎𝑣𝑔 acts on a mass 𝑚 moving with an initial velocity 𝑣 0

Impulse-Momentum Impulse-Momentum Theorem states: 𝐽 𝑎𝑣𝑔 =Δ 𝑝

Conservation of Momentum Consider a mid-air collision of two masses: 𝑚 1 𝑚 2 𝑣 1,0 𝑣 2,0 Δ𝑡 𝐹 2→1 𝐹 1→2 Δ𝑡 𝑚 2 𝑚 1 𝑣 1,𝑓 𝑣 2,𝑓

Conservation of Momentum In general:  ∑ 𝐽 𝑒𝑥 =Δ p Where 𝐽 𝑒𝑥 is any impulse caused by a mass not included in Δ𝑝.

Conservation of Momentum A corollary of this result is: In the absence of external forces Δ p =0 kg m s This result is called Conservation of Momentum.

Conservation of Momentum & Collisions Because the forces between two objects during a collision are often very large and extremely short in duration, we can apply conservation of momentum to find the velocities immediately before/after a collision.

Conservation of Momentum - Example A 60 𝑘𝑔 archer stands at rest on frictionless ice and fires a 0.030 𝑘𝑔 arrow horizontally with a velocity of 85 𝑚 𝑠 𝑖 . With what velocity does the archer move after firing the arrow?

𝑚 𝑎 =0.030 𝑘𝑔 𝑚 𝑚 =60 𝑘𝑔 𝑣 0,𝑎 =0 𝑚 𝑠 𝑣 0,𝑚 =0 𝑚 𝑠 𝑣 𝑓,𝑎 = 85 𝑚 𝑠 𝑖 𝑣 𝑓,𝑚 = ?

Conservation of Momentum - Example Two blocks of mass 𝑀 and 4𝑀 sit side-by-side on opposite ends of a compressed massless spring. The spring has a spring constant of 𝑘 and is compressed Δ𝑥. How fast does the block of mass 𝑀 move when the spring is released?

Conservation of Momentum The previous examples involved conservation of momentum for one-dimensional motion. However, conservation of momentum also holds for 2 and 3 dimensional motion as well: ∑ 𝑝 0 =∑ 𝑝 𝑓 ∑ 𝑝 𝑥,0 =∑ 𝑝 𝑥,𝑓 ∑ 𝑝 𝑦,0 =∑ 𝑝 𝑦,𝑓

Conservation of Momentum-Example A firework is initially moving vertically upwards with a speed of 45 𝑚/𝑠. The firework suddenly breaks into two equal pieces with velocities 𝑣 𝑓,1 and 𝑣 𝑓,2 as shown in the figure below. Determine the speed of the two parts of the rocket. 𝑣 0 =45 𝑚/𝑠 𝑣 1 30° 60° V2 = 45 m/s V1 = 77.9 m/s

30° 𝑣 0 =45 𝑚/𝑠 𝑣 1 60° 𝑣 1

Collisions There are two types of collisions: Elastic Collisions – both energy and momentum are conserved Inelastic Collisions – only momentum is conserved Perfectly Inelastic Collisions – inelastic collision where the masses stick together

Collisions Elastic Collisions are not an idealized approximation. They occur anytime two objects “collide” without touching. 𝑒 −

Collisions An example of an inelastic collision is a collision between two cars. When two cars collide, the bumper and body of the car deform. During this deformation the bonds between atoms a stretched and rotated, increasing the potential energy stored on the atomic scale, and reducing total energy on the macroscopic scale.

Collisions A perfectly inelastic collision occurs when the two objects stick together after colliding. Examples of perfectly inelastic collisions are: Meteor hitting Earth A bullet or arrow hits a target and gets stuck in it

Collisions Comparison of Elastic and Perfectly Inelastic Collisions: Variables: 2 Initial velocities 2 Final velocities Equations: Conservation of Momentum Conservation of Energy Perfectly Inelastic Collisions: Variables: 2 Initial velocities 1 Final velocities Equations: Conservation of Momentum

Collisions - Example A 0.2 𝑘𝑔 bullet is fired at a target which has a mass of 2.5 𝑘𝑔. The bullet gets stuck inside the target. What is the ratio of the final to initial kinetic energies: 𝐾 𝐸 𝑓 𝐾 𝐸 0 ?

Collisions - Example 𝑚 𝑏 =0.2 𝑘𝑔 𝑚 𝑡 =2.5 𝑘𝑔 𝑣 0 𝑣 𝑓

Collisions - Example Two masses collide elastically in one-dimension. Mass one has a mass, m 1  and an initial velocity  𝑣 0 . Mass two has a mass  𝑚 2  and is initially at rest. Find  𝑣 1,𝑓  and  𝑣 2,𝑓 . 𝑣 0 𝑚 1 𝑚 2 𝑣 1,𝑓 𝑣 2,𝑓 𝑚 1 𝑚 2

Time-Dependent Force 𝐹 𝑡 Suppose 𝐹 varies in times, and we are given 𝐹 𝑡 , how do we calculate the impulse 𝐽?

Time-Dependent Force 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡 We can approximate the impulse by breaking the interval into small segments during which 𝐹 𝑡 ≈𝑐𝑜𝑛𝑠𝑡. 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡

Time-Dependent Force 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡 We can approximate the impulse by breaking the interval into small segments during which 𝐹 𝑡 ≈𝑐𝑜𝑛𝑠𝑡. 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡

Time-Dependent Force 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡 We can approximate the impulse by breaking the interval into small segments during which 𝐹 𝑡 ≈𝑐𝑜𝑛𝑠𝑡. 𝐽≈∑𝐹(𝑡) 𝛿𝑡 𝐹 𝑡

𝐽= 𝑡 0 𝑡 𝑓 𝐹 𝑡 𝑑𝑡 Time-Dependent Force In the limit that 𝛿𝑡, the duration of the time-intervals, goes to zero: 𝐽= 𝑡 0 𝑡 𝑓 𝐹 𝑡 𝑑𝑡

Average of a Time-Dependent Force The impulse of an average force was defined as: 𝐽= 𝐹 𝑎𝑣𝑔 Δ𝑡 Combining this with 𝐽= 𝑡 0 𝑡 𝑓 𝐹 𝑡 𝑑𝑡 gives: 𝐹 𝑎𝑣𝑔 = 1 Δ𝑡 𝑡 0 𝑡 𝑓 𝐹 𝑡 𝑑𝑡

Time-Dependent Force 𝐹 𝐹 𝑎𝑣𝑔 𝑡 𝑡 0 𝑡 𝑓

Force-Momentum Relationship Applying the Fundamental Theorem of Calculus to: Δ𝑝=𝐽= 𝑡 0 𝑡 𝑓 𝐹 𝑡 𝑑𝑡 Gives: 𝐹 = 𝑑 𝑝 𝑑𝑡

Force-Momentum Relationship For an object who’s mass is constant, this is obvious: 𝐹 = 𝑑 𝑝 𝑑𝑡 = 𝑑 𝑑𝑡 (𝑚 𝑣 ) =𝑚 𝑑 𝑣 𝑑𝑡 =𝑚 𝑎

However, if the mass is not constant: 𝐹 = 𝑑 𝑝 𝑑𝑡 = 𝑑 𝑑𝑡 (𝑚 𝑣 )

Force-Momentum Relationship Moving v (𝑑𝑚/𝑑𝑡) to the other side gives: 𝐹 − 𝑣 𝑑 𝑚 𝑑𝑡 =𝑚 𝑎 Thrust

Impulse-Momentum - Example A dump truck is being filled with sand at a rate of 55 𝑘𝑔 per seconds. The sand falls from rest, 2.0 𝑚 into the bed of the truck. How much does the normal force, 𝐹 𝑁 , exceed the actual weight of the truck?

d𝑚 d𝑡 =55 𝑘𝑔 𝑠 Δℎ=2.0 𝑚 𝐹 𝑁 55 𝑘𝑔 𝑠 2.0 𝑚 𝐹 𝑔

Center of Mass Until now, we have treated objects like points of mass. Real objects have volume and shape. Because of rotation, the actual motion of objects can be very complicated. Does this mean everything we’ve learned is a useless approximation?

Center of Mass 𝑣 0

Center of Mass 𝑣 0

Center of Mass 𝑣 0

Center of Mass The center of mass is the point on an extended object, that follows the trajectory of a point mass.

Center of Mass Center of mass 𝑣 0

Center of Mass 𝑣 0

Center of Mass Link: https://www.youtube.com/watch?v=DY3LYQv22qY

Center of Mass The center of mass is the point on an extended object, that follows the trajectory of a point mass. The center of mass is also the point that you could hang an object from, so that it balances.

Center of Mass The center of mass is the point on an extended object, that follows the trajectory of a point mass. The center of mass is also the point that you could hang an object from, so that it balances. For a collection of masses, the center of mass is a weighted average of positions of the masses

Center of Mass The center of mass is the point on an extended object, that follows the trajectory of a point mass. The center of mass is also the point that you could hang an object from, so that it balances. For a collection of masses, the center of mass is a weighted average of positions of the masses The center of mass is NOT the point at which there is equal mass on either side.

𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖 Center of Mass For point masses lying along a straight line, the center of mass is given by: 𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖

A set of masses are attached to a massless rod A set of masses are attached to a massless rod. Find the center of mass, as measured from the left side of the rod. 𝑚 1 =10 𝑘𝑔 𝑚 2 =5 𝑘𝑔 𝑚 3 =40 𝑘𝑔 2 𝑚 6 𝑚

Find the center of mass of the triangular wedge shown below Find the center of mass of the triangular wedge shown below. The wedge has a constant mass per unit area of 𝜎. 6 𝑚 12 𝑚

𝑥 𝑐.𝑚. = ∫𝜆 𝑥 𝑥 𝑑𝑥 ∫𝜆 𝑥 𝑑𝑥 Center of Mass For any continuous mass, the center of mass is given by: 𝑥 𝑐.𝑚. = ∫𝜆 𝑥 𝑥 𝑑𝑥 ∫𝜆 𝑥 𝑑𝑥 Where 𝜆 is the mass density per unit length

; 𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖 𝑦 𝑐.𝑚. = 𝑚 𝑖 𝑦 𝑖 ∑ 𝑚 𝑖 Center of Mass For a two dimensional object, we break the equation into two parts (𝑥 and 𝑦): 𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖 𝑦 𝑐.𝑚. = 𝑚 𝑖 𝑦 𝑖 ∑ 𝑚 𝑖 ;

A set of masses are attached to four corners of a massless box A set of masses are attached to four corners of a massless box. Find the center of mass, as measured from the bottom left corner. 8 𝑚 𝑚 3 =15 𝑘𝑔 𝑚 4 =5 𝑘𝑔 4 𝑚 𝑚 1 =2 𝑘𝑔 𝑚 2 =8 𝑘𝑔

Center of Mass & Momentum Consider the collection of objects that are all moving. How will the center of mass change as the object move? 𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖

Center of Mass & Momentum How will the center of mass change as the object move? 𝑥 𝑐.𝑚. = 𝑚 𝑖 𝑥 𝑖 ∑ 𝑚 𝑖 𝑣 𝑐.𝑚. = 𝑚 𝑖 𝑣 𝑖 ∑ 𝑚 𝑖

Center of Mass & Momentum How will the center of mass change as the object move? Any time momentum is conserved, the velocity of the center of mass is constant. 𝑣 𝑐.𝑚. = 𝑝 𝑖 ∑ 𝑚 𝑖

Center of Mass & Momentum Two distant stars of masses 2𝑀 and 3𝑀 are initially 10 𝑙𝑦 apart. Both stars are initially at rest relative to one another. Gravity pulls the two stars towards each other. How far does the center of mass of the two stars move?

Center of Mass & Momentum Two distant stars of masses 2𝑀 and 3𝑀 are initially 10 𝑙𝑦 apart. Both stars are initially at rest relative to one another. Gravity pulls the two stars towards each other. How far does each star move before they collide?

10 𝑙𝑦 2𝑀 3𝑀