## Conservation the momentum

Consider two interacting objects. If thing 1 pushes on thing 2 v a force F = 10 N because that 2 s come the right, climate the inert of thing 2 transforms by 20 Ns = 20 kgm/s to the right. Through Newton"s 3rd law object 2 pushes on thing 1 through a force F = 10 N for 2 s to the left. The momentum of object 1 transforms by 20 Ns = 20 kgm/s to the left. The complete momentum the both objects does no change. For this reason we say the the total momentum that the communicating objects is conserved.

Newton"s 3rd law implies that the full momentum of a system of interacting objectsnot action on by exterior forces is conserved.

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The total momentum in the world is conserved. The momentum of a single object, however, alters when a network force acts on the object for a finite time interval. Conversely, if no net force acts upon an object, its momentum is constant.

For a system of objects, a component of the momentum follow me a favored direction is constant, if no net external force with a ingredient in this preferred direction acts on the system.

### Collisions

In collisions in between two secluded objects Newton"s 3rd law indicates that inert is always conserved. In collisions, that is assumed the the colliding objects connect for together a brief time, the the impulse due to external forces is negligible. Thus the complete momentum that the system just before the collision is the very same as the full momentum simply after the collision. Collisions in which the kinetic power is also conserved, i.e.in which the kinetic power just ~ the collision equals the kinetic energy just before the collision, are called elastic collision. In this collisions no ordered power is converted into thermal energy. Collisions in i m sorry the kinetic energy is no conserved, i.e. In which some ordered energy is convert into internal energy, are dubbed inelastic collisions. If the two objects stick together after the collision and also move through a usual velocityvf, climate the collision is stated to beperfectly inelastic.

Note:In collisions between two isolated objects inert is always conserved.Kinetic energy is only conserved in elastic collisions.

We constantly have m1v1i + m2v2i= m1v1f + m2v2f.Only for elastic collisions perform we additionally have ½m1v1i2+ ½m2v2i2 = ½m1v1f2+ ½m2v2f2.

Problem:

If 2 objects collide and also one is at first at rest, is it possible for both to be at rest after the collision? Is it possible for one to it is in at rest after the collision? Explain!

Solution:

Reasoning:In collisions between two objects inert is conserved. Because the initial inert is no zero, the last momentum is not zero. Both objects cannot be in ~ rest.It is possible for one of the objects to it is in at remainder after the collision. For example, if the masses of the 2 objects room equal, then after a head-on elastic collision the object originally at rest is moving and also the object at first moving is at rest.Problem:

A 10 g cartridge is stopped in a block of lumber (m = 5 kg). The speed of the bullet-wood combination immediately ~ the collision is 0.6 m/s. What to be the original speed that the bullet?

Solution:

Reasoning:This is a perfectly inelastic collision. The 2 objects stick with each other after the collision and also move with a common velocity v
f. I think the motion is follow me the x-direction. We then havem1v1 + m2v2 = (m1 + m2)vf.m1 is the fixed of the bullet, v1 its early stage velocity, m2 is the fixed of the block, v2 its early velocity.Details the the calculation:Initially the block is in ~ rest, v2 = 0. Therefore(0.01 kg)v1 = (5.01 kg)(0.6 m/s). V1 = 300.6 m/s.Problem:

Two car of equal mass and equal speeds collide head on. Execute they experience a greater force if the collision is elastic or perfect inelastic and they stick together?

Solution:

Reasoning:Let the cars move along the z-axis. When the collision is elastic, the readjust in momentum of the auto moving at first in the optimistic x-direction is pfinal - pinitial = -2 pinitial, since the auto bounces ago and pfinal = -pinitial. Once the collision is inelastic, the cars involved rest and the adjust in inert of the automobile moving initially in the confident x-direction is pfinal - pinitial = pinitial, due to the fact that pfinal = 0. So the impulse has actually twice the size in the elastic collision. If the collisions last around the same amount that time, climate the average pressure experienced by the automobile is twice as large in the elastic collision.Problem:

A 90 kg fullback running eastern with a rate of 5 m/s is tackled through a 95 kg opponent running north v a speed of 3 m/s. If the collision is perfect inelastic, calculate the speed and the direction of the players just after the tackle.

Solution:

Reasoning:Momentum is a vector. In the collision, the full momentum is conserved.Details that the calculation:The initial momentum of player 1 is p
1 =(90 kg)(5 m/)s ns = 450 kgm/s i.The initial momentum of player 2 is p2 =(95 kg)(3m/s)j = 285 kgm/s j.Momentum is conserved, the final momentum p that both players is p= p1 + p2.p = (m1 + m2)v.v = 2.432 m/si + 1.54 m/s j.v2 = 8.29 (m/s)2, v = 2.88 m/s.The rate of the football player after the collision is 2.88 m/s.tanθ = py/px = 285/450 = 0.63, θ = 32.34o.Their direction of travel makes an angle θ = 32.34o with the x-axis. (The x-axis is pointing east.)Problem:

A 30,000 kg freight vehicle is coasting in ~ 0.850 m/s through negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. What is the final velocity the the loaded freight car?

Solution:

Reasoning:For a device of objects, a ingredient of the momentum follow me a chosen direction is constant, if no net external force v a ingredient in this preferred direction acts on the system.Details of the calculation:Let the direction of movement of the vehicle be the positive x-direction.Before the intake the momentum of the car was in the x-direction v magnitude p = mv = 2.55*104 kgm/s.The x- ingredient of the inert of the scrap metal was zero, so the total initial inert in the x-direction to be pi = 2.55*104 kgm/s.No outside force acts on the objects in the x-direction, as such the x-component that the complete momentum is conserved.pf = (mcar + mmetal)vf = (140,000 kg)vf = pi = 2.55*104 kgm/s.vf = (2.55*104 kgm/s)/(140,000 kg)= 0.182 m/s.Problem:

After a fully inelastic collision in between two objects of same mass, each having actually initial rate v, the two relocate off along with speed v/3. What was the angle in between their early directions?

Solution:

Reasoning:The exterior force exhilaration on the mechanism is zero, the complete momentum is conserved.Assume the the last velocity of the objects points in the +x-direction. Before the collision the velocity vector that both objects renders an angle θ v the x-axis together shown.
Details that the calculation:p1x + p2x = pfx = pf, p1y+ p2y = 0.2mv cosθ = 2mv/3, cosθ = 1/3, θ = 70.5o.The angle in between their initial directions is 2θ = 141o.Problem:

The mass of the blue puck is 20% better than the fixed of the environment-friendly one. Prior to colliding, the pucks strategy each other with equal and also opposite momenta, and the environment-friendly puck has actually an initial rate of 10 m/s. Discover the speed of the pucks after ~ the collision, if fifty percent the kinetic energy is lost throughout the collision.

Solution:

Reasoning:The collision is inelastic, since energy is not conserved.The total momentum of the 2 pucks is zero before the collision and after the collision.Let fragment 1 it is in the eco-friendly puck and also particle 2 it is in the blue puck. Before and also after the collision the proportion of the speed is v2/v1= m1/m2 = 1/1.2.The final kinetic energy of the system equates to ½ time its early kinetic energy.½m1v1i2+ ½m2v2i2 = 2*(½m1v1f2+ ½m2v2f2).Details of the calculation:½m1v1i2+ ½m2v2i2 = 2*(½m1v1f2+ ½m2v2f2).Algebra:m1v1i2+ m2v2i2 = 2 (m1v1f2 + m2v2f2).m1v1i2 + 1.2 m1(v1i/1.2)2= 2 (m1v1f2 + 1.2 m1(v1f/1.2)2).m1(v1i2 + v1i2/1.2) = 2 m1(v1f2 + v1f2/1.2).v1i2 = 2 v1f2, v1f = 0.707 v1i.v1f = 7.07 m/s, v2f = 5.89 m/s.Module 5: inquiry 1

During a visit come the International room Station, an astronaut to be positioned motionless in the center of the station, out of with of any solid thing on which he could exert a force. Suggest a technique by i m sorry he might move self away from this position, and also explain the rebab.netics involved.

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Module 5: concern 2

Football coaches recommend players come block, hit, and also tackle v their feet ~ above the ground quite than by leaping through the air. Utilizing the ideas of momentum, work, and also energy, describe how a football player can be an ext effective through his feet on the ground.