#### You've reached the end of your free preview.

Want to read all 11 pages?

**Unformatted text preview: **Homework 8: Momentum
Due: 11:59pm on Sunday, March 19, 2017
To understand how points are awarded, read the Grading Policy for this assignment. Impulse and Momentum Ranking Task
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all
automobiles are brought to rest. Part A
Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below.
ANSWER: Reset largest Help smallest The correct ranking cannot be determined. Correct Part B
Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. Hint 1. Relating impulse to momentum
The impulse applied to an object is equal to the object’s change in momentum. Therefore, the impulse needed to stop them should be equal to the
difference between the initial momentum and the final momentum. (The final momentum is zero since the car is brought to a stop.) Hint 2. Apply the impulsemomentum theorem
All of the cars are brought to rest. What is the final momentum \texttip{p_{\rm final}}{p_final} of each automobile?
Enter your answer in kilogram meters per second to three significant figures. Hint 1. How to find momentum
Recall that an object's momentum is equal to the product of its mass and velocity.
ANSWER:
\texttip{p_{\rm final}}{p_final} = 0 \rm {kg \cdot m/s} Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js ANSWER: Reset largest Help smallest The correct ranking cannot be determined. Correct Part C
Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. Hint 1. How to approach the problem
You know the impulse needed to stop the cars. However, this impulse could be a very large force exerted over a fraction of a second, a very small
force exerted over several minutes, or any situation in between, so long as the force multiplied by the time gives the proper impulse. You must
know the time intervals over which the stopping forces are exerted to determine the magnitudes of the stopping forces from the impulses.
ANSWER: Reset largest The correct ranking cannot be determined. Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js Help smallest Correct
The more momentum an object has, the more impulse is needed to stop it. However, this impulse can be provided via a large force acting over a
short time interval or a relatively small force acting over a relatively long time interval. If you are driving down the highway at 55 \rm {mph}, you can
stop your car by either lightly pressing on the brakes and traveling a long time before stopping, or pressing more firmly on the brakes and stopping
more quickly. In both cases, your braking system has applied the same amount of impulse to your car. Impulse and Momentum I
Force of a Baseball Swing. A baseball has mass 0.149 {\rm kg} . At some time, the ball comes in contact with a bat. Part A
If we consider just the ball as the system, was there an external force acting on the system? What does that tell you about the change in momentum of
the ball? Express your answer to three significant figures and include the appropriate units.
ANSWER:
An external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum did change.
No external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum did change. Correct Part B
If the velocity of a pitched ball has a magnitude of 45.0 {\rm m/s} to the left and the batted ball's velocity is 57.0 {\rm m/s} in the opposite direction, find
the change in momentum of the ball.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
\Delta p = 15.2 \large{{\rm \frac{kg{\cdot}m}{s}}} Correct Part C
If the ball remains in contact with the bat for 2.5 {\rm ms} , find the magnitude of the average force applied by the bat. Express your answer to two significant figures and include the appropriate units.
ANSWER:
F = 6.1 {\rm kN} Correct Impulse and Momentum II
A 2.00 {\rm kg} stone is sliding to the right on a frictionless horizontal surface at 4.50 {\rm m/s} when it is suddenly struck by an object that exerts a large
horizontal force on it for a short period of time. The graph in the figure shows the magnitude of this force as a function of time.
Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js Part A What impulse does this force exert on the stone? ANSWER:
p = 2.50 {\rm kg \cdot m/s} Correct Part B Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the right. ANSWER:
v = 5.75 {\rm m/s} Correct Part C
Just after the force stops acting, find the direction of the stone's velocity if the force acts to the right.
ANSWER:
to the right
to the left Correct Part D
Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the left.
ANSWER:
v = 3.25 {\rm m/s} Correct Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js
Part E Just after the force stops acting, find the direction of the stone's velocity if the force acts to the left.
ANSWER:
to the right
to the left Correct Conservation of Momentum I
An astronaut in her space suit has a mass of 87.0 kg. Her tether loses its attachment to her spacecraft while she is on a spacewalk. Initially at rest (with
respect to her spacecraft) she throws her oxygen tank, which has a mass of 12.0 kg, away from her spacecraft with a speed of 8.00 m/s to propel her back
towards it. (Note: The astronaut’s mass after the throw is 87.0 kg – 12.0 kg = 75.0 kg). Part A
During the throw, is there a net force acting on the system of astronaut+tank? What does this tell you about the total momentum of the system?
ANSWER:
No external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum changed.
An external force was exerted on the system, and as a result the system's momentum did not change. Correct Part B
Find the speed of the astronaut after she throws the tank.
ANSWER:
v = 1.28 m/s Correct Part C
If her oxygen will run out in 2.00 min, how far away from the spacecraft can she be and still reach it before her oxygen runs out?
ANSWER:
d_{max} = 154 m Correct One Dimensional Collisions I
You and your friend are standing on a frictionless sheet of ice. Your friend throws you a 0.400 kg ball that is traveling horizontally at 10.0 m/s. Your mass is
70.0 kg and you are initially at rest. The ball collides with your body. Part A
Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js During the collision, is there a net force acting on the system of you+ball? What does this tell you about the total momentum of the system?
ANSWER:
An external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum changed. Correct Part B
Assume that you catch the ball. Select the choice that classifies this collision
ANSWER:
This is an elastic collision
This is a perfectly inelastic collision
Not enough information to classify this collision
This in an inelastic collision Correct Part C
Determine the speed of you and the ball after the catch.
ANSWER:
v = 5.68×10−2 m/s Correct Part D
Now, assume that you don’t catch the ball, but instead it bounces off your chest with the exact speed that it hit you with (but in the opposite direction of
course). Determine your speed after the collision.
ANSWER:
v = 0.114 m/s Correct Part E
You should have found that your speed was greater when the ball hit you and bounced off than it was when you caught the ball. Select any of the
following choices which explains why this is the case.
ANSWER: Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js When the ball bounced off your chest, you exerted a larger force on the ball than when you caught it.
When the ball bounced off your chest, you exerted a smaller force on the ball than when you caught it.
When bouncing off your chest, the ball's momentum changed less than it did when you caught it.
When bouncing off your chest, the ball's momentum changed more than it did when you caught it. All attempts used; correct answer displayed One Dimensional Collisions II
A 1200 kg car traveling initially with a speed of 25.00 m/s to the right crashes into the back of a 9000 kg truck initially moving in the same direction with a
speed of 20.00 m/s. After the collision, the velocity of the car is 18.00 m/s to the right. Part A
During the collision, is there a net force acting on the system of the car + truck? What does this tell you about the total momentum of the system?
ANSWER:
No external force was exerted on the system, and as a result the system's momentum changed.
An external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum did not change. Correct Part B
Find the xcomponent of the velocity of the truck after the collision.
ANSWER:
v_x = 20.9 m/s Correct Part C
Find the total kinetic energy of the system (car + truck) before the collision.
ANSWER:
K_i = 2.18×106 J Correct Part D
Find the total kinetic energy of the system (car + truck) after the collision.
ANSWER: Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js K_f = 2.17×106 J Correct Part E
What would this collision be classified as?
ANSWER:
An elastic collision
An inelastic collision
Not enough information
A perfectly inelastic collision Correct Two Dimensional Collisions I
A billiard ball (ball A) moving along the positive xaxis at 5.00 m/s strikes a stationary ball (ball B) of the same mass. Immediately after the collision, ball B
moves at 4.33 m/s at an angle of 30° with respect to the original line of motion of ball A. Part A
During the collision, is there a net force acting on the system of the two balls during the collision? What does this tell you about the total momentum of
the system?
ANSWER:
No external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum changed. Correct Part B
Find the xcomponent of the velocity of ball A after the collision.
ANSWER:
v_x = 1.25 m/s Correct Part C
Find the ycomponent of the velocity of ball A after the collision.
ANSWER:
Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js v_y = 2.17 m/s Correct Part D
Find the speed of ball A after the collision.
ANSWER:
v = 2.50 m/s Correct Part E
Find the direction of the velocity with respect to the positive xaxis of ball A after the collision. Choose the positive angle (counterclockwise).
ANSWER:
\theta = 300 degrees counterclockwise from the positive xaxis Correct Two Dimensional Collisions II
Two cars collide at an intersection. Car A, with mass 2000 kg, is going from east to west traveling at unknown speed, while car B, of mass 1500 kg, is going
from north to south at 15 m/s. As a result of the collision, the two cars stick together. You determine that after the collision, the cars moved at an angle of 65°
south of west (i.e. 65° counterclockwise from the negative xaxis) from the point of impact. However, you do not know the speed at which they traveled. Part A
Is there a net force acting on the system of the cars during the collision? What does this tell you about the total momentum of the system during the
collision?
ANSWER:
An external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum changed.
An external force was exerted on the system, and as a result the system's momentum changed. Correct Part B
Determine the speed of the cars (which, remember, are stuck together) after the collision.
ANSWER:
v_{\rm after} = 7.09 m/s Correct
Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js Part C
Determine the initial speed of car A.
ANSWER:
v_{\rm A, i} = 5.25 m/s Correct Elastic Collision in One Dimension
Glider A has mass 0.150 kg and is moving to the right on a horizontal air track with a speed of 0.80 m/s. It has a headon collision with glider B, a 0.300 kg
glider moving to the left with a speed of 2.20 m/s. (A glider is a cart on a frictionless surface). Assume the collision is elastic, and that the gliders move only
along the xaxis. Part A
During the collision, is there a net force acting on the system of the two gliders during the collision? What does this tell you about the total momentum of
the system?
ANSWER:
An external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum changed.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum did not change. Correct Part B
What does the fact that the collision was elastic tell you?
ANSWER:
The momentum of the system was conserved.
The kinetic energy was conserved.
The total energy was conserved. Correct Part C
Find the xcomponent of the velocity of each glider after the collision.
ANSWER:
v_{\rm A, f}, v_{\rm B,f} = 3.20,0.200 m/s Correct Using Conservation of Energy and Conservation of Momentum
Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js In a broadway performance, an 80.0 kg actor swings from a 3.75 m long cable that is horizontal when he starts. At the bottom of his arc, he picks up his 55.0
kg costar in a perfectly inelastic collision. Ignore air resistance. Part A
Determine the speed of the actor just before he collides with his costar.
ANSWER:
v_{\rm before} = 8.57 m/s Correct Part B
During the instant of the collision, is there a net force acting on the system of the star+costar? What does this tell you about the total momentum of the
system?
ANSWER:
An external force was exerted on the system, and as a result the system's momentum did not change.
An external force was exerted on the system, and as a result the system's momentum changed.
No external force was exerted on the system, and as a result the system's momentum did not change.
No external force was exerted on the system, and as a result the system's momentum changed. Correct Part C
Find the speed of the system (star+costar) just after the collision.
ANSWER:
v_{\rm after} = 5.08 m/s Correct Part D
Determine the maximum height to which the star+costar rise in their swing after the collision.
ANSWER:
h_{\rm max} = 1.32 m Correct
Score Summary:
Your score on this assignment is 99.0%.
You received 114.8 out of a possible total of 116 points. Loading [MathJax]/jax/output/HTMLCSS/autoload/maction.js ...

View
Full Document

- Fall '19