3rd Quarter Review

Momentum

Circular Motion

1. Determine the momentum of a ...

• 60-kg halfback moving eastward at 9 m/s.

• 1000-kg car moving northward at 20 m/s.

• 40-kg freshman moving southward at 2 m/s.

1 Answers

A) p = m * v (60 kg)(9m/s) = 540 kg * m /s

B) p = m * v(1000 kg)(20 m/s) 20,000 kg* m/s

C) p = m * v40 kg * 2 m/s80 kg * m/s

2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...

• its velocity were doubled.

• its velocity were tripled.

• its mass were doubled (by adding more passengers and a greater load)

• both its velocity were doubled and its mass were doubled.

2 answers

A) 40, 000 units (doubling the velocity will double the momentum)

B) 60, 000 units (tripling the velocity will triple the momentum

C) 40, 000 units (doubling the mass will double the momentum)

D) 80, 000 units (doubling the mass and doubling the velocity will quadruple the momentum)

3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m

= 120 kg) are running down the football field. Consider their ticker tape patterns below.

• Compare the velocities of these three players. How many times greater is the velocity of the halfback and the velocity of the tight end than the velocity of the lineman?

Vector Diagrams

Tabulate your Answers

Greatest velocity change?

Greatest acceleration?

Greatest momentum change?

Greatest Impulse?

Answers

Greatest velocity change? B (changes from +30 m/s to -28 m/s which is a change of 58 m/s and A only changes -15 m/s)

Greatest acceleration? B, because it has the greatest velocity change and acceleration is dependent on velocity change.

Greatest momentum change? B, because momentum is dependent on velocity and the change in velocity is greatest in B.

Greatest Impulse? B, impulse is momentum change and the momentum change is greatest in B.

Graphs

Setting up your answers

Greatest velocity change?

Greatest acceleration?

Greatest momentum change?

Greatest Impulse?

AnswersGreatest velocity change?

Greatest in A because it changes from +5 m/s to -3 m/s which is a change of 8 whereas B only changes 4 m/s

Greatest acceleration?

A, acceleration is greatest because velocity change is greatest in A

Greatest momentum change?

A, momentum is dependent on velocity change and that is greatest in A

Greatest Impulse?

A, Impulse equals the change in momentum

Example 4• 2. In a phun physics demo, two

identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below.

Question 4a

• Which balloon (A or B) has the greatest acceleration? Explain.

Answer 4a

• Balloon B has the greatest acceleration The rate at which the velocity changes is greatest for balloon B, this is shown by the fact that the speed (distance/time) changes most rapidly.

Example 4B

• Which balloon (A or B) has the greatest final velocity? Explain.

Answer 4B

• Balloon B has the greatest final velocity. At the end of the diagram, the distance traveled in the latest interval is greatest for Balloon B.

Question 4C

• Which balloon (A or B) has the greatest momentum change? Explain.

Answer 4C

• Balloon B has the greatest momentum change. The final velocity is greatest for Balloon B, its velocity change is also the greatest. Momentum change depends on velocity change. The balloon with the greatest velocity change will have the greatest momentum change.

Question 4D

• Which balloon (A or B) experiences the greatest impulse? Explain.

Answer 4D

• The impulse is the same for each car. The impulse equals the momentum change. If the momentum change is the same for each car, then so must be the impulse.

Question 5

• The diagram to the right depicts the before- and after-collision speeds of a car which undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car "sticks" to the wall.

Question 5A

• In which case (A or B) is the change in velocity the greatest? Explain.

Answer 5A

• Case A has the greatest velocity change. The velocity change is -9 m/s in case A and only -5 m/s in case B.

Question 5B

• In which case (A or B) is the change in momentum the greatest? Explain.

Answer 5B

• Case A has the greatest momentum change. The momentum change is dependent on the velocity change; the object with the greatest velocity change has the greatest momentum change.

Question 5C

• In which case (A or B) is the impulse the greatest? Explain.

Answer 5C

• The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for car A then the impulse is greatest.

Question 5D

• In which case (A or B) is the force which acts upon the car the greatest (assume contact times are the same in both cases)? Explain.

Answer 5D

• The impulse is greatest for car A. The force is related to impulse (I = F * t). The bigger impulse for car A is attributed to the greatest force upon car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds.

• 5. If a 5-kg bowling ball is projected upward with a velocity of 2.0 m/s, then what is the recoil velocity of the Earth (mass = 6.0 x 10^24 kg).

Answer 5

• Since the ball has an upward momentum of 10 kg m/s, the Earth must have a downward momentum of 10 kg m/s. To find the velocity of the Earth, use the momentum equation p = m * v. This equation rearranges to v = p/m. By substituting into this equation

• v = (10 kg m/s) / (6 x 10 24 kg). • V = 1.67 * 10 -24 m/s (downward)

• 7. Would you care to fire a rifle that has a bullet ten times as massive as the rifle? Explain.

Answer 7

• Absolutely not! In a situation like this, the target would be a safer place to stand than the rifle. The rifle would have recoil velocity that is ten times larger than the bullet’s velocity. This would produce the effect of “the rifle actually being the bullet.”

• A railroad diesel engine has four times the mass of a flatcar. If a diesel coasts at 5 km/hr into a flatcar that is initially at rest, how fast do the two coast if they couple together?

Train Answer

• 4 km/hr

• 5. A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign on the velocity of the two vehicles.)

• 5. The problem can be solved using a momentum table:

•

• For questions #5-#8: An object is moving in a clockwise direction around a circle at constant speed. Use your understanding of the concepts of velocity and acceleration to answer the next four questions. Use the diagram shown at the right.

• 5. Which vector below represents the direction of the velocity vector when the object is located at point B on the circle?

Answer 5

• Answer D

• The velocity vector is directed tangent to the circle that would be downward when at point B.

• 6. Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle?

Answer 6

• B The acceleration vector would be directed towards the center that would be up and to the right when at point C

• 7. Which vector below represents the direction of the velocity vector when the object is located at point C on the circle?

Answer 7

• A The velocity vector would be directed tangent to the circle and that would be upwards at point C.

• 8. Which vector below represents the direction of the acceleration vector when the object is located at point A on the circle?

Answer 8

• D the acceleration vector would be directed towards the center and that would be straight down when the object is at point A

Example 2

• 2. A 1.5-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.0 m. At the bottom of the circular loop, the speed of the bucket is 6.0 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.

• m = 1.5 kg

• a = ________ m/s/s

• Fnet = _________ N

Answer 2

• F grav = 15 N

• a = 36 m/s/s

• F net = 54 N Up

• F tens = 69 N

• Anna Litical is riding on The Demon at Great America. Anna experiences a downwards acceleration of 15.0 m/s2 at the top of the loop and an upwards acceleration of 20.0 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 1000 kg roller coaster car.

• Steps 1 and 2 involve the construction of a free body diagram and the identification of known and unknown quantities. This is shown in below.

• Given Info:

• m = 1000 kg

• atop = 15.0 m/s2 , down

• bottom = 20.0 m/s2 , up

• Find:

• Fnorm at top and bottom

• Step 3 of the suggested method would not apply to this problem since there are no forces directed "at angles" (that is, all the forces are either horizontally or vertically directed). Step 4 of the suggested method involves the determination of any known forces. In this case, the force of gravity can be determined from the equation Fgrav = m * g. An approximate g value of 10 m/s2 will be used in this problem in order to simplify some of the math and highlight the concepts. So the force of gravity acting upon the 1000-kg car is approximately 10 000 N. Step 5 of the suggested method would be used if the acceleration were not given. In this instance, the acceleration is known. If the acceleration were not known, then it would have to be calculated from speed and radius information.

• Step 6 of the suggested method involves the determination of an individual force - the normal force. This will involve a two-step process: first the net force (magnitude and direction) must be determined; then the net force must be used with the free body diagram to determine the normal force. This two-step process is shown below for the top and the bottom of the loop.

• Bottom of Loop• Fnet = m * a Fnet =

(1000 kg) * (20 m/s2, up)

• Fnet = 20 000 N, up• From FBD:• • Fnorm must be greater

than the Fgrav by 20000 N in order to supply a net upwards force of 20000 N. Thus,

• Fnorm = Fgrav + Fnet• Fnorm = 30 000 N

• Top of Loop• Fnet = m * a Fnet = (1000

kg) * (15 m/s2, down)• Fnet = 15 000 N, down• From FBD:• • Fnorm and Fgrav

together must combine together (i.e., add up) to supply the required inwards net force of 15000 N. Thus,

• Fnorm = Fnet - Fgrav• Fnorm = 5 000 N

• 3. The Cajun Cliffhanger at Great America is a ride in which occupants line the perimeter of a cylinder and spin in a circle at a high rate of turning. When the cylinder begins spinning very rapidly, the floor is removed from under the riders' feet. What effect does a doubling in speed have upon the centripetal force? Explain.

Answer

• Doubling the speed of the ride will cause the force to be four times greater than the original force.

• According to the eqn F = m (v2/R), force and velocity (v2) are directly proportional.

• So 2x speed would equal 4x force.

• 4. Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters.

answer

• Fnet = 533 N

• T = 2.93 s since 10 cycles take 29.3 s

• Find speed (6.22 m/s) then find acceleration (13.3 m/s2)

• Sample Speed Skater Problem

• Bonny Blair is ice skating at the Olympic games. She is making a sharp turn with a radius of 20.0 m and with a speed of 16.0 m/s. Use Newton's second law to determine the acceleration and the angle of lean of Bonnie's 55.0-kg body.

• Steps 1 and 2 involve the construction of a free body diagram and the identification of known and unknown quantities. This is shown in below.

Given Info:m = 55.0 kgv = 16.0 m/sr = 20.0 m

Find:a = ???

Angle of lean = ???

• Step 3 of the suggested method involves resolving any forces which act at angles into horizontal and vertical components. This is shown in the diagram at the right. The contact force can be broken into two components - Fhoriz and Fvert. The vertical component of force would balance the force of gravity; and as such, the vertical component will be equal in magnitude to the force of gravity.

• The horizontal component of force remains unbalanced. As mentioned in the above discussion, this horizontal component is the net inward force; and as such, Fhoriz is equal to m*a. Finally, the two components are related to the angle of lean by the tangent function. Simple algebraic manipulation would yield the relationship shown in the graphic at the right. So the angle of lean can be found if the vertical and horizontal component of force are known.

• Step 4 of the suggested method involves the determination of any known forces. In this case, the force of gravity can be determined from the equation Fgrav = m * g. An approximate g value of 10 m/s2 will be used in this problem in order to simplify some of the math and highlight the concepts. So the force of gravity acting upon Bonnie's 55.0-kg body is approximately 550 N. And since this force is balanced by the vertical component of the contact force, the Fvert is also 550 N. Step 5 involves determination of Bonnie's acceleration as she makes the turn. This can be accomplished by using the acceleration equation for circular motion.

• Step 4 of the suggested method involves the determination of any known forces. In this case, the force of gravity can be determined from the equation Fgrav = m * g. An approximate g value of 10 m/s2 will be used in this problem in order to simplify some of the math and highlight the concepts. So the force of gravity acting upon Bonnie's 55.0-kg body is approximately 550 N. And since this force is balanced by the vertical component of the contact force, the Fvert is also 550 N. Step 5 involves determination of Bonnie's acceleration as she makes the turn. This can be accomplished by using the acceleration equation for circular motion.

• Circular Motion in Football • A 90-kg GBS fullback is running a sweep around the left

side of the line. The fullback's path as seen from above is shown in the diagram. As he rounds the turn, he is momentarily moving in circular motion, sweeping out a quarter-circle with a radius of 4.0 meters. The fullback makes the turn with a speed of 5.0 m/s. Use a free-body diagram and your understanding of circular motion and Newton's second law to determine

• acceleration • Fgrav • Fnorm • Ffrict • Angle of lean

Answers

• a = v2 / R

• a = 6.25 m/s2

• Fnet = 563 N

• F grav = 900 N

• F horiz = F net = 563 N

• Theta = invtan (Fvert/Fhoriz) = 58 degrees

• Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.37 x 106 m from earth's center.

• The solution of the problem involves substituting known values of G (6.67 x 10-11 N m2/kg2), m1 (5.98 x 1024 kg ), m2 (70 kg) and d (6.37 x 106 m) into the universal gravitation equation and solving for Fgrav. The solution is as follows:

• The planet Jupiter is more than 300 times as massive as Earth, so it might seem that a body on the surface of Jupiter would weigh 300 times as much as on Earth. But it so happens a body would scarcely weigh three times as much on the surface of Jupiter as it would on the surface of the Earth. Explain why this is so.

Answer

• The effect of the greater mass of Jupiter is partly offset by the fact that the radius of Jupiter is larger. An object on Jupiter’s surface is 10 x farther from Jupiter’s center than it would be on Earth’s surface. So the 300-fold increase in the force (due to the greater mass) must be divided by 100 since the distance is 10x greater.