PHYS 1010Q Lecture 6: Momentum

57 views2 pages

Verified Note

scarletyak10

5 Feb 2019

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

PHYS 1010Q verified notes

6/14View all

PHYS 1010Q Lecture Notes - Lecture 5: Parachuting, Vtt Technical Research Centre Of Finland, Net Force

PHYS 1010Q Lecture 6: Momentum

PHYS 1010Q Lecture 7: Energy, Work, Work and Kinetic Energy

Document Summary

Example: now suppose he is standing on the scales while riding in an elevator that is accelerating upward at 2m/s . When you catch the ball, the impact depends on both the mass of the ball and on its velocity. We thus de ne the momentum of an object as momentum=mass x velocity p=mv. A bowling ball moving at a low speed and a golf ball moving at a high speed can have the same momentum. Suppose the momentum of an object changes by an amount p. If the velocity change v takes place over a time interval t, then v=a t where a is the acceleration. Then if we substitute this result into our equation for p, we have. Now newton tells us that the acceleration is due to a net force f= ma. So putting this together with the equation for p, we have.

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
6
7
8

puceworm982

a.) A hockey puck is observed to be sliding along a flatfrictionless surface at a speed of 4 cm/min. There is no net forceacting on the puck. Assuming it doesn't smash into anything, howfast will the puck be going 17 min later?
cm/min

b.) An object is accelerating. Which one of the followingstatements is true?
The object must be speeding up.
Gravity must be causing the object to accelerate.
The object must be changing directions.
The object must be slowing down.
There is a net force acting on the object.

c.) A 100 kg block weighs 220 lb on Earth. If the block weretransported to a planetary body that had a smaller free fallacceleration (as compared to Earth), which one of the followingwould be true?
The block would have the same weight and a smaller mass.
The block would have the same mass and a bigger weight.
The block would have the same mass and the same weight.
The block would have the same weight and a bigger mass.
The block would have the same mass and a smaller weight.
d.) What does it mean (in physics) to say that a quantity isconserved?
It means that the quantity is not being used.
It means the total amount is zero.
It means the total amount never changes.
It means it's being saved up.

e.) At a certain point in time, the total energy of a closed systemis observed to be 176 J. Which one of the following statements istrue?
Three hours later the energy of the system is 352 J.
Four hours later the energy of the system is 88 J.
All of the energy must be kinetic if the objects are moving.
Twelve hours later the energy of the system is 176 J.
Some of the energy must be potential.

g.) Which of the following statements are impossible? Choose allthat apply.
The parked truck had 8770 J of KE.
The scooter had 0 J of KE and 0 kgÂ·m/s of momentum.
A ball moving in a circle had 0 N of net force acting on it.
A box had an acceleration of 0 m/s2 while a net force of 311 Nacted on it.
Jack exerted 18 N of force due east on a cart. As a direct result,the cart exerted 18 N of force due west upon Jack.
A toothpick exerts a smaller gravitational force upon the Moon thanthe Moon exerts upon it.

a.) Two balls make up a closed system. Ball A has 701 J of energyand ball B has 78 J. After they collide, ball A has 665 J. How muchenergy does ball B have after they collide?
J

b.) An asteroid and a comet collide head on. Prior to thecollision, the asteroid has a momentum of 700,000 kgÂ·m/s up and thecomet has a momentum of 800,000 kgÂ·m/s down. After the collisionthe asteroid has a momentum of 750,000 kgÂ·m/s down. After thecollision, what is the comet's momentum?
kgÂ·m/s
North
South
East
West
Up
Down
Left
Right
No direction because the comet has stopped

c.) How much does a 28 kg object weigh on Moon? Please feel free toreference the Solar System Data table located at the beginning ofUnit II.
______N

d.) A 660 N net force acts on a 42 kg object. Determine theobject's acceleration.
_____m/s2

e.) Calculate the gravitational force between a 657 kg object(resting on Earth) and the Moon. The mass of the Moon is 7.347Ã1022kg and the distance between the object & the Moon is 3.844Ã108m.
____N

Question 1 of 7 0.5 Points
A 6 N force acts on a 0.3-kg object for 0.3 seconds. By how muchdoes the object's momentum change (in kg-m/s)?

Question 2 of 7 0.5 Points
A 0.4-kg object is traveling to the right (in the positivedirection) with a speed of 1 m/s. After a 0.2 s collision, theobject is traveling to the left at 3 m/s. What is the magnitude ofthe impulse (in N-s) acting on the object? The answer must bepositive.

Question 3 of 7 1.0 Points
The force shown in the attached figure is the net eastward forceacting on a ball. The force starts rising at t = 0.020 s, fallsback to zero at t = 0.062 s, and reaches a maximum force of 55 N atthe peak. Determine with an error no bigger than 25% (high or low)the magnitude of the impulse (in N-s) delivered to the ball. Hint:Do not use J = F?t. Look at the figure. Find the area of a nearlyequally sized triangle.

Question 4 of 7 1.0 Points
A bat hits a moving baseball. If the bat delivers a net eastwardimpulse of 0.6 N-s and the ball starts with an initial horizontalvelocity of 3.8 m/s to the west and leaves with a 5.2m/s velocityto the east, what is the mass of the ball (in grams)?
Part 5 of 7 -

Question 5 of 7 0.5 Points
The force sensor measures the force on the sensor due to thebumper, but the cart's momentum change arises from the force on thecart due to the bumper. Which of the following facts are needed toassert that the magnitude of these two forces are nearly equal atall times.
1. Their magnitudes differ by the magnitude of the net force on thebumper.
2. The force of friction is small.
3. The net force on the bumper is small.
4. The cart's weight is canceled by the the normal force exerted bythe track.

A. 1, 3

B. 2, 4

C. 3, 4

D. 1, 2

E. 1, 4

F. 1, 2, 3, 4

G. 2, 3

Part 6 of 7 -

Question 6 of 7 1.0 Points
A cart traveling at 0.3 m/s collides with stationary object. Afterthe collision, the cart rebounds in the opposite direction. Thesame cart again traveling at 0.3 m/s collides with a differentstationary object. This time the cart is at rest after thecollision. In which collision is the impulse on the cartgreater?

A. The second collision.

B. Cannot be determined without knowing the mass of the cart.

C. The impulses are the same.

D. The first collision.

E. Cannot be determined without knowing the rebound speed of thefirst collision. Reset Selection

Part 7 of 7 -

Question 7 of 7 0.5 Points
The force vs. time graphs for two collisions are shown. Whichcollision has the largest impulse?

A. The impulse for collision 2 since it lasts longer.

B. Cannot be determined since the masses are unknown.

C. The impulse for collision 1 since it has the larger maximumforce.

D. Cannot be determined since the initial an final speeds areunknown.

E. The impulses are the same.

PHYS 1010Q Lecture 6: Momentum

PHYS 1010Q Lecture Notes - Lecture 5: Parachuting, Vtt Technical Research Centre Of Finland, Net Force

PHYS 1010Q Lecture 6: Momentum

PHYS 1010Q Lecture 7: Energy, Work, Work and Kinetic Energy

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYS 1010Q Lecture Notes - Lecture 7: Seat Belt, Stopping Time, Tennis Ball

PHYS 1010Q Lecture Notes - Lecture 6: Tennis Ball, Seat Belt, Bungee Cord

PHYS 1010Q Lecture Notes - Lecture 7: Escape Velocity

Related Questions