PHY 101 Chapter Notes - Chapter 9: Weighted Arithmetic Mean, Point Particle, Inelastic Collision

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16 Nov 2016

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9-1 linear momentum linear momentum (p): the product of the mass and the velocity of an object; p=mv; units= kg * m/s. A constant linear momentum is the momentum of an object of mass m that is moving in a straight line with velocity v p = pf pi. P is a vector quantity with both magnitude and direction. Newton"s 2nd law is only valid for objects with constant mass for the total momentum of a system of objects (sum of the forces f= ma) Newton"s second law: f = p/ t. This form of newton"s law is the equivalent of the law with m*a as long as the mass remains constant. 9-3 impulse impulse: the average force f times the length of time t that two surfaces are in contact, which is the area under the force vs. time curve i= fav impulse is a change in momentum. A typical time for impulse is milliseconds.

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)

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puceworm982

Kinetic energy KE has dimensions kg · m²/s². It can be written in terms of the momentum p (Topic 6) and mass m as

$KE= \frac{p^{2}}{2 m}$

(a) Determine the proper units for momentum using dimensional analysis. (b) Refer to Problem 5. Given the units of force, write a simple equation relating a constant force F exerted on an object, an interval of time t during which the force is applied, and the resulting momentum of the object, p.

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