Kinesiology 2241A/B Lecture Notes - Lecture 18: Momentum, Mattress, Lincoln Near-Earth Asteroid Research

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greenkangaroo690

20 Mar 2018

School

Western University

Department

Kinesiology

Course

Kinesiology 2241A/B

Professor

Thomas Richard Jenkyn

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Document Summary

Linear momentum is the mass of an object multiplied by velocity. Isaac newton called this the "quantity of motion" that an object had. Linear momentum can come from the mass or the velocity. Linear momentum will tell you where you are going to go (exc. When you lose traction on the road linear momentum is going to tell you where your car is going to go: velocity always goes tangent to the pathway that you are following. F= (m2-m1)/delta t can be rearranged to: the quantity of force times time is called "linear impulse, rearranged this would look like: Landing on concrete vs. landing on a mattress- mattress lengthens the time of impact) Slowing impulse: to slow the ball down to zero momentum, she needs to apply an impulse in the opposite direction of 10 kgm/s. Land with straight knees and f is large at delta t very short (ouch)

Related Questions

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
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7
8

puceworm982

Object A collides on a horizontal frictionless surface with an initially stationary target, object X. The initial and final velocities of object A are shown. The final velocity of object X is not given.

a). At an instant during the collision, is the net force on object A zero or non-zero?

b) [During the collision, is the momentum of object A conserved? Explain. Is the momentum of the system consisting of objects A and X conserved? Explain. On the same horizontal surface, object C collides with an initially stationary target, object Z. The initial speeds of objects C and A are the same, and m_x = m_z > m_A = m_c. After the collisions, object C moves in the direction shown and has the same final speed as object A. In the space below, copy the vectors v_C1 and v_cr with their tails together. Use these vectors to draw the change in velocity vector for glider C, Delta v_c. Is the magnitude of the change in velocity vector of object A greater than, less than, or equal to the magnitude of the change in velocity vector of object C? Explain? Is the magnitude of the change in the momentum vector of object A greater than, less than, or equal to the magnitude of the change in the momentum vector of object C? Explain? Is the final speed of object X greater than, less than, or equal to the final speed of object Z? Explain. Consider the following incorrect statement: Gliders A and C have some change in momentum. They have the same mass, and because they have the same initial speed and same final speed. Delta-v is the same for each of.them." Discuss the error(s) in the reasoning.

sangriaporcupine175

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.

ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.

(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgÂ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgÂ·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is Â½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.

MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgÂ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgÂ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

Kinesiology 2241A/B Lecture Notes - Lecture 18: Momentum, Mattress, Lincoln Near-Earth Asteroid Research

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