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13 Dec 2019

A small space probe, of mass 240kg, is launched from a spacecraftnear mars. It travels toward the surface of Mars, where it willland. At a time 20.7 seconds after it was launched, the probe is atlocation <4.30 x 10^3, 8.70 x 10^2, 0> m, and at his smaetime it momentum is <4.40 x 10^4, -7.60 x 10^3, 0> kg m/s. Atthis instant, the net force on the probe due to the gravitationalpull of Mars plus the air resistance acting on the probe is ,<-7x 10^3, -9.2 x 10^2, 0> N.

a) Assuming that the net force on the probe is approximatelyconstant over this time interval, what its the momentum of theprobe and its location 20.9 seconds after it is launched?

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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.)

Learning Goal: To learn about the impulse-momentum

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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A ball is kicked from a location < 9, 0, -6 > (on the ground)with initial velocity < -10, 17, -3 > m/s. The ball's speedis low enough that air resistance is negligible.

velocity of the ball = <-10, 14,06, -3>

What is the average velocity of the ball over this timeinterval?

Use the average velocity to find the location of the ball 0.3seconds after being kicked:



Now consider a different time interval: the interval between theinitial kick and the moment when the ball reaches its highestpoint. We want to find how long it takes for the ball to reach thispoint, and how high the ball goes.

What is the y-component of the ball's velocity at the instant whenthe ball reaches its highest point (the end of this timeinterval)?

Fill in the missing numbers in the equation below (update form ofthe Momentum Principle):
mvyf = mvyi + Fnet,y?t
m( ) = m( )+ -mg?t

How long does it take for the ball to reach its highestpoint?
?t = ( ) s.
Knowing this time, first find the y-component of the averagevelocity during this time interval, then use it to find the maximumheight attained by the ball:
ymax = ( ) m.

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