Physics - OAK PARK USD
AP Physics 4: Linear Momentum and Energy Name __________________________
A. Momentum and Energy formulas
1. impulse changes object's momentum: FΔt = mΔv = Δp
2. quantity of motion (vector quantity): p = mv (kg•m/s)
3. work changes object's energy: W = F||d (J)
a. measured in joules (1 J = 1 N•m)
a. Only component of F parallel to d does work
F
θ
d
1. F|| = Fcosθ ∴ W = (Fcosθ)d
2. include sign ∴ W > 0 when F → d →
3. F ↑↓ when d → (orbit)
b. variable force—stretching a spring
1. graph spring force (Fs) vs. position (x)
| Fs = kx | |
| | slope = k |x |
| | | |
| |Area = W | |
2. Fs = kx ∴ slope = ΔFs/Δx = k
3. W = Fsx ∴ area = ½x(kx) = ½kx2 = W
c. power is the rate work is done: P = W/t (W)
1. measured in watts (1 W = 1 J/s)
2. P = W/t = F(d/t) = Fvav (v is average)
3. graphing
a. P = W/t ∴ slope of W vs. t graph
b. P = Fvav ∴ area under F vs. v graph
4. kilowatt-hour, 1KWh = 3.6 x 106 J
4. mechanical energy
a. work-energy theorem: work done to an object increases mechanical energy; work done by an object decreases mechanical energy
b. scalar quantity, like work
c. kinetic energy—energy of motion
1. positive only
2. K = ½mv2 = p2/2m
|Steps |Algebra |
|start with |v2 = v02 + 2ad |
|assume Ko = 0 ∴ vo = 0 |v2 = 2ad |
|solve for ad |ad = ½v2 |
|start with |K = W = Fd |
|substitute ma for F |K = (ma)d = m(ad) |
|substitute ½v2 for ad |K = m(½v2) |
|rearrange |K = ½mv2 |
|start with |mv = p |
|square both sides |m2v2 = p2 |
|divide both sides by 2m |m2v2/2m = p2/2m |
|substitute K for ½mv2 |K = p2/2m |
d. potential energy—energy of relative position
1. gravitational potential energy
a. based on arbitrary zero
(usually closest or farthest apart)
b. Ug = mgh (near the Earth's surface)
|Steps |Algebra |
|start with |Ug = W = Fd |
|substitute mg for F |Ug = (mg)d |
|substitute h for d |Ug = mgh |
c. Ug = -GMm/r (orbiting system)
1. G = 6.67 x 10-11 N•m2/kg2
2. r = distance from center to center
3. Ug = 0 when r is ∞ ∴ Ug < 0 for all values of r because positive work is needed reach Ug = 0
2. spring (elastic) potential energy, Us = ½kx2
a. Us = W to stretch the spring
b. see work by a variable force above
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B. Solving Work-Energy Problems
1. work done on object A by a "nonconservative" force (push or pull, friction) results in the change in amount of mechanical energy
2. work done on object A by a "conservative" force (gravity, spring) results in the change in form of mechanical energy (U Δ K) for object A, but no loss in energy
a. conservative forces (Fg and Fs)
1. Fg ↓ d ↓: Ug → K, Fg ↓ d ↑: K → Ug
2. Fs ↓ d ↓: Us → K, Fs ↓ d ↑: K → Us
b. process isn't 100 % efficient
1. friction (W = Ffd) reduces mechanical energy
2. mechanical energy is converted into random kinetic energy of the object's atoms and the temperature increases = heat energy—Q
3. total energy is still conserved
3. examples
| | |Process |Energy |
|1 |Work done to |pull a pendulum bob off center |W → Ug |
| |Transformation |release pendulum |Ug → K |
| |Work done by |hit a stationary object |K → W |
|2 |Work done to |throw a ball into the air |W → K |
| |Transformation |ball rises and falls |K Δ Ug |
| |Work done by |falling ball dents ground |K → W |
|3 |Work done to |load a projectile in spring-gun |W → Us |
| |Transformation |release projectile |Us → K |
| |Work done by |projectile penetrates target |K → W |
4. general solution for work-energy problems
|determine initial energy of the object, Eo |
|if elevated h distance: Ug = mgh |
|if accelerated to v velocity: K = ½mv2 |
|if spring compressed x distance: Us = ½kx2 |
|determine energy added/subtracted due to an external push or pull: Wp = |
|±F||d |
|determine energy removed from the object by friction: Wf = Ffd = |
|(μmgcosθ)d |
|d is the distance traveled |
|θ is the angle of incline (0o for horizontal) |
|determine resulting energy, E' = Eo ± Wp – Wf |
|determine d, h, x or v |
|if slides a distance d: 0 = Eo ± Wp – μmgcosθd' |
|if elevates a height h: E' = mgh' |
|If compresses a spring x: Us: E' = ½kx'2 |
|if accelerated to velocity v: E' = ½mv'2 |
|general equation (not all terms apply for each problem) |
|K + Ug + Us ± Wp – Wf = K' + Ug'+ Us' |
|½mv2 + mgh + ½kx2 ± Fpd – Ffd = ½mv'2 + mgh' + ½kx'2 |
C. Solving Collision Problems
1. collision between particles doesn't change the total amount of momentum because the impulse on A equals the impulse on B, but in the opposite directions
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a. mAvA + mBvB = mAvA’ + mBvB’
|Steps |Algebra |
|start with Newton's Law |FA = -FB |
|multiply both side by Δt |FAΔt = -FBΔt |
|substitute mΔv for FΔt |mAΔvA = -mBΔvB |
|substitute v' – v for Δv |mA(vA’ – vA) = -mB(vB’ – vB) |
|collect like v terms |-mAvA – mBvB = -mAvA’ – mBvB’ |
|multiply by -1 |mAvA + mBvB = mAvA’ + mBvB’ |
b. two particles collide and stick together
1. inelastic collisions (vA' = vB')
2. mAvA + mBvB = (mA + mB)v’
|Steps |Algebra |
|start with |mAvA + mBvB = mAvA’ + mBvB’ |
|substitute v' for vA' and vB' |mAvA + mBvB = mAv’ + mBv’ |
|simplify |mAvA + mBvB = (mA + mB)v’ |
c. two particles collide and bounce off
1. elastic collisions (vA' ≠ vB')
2. both energy and momentum are conserved
a. two unknowns ∴ need two equations
b. mAvA + mBvB = mAvA’ + mBvB’
c. ½mAvA2 + ½mBvB2 = ½mAvA’2 + ½mBvB’2
d. simpler equation: vA + vA' = vB + vB’
|mAvA + mBvB = |½mAvA2 + ½mBvB2 = |
|mAvA’ + mBvB’ |½mAvA’2 + ½mBvB’2 |
|mA(vA – vA') = mB(vB' – vB) |mA(vA2 – vA'2) = mB(vB'2 – vB2) |
|mA(vA2 – vA'2) = mB(vB'2 – vB2) |
|mA(vA – vA') = mB(vB' – vB) |
|(vA – vA')(vA + vA') = (vB – vB')(vB + vB') |
|(vA – vA') = (vB' – vB) |
3. solving two equations and two unknowns
|fill in vA and vB into vA + vA' = vB + vB' |
|write expression for vA’ in terms of vB' |
|substitute vA' expression in equation: |
|mAvA + mBvB = mAvA’ + mBvB’ |
|solve for vB’ |
|solve for vA’ using the expression for vA' above |
4. collisions in two dimensions
a. px is conserved independently of py
b. elastic collision
1. mAvAx + mBvBx = mAvAx’ + mBvBx’
2. mAvAy + mBvBy = mAvAy’ + mBvBy’
3. solve two equations & two unknowns
c. inelastic collision
1. mAvAx + mBvBx = (mA + mB)vx'
2. mAvAy + mBvBy = (mA + mB)vy'
3. solve two equations & two unknowns
d. object explodes into two pieces mA and mB
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1. (mA + mB)v = mAvA’ + mBvB’
2. opposite inelastic collision equation
2. solve ballistics problems
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|? |
|m vm |
|bullet collides inelastically with block: mvm = (M + m)v' |
|block swings or slides (conservation of energy) |
|block swings like a pendulum to height h |
|K = Ug ∴ ½(M + m)v'2 = (M + m)gh ∴ h = v'2/2g |
|block slides a distance d along a rough surface |
|K = Wf ∴ ½(M + m)v'2 = μ(M + m)gd ∴ d = v'2/2μg |
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A. Momentum and Energy Formulas
Human Power Lab
Run up a flight of stairs at the football stadium and calculate the power that you generated.
a. Collect the following data.
|weight Fg-lbs | |Number of steps N | |
|time t | |height of step y | |
| | |width of step x | |
b. Calculate the following from the data.
| |Formula |Calculation |
|dy |dy = Ny | |
|dx |dx = Nx | |
|d |d = (dx2 + dy2)½ | |
|v |v = d/t | |
|Fg |Fg = Fg-lbs x 4.45 | |
|m |Fg = mg | |
|K |K = ½mv2 | |
|Ug |Ug = Fgdy | |
|P |P = (Ug + K)/t | |
c. Calculate the power in horse power (1 hp = 746 W).
| |
Questions 1-24 Briefly explain your answer.
1. An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it?
(A) increase (B) the same (C) decrease
|C—rain increases mass, but momentum is conserved and p = mv ∴ increase |
|mass = decrease v |
Questions 2-3 Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each for 1 s.
2. Which box has more momentum after the force acts?
(A) light (B) heavy (C) tie
|C—impulses are equal ∴ change in momentum is equal |
3. Which box has the greater velocity after the force acts?
(A) light (B) heavy (C) tie
|A—equal change in momentum: Δp = Δ(mv), but smaller mass would produce |
|greater change in v |
Questions 4-6 Use the following options
(A) W > 0 (B) W = 0 (C) W < 0
4. Work done by a person holding a 10-kg box.
|B—without displacement, work = 0 (W = Fd) |
5. Work done by a person pulling a 10-kg box across a rough floor at constant speed.
|A—force and distance are in opposite directions |
6. Work done by friction when a 10-kg box is pulled across a rough floor at constant speed.
|C—force and distance are in the opposite directions |
7. A box is being pulled up a rough incline by a rope. How many forces are doing work on the box?
(A) 1 (B) 2 (C) 3 (D) 4
|C—friction, gravity and tension |
Questions 8-9 Alice applied 10 N of force over 3 m in 10 s. Bill applies the same force over the same distance in 20 s.
8. Who did more work?
(A) Alice (B) Bill (C) They did the same
|C—W = fd, which is the same for Alice and Bill |
9. Who produced the greater power?
(A) Alice (B) Bill (C) They did the same
|A—P = W/t, since Alice did the work in less time she produced greater |
|power |
10. Car A has twice the mass of car B, but they both have the same kinetic energy. How do their speeds compare?
(A) 2vA = vB (B) √2vA = vB (C) 4vA = vB
|B—½mAvA2 = ½mBvB2 |
|2mBvA2 = mBvB2 → 2vA2 = vB2 ∴ √2vA = vB |
11. A system of particles has total kinetic energy of zero. What can you say about the total momentum of the system?
(A) p > 0 (B) p = 0 (C) p < 0
|B—kinetic energy is zero only when v = 0 ∴ p = mv = 0 |
12. A system of particles has total momentum of zero. Does it necessarily follow that the total kinetic energy is zero?
(A) yes (B) no
|B—momentum is a vector quantity, a system of particles could be moving, |
|but in opposite directions |
13. Two objects have the same momentum. Do these two objects also have the same kinetic energy?
(A) yes (B) no
|B—K = ½mv2, where p = mv; since v is squared in K, but not p, then |
|differences in v will not have the same effect |
Questions 14-15 Stone A has twice the mass as stone B. They are dropped from a cliff and reach a point just above ground.
14. What is the speed of stone A compared to stone B?
(A) vA = vB (B) vA = ½vB
(C) vA = 2vB (D) vA = 4vB
|A—acceleration due to gravity is equal and they fall the same distance ∴|
|velocities must be the same |
15. What is the kinetic energy of stone A compared to stone B?
(A) KA = KB (B) KA = ½KB
(C) KA = 2KB (D) KA = 4KB
|C—stone A has twice the mass as B ∴ it has twice the K |
Questions 16-17 Car A and Car B are identical, but car A has twice the velocity of car B.
16. What is the kinetic energy of car A compared to car B?
(A) KA = KB (B) KA = ½KB
(C) KA = 2KB (D) KA = 4KB
|D—K = ½mv2 ∴ car A has 4 times the kinetic energy |
17. Car A takes d meters to stop. What is the stopping distance for car B?
(A) 2d (B) 4d (C) ½d (D) ¼d
|D—K = Wf = μmgd; car B has ¼ the kinetic energy ∴ it takes ¼ the |
|distance |
18. The work Wo accelerates a car for 0 to v. How much work is needed to accelerate the car from v to 3v?
(A) 2Wo (B) 3Wo (C) 8Wo (D) 9Wo
|C—W = ΔK: ΔK 0 → v ∝ v2 - 02 = v2 |
|ΔK from v to 3v ∝ (3v)2 – v2 = 8v2 |
19. A golfer making a putt gives the ball an initial velocity of vo, but he has misjudged the putt, and the ball only travels one-quarter of the distance to the hole. What speed should he have given the ball?
(A) 2vo (B) 3vo (C) 4vo (D) 8vo
|A—he needs 4 x the kinetic energy, which requires 2 x v (K = ½mv2) |
20. Which can never be negative?
(A) W (B) K (C) U (D) p
|B—mass can't be negative and v is squared ∴ K = ½mv2 is always positive |
21. You and your friend both solve a problem involving a skier going down a slope, starting from rest. The two of you have chosen different levels for h = 0 in this problem. Which of the quantities will you and your friend agree on?
I. Ug II. ΔUg III. K
(A) I only (B) II only (C) III only (D) II and III
|D—your absolute value of Ug is different, but the change is the same and|
|K depends on v not h |
22. Two paths lead to the top of a big hill. A is steep and direct, while the B is twice as long but less steep. The change in potential energy on path A compared to path B is
(A) UA < UB (B) UA = UB (C) UA > UB
|B—the beginning and end points are the same ∴ Δh is the same |
23. How does the work required to stretch a spring 2 cm compare with the work required to stretch it 1 cm?
(A) W2 = W1 (B) W2 = 2W1
(C) W2 = 4W1 (D) W2 = 8W1
|C—W = ΔUs = ½k(Δx)2, since Δx is 2 x then W = 22 x |
24. A mass attached to a vertical spring causes the spring to stretch and the mass to move downward. Which is true about the sign for ΔUs and ΔUg?
(A) +ΔUs, +ΔUg (B) +ΔUs, –ΔUg
(C) –ΔUs, +ΔUg (D) –ΔUs, –ΔUg
|B—ΔUs = ½k(Δx)2: since Δx is positive ΔUs is positive |
|ΔUg = mgΔh: since Δh is negative ΔUg is negative |
25. What is the momentum of a 0.5-kg ball traveling at 18 m/s?
|p = mv = (0.5 kg)(18 m/s) p = 9.0 kg•m/s |
26. What force is generated by a racket, which strikes a 0.06-kg tennis ball that reaches a speed of 65 m/s in 0.03 s?
|FΔt = mΔv |
|F(0.03 s) = (0.06 kg)(65 m/s)F = 130 N |
27. How much work is done to move a 50-kg crate horizontally 10 m against a 150-N force of friction?
|W = F||d = (150 N)(10 m) = 1500 J |
28. How much work is done to pull a 100-kg crate horizontally 10 m using a force of 100 N at 30o?
|W = F||d = (100 N)(cos30)(10 m) = 866 J |
29. How much work is done to carry a 100-kg crate 10 m up a 30o ramp?
|W = F||d = (1000 N)(sin30)(10 m) = 5000 J |
30. Why is work not needed to keep the earth orbiting the sun?
|Work is not needed to keep the Earth orbiting the sun because there is |
|no force in the direction of motion. |
31. How much power is needed to change the speed of a 1500-kg car from 10 m/s to 20 m/s in 5 s?
|P = W/t = ½m(v2 – vo2)/t |
|P = ½(1500)(202 – 102)/5 = 45,000 W |
32. How much power does a 75-kg person generate when climbing 50 steps (rise of 25 cm per step) in 12 s?
a. in Watts
|P = W/t = ΔU/t = mgh/t |
|P = (75 kg)(10 m/s2)(50 x 0.25 m)/12 s = 780 W |
b. in horse power (1hp = 746 W)
|780 W x 1 hp/746 W = 1.05 hp |
33. How much power is needed to maintain a speed of 25 m/s against a total friction force of 200 N?
|P = Fvav = (200 N)(25 m/s) = 5000 W |
34. How long will it take a 1750-W motor to lift a 285-kg piano to a sixth-story window 16 m above?
|P = W/t = mgh/t |
|1750 W = (285 kg)(10 m/s2)(16 m)/t ∴ t = 26 s |
35. A 1000-kg car travels at 30 m/s against 600-N of friction.
a. How much power does the car engine deliver?
|P = Fvav = (600 N)(30 m/s) = 18,000 W |
b. The car accelerates at 2 m/s2. How much power does the car engine deliver now?
|P = Fvav = (ma + Ff)vav |
|P = [(1000 kg)(2 m/s2) + 600 N](30 m/s) = 78,000 W |
c. The car goes up a 10o incline at 30 m/s. How much power does the car engine deliver now?
|P = Fvav = (mgsinθ + Ff)vav |
|P = [(1000 kg)(10 m/s2)(sin10) + 600 N](30 m/s) = 70,000 W |
36. What is the kinetic energy of a 2-kg block moving at 9 m/s?
|K = ½mv2 = ½(2 kg)(9 m/s)2 = 81 J |
37. What is the gravitational potential energy of a 2-kg block that is 6 m above zero potential energy?
|Ug = mgh = (2 kg)(10 m/s2)(6 m) = 120 J |
38. What is the gravitational potential energy of the earth-moon system? (MEarth = 5.97 x 1024 kg, Mmoon = 7.35 x 1022 kg, distance between Earth and moon, r = 3.84 x 108 m)
|Ug = -GMm/r |
|Ug = -(6.67 x 10-11)(5.97 x 1024)(7.35 x 1022)/3.84 x 108 |
|Ug = -7.6 x 1028 J |
39. Consider the following spring (k = 100 N/m).
a. Calculate the force (F = kx) needed to stretch a spring from 0.0 m to 0.5 m.
|x (m) |0.0 |0.1 |0.2 |0.3 |0.4 |0.5 |
|F (N) |0 N |10 N |20 N |30 N |40 N |50 N |
b. Graph the data
|40 N | | | | | |
| | | | | | |
|20 N | | | | | |
| | | | | | |
|0 N | | | | | |
| | |0.2 m |0.4 m |
c. Calculate the area under the graph.
|A = ½BH = ½(0.50)(50) = 12.5 N•m |
d. How much potential energy is stored in the stretched spring?
|Us = ½kx2 = ½(100)(0.50)2 = 12.5 J |
B. Solving Work-Energy Problems
Questions 40-47 Briefly explain your answer.
40. Three balls of equal mass start from rest and roll down different ramps. All ramps have the same height. Which ball has the greatest speed at the bottom of the ramp?
? ? ?
A B C
(A) A (B) B (C) C (D) All the same
|D—the same amount of Ug is converted to K ∴ same speed |
41. A stationary block slides down a frictionless ramp and attains a speed of 2 m/s. To achieve a speed of 4 m/s, how many times higher must the block start from?
(A) 2 times (B) 4 times (C) 8 times
|B—a speed of 4 m/s requires 4 x as much K ∴ the ball must lose 4 x Ug = |
|4h |
42. A box sliding on a frictionless flat surface runs into a fixed spring, which compresses a distance x to stop the box. If the initial speed of the box is doubled, how much would the spring compress?
(A) ½x (B) x (C) 2x (D) 4x
|C—Us = K → ½kx2 = ½mv2 ∴ x ∝ v (double v = double x) |
Questions 43-44 Alice and Bill start from rest at the same elevation on frictionless water slides with different shapes.
? Bill
? Alice
43. At the bottom whose velocity is greater?
(A) Alice (B) Bill (C) tie
|C—Ug = K → mgh = ½mv2 ∴ v = (2gh)½, since h is the same v is the same |
44. Who makes it to the bottom in the least amount of time?
(A) Alice (B) Bill (C) both are the same
|A—Alice is faster at all times except the very end (she has fallen a |
|greater h at each instance) |
45. A cart starting from rest rolls down a hill and at the bottom has a speed of 4 m/s. If the cart were given a push, so its initial speed at the top of the hill was 3 m/s, what would be its speed at the bottom?
(A) 4 m/s (B) 5 m/s (C) 6 m/s (D) 7 m/s
|B—K' = Ko + Ug |
|½mv2 = ½m(3)2 + ½m(4)2 → v2 = 32 + 42 = 25 ∴ v = 5 m/s |
46. You see a leaf falling to the ground with constant speed. When you first notice it, the leaf has initial total mechanical energy Ei. You watch the leaf until just before it hits the ground, at which point it has final total mechanical energy Ef. How do these total energies compare?
(A) Ei < Ef (B) Ei = Ef (C) Ei > Ef
|C—energy is lost due to air resistance |
47. You throw a ball straight up into the air. In addition to gravity, the ball feels a force due to air resistance. Compared to the time it takes the ball to go up, the time to come back down is
(A) less (B) equal (C) greater
|C—at each elevation, the ball is traveling faster on the way up compared|
|to the way down |
48. A rock is dropped from 20 m. What is the final velocity?
a. Use kinematics to solve this problem.
|v2 = vo2 + 2ad |
|v2 = 2(-10 m/s2)(-20 m) = ±20 m/s |
b. Use energy to solve this problem.
|Ug-0 + K0 = Ug-t + Kt → mgh = ½mv2 |
|(10 m/s2)(20 m) = ½v2 ∴ v = ±20 m/s |
49. A pendulum bob reaches a maximum height of 0.6 m above the lowest point in the swing, what is its fastest speed?
|Ug-0 + K0 = Ug-t + Kt → mgh = ½mv2 |
|(10 m/s2)(0.6 m) = ½v2 ∴ v = 3.5 m/s |
50. How far must a 1 kg ball fall in order to compress a spring 0.1 m? (k = 1000 N/m)
|Ug-0 + Us-0 = Ug-0 + Us-0 → mgh = ½kx2 |
|(1)(10 m/s2)h = ½(1000 kg)(0.1 m/s)2 ∴ h = 0.50 m |
51. A 10-kg box is initially at the top of a 5-m long ramp set at 53o. The box slides down to the bottom of the ramp. The force of friction is 31 N. Determine the
a. potential energy at the top of the ramp.
|Ug = mgh = mgLsinθ |
|Ug = (10 kg)(10 m/s2)(5 m)sin53 = 400 J |
b. work done by friction during the slide.
|Wf = Ffd = (-31 N)(5 m) = -155 J |
c. velocity of the box at the bottom of the ramp.
|Ug + Wf = Kt = ½mv2 |
|400 J – 155 J = ½(10 kg)v2 ∴ v = 7 m/s |
52. A spring (k = 500 N/m) is attached to the wall. A 5-kg block on a horizontal surface (μ = 0.25) is pushed against the spring so that the spring is compressed 0.2 m. The block is released and propelled across the surface.
a. Determine the potential energy of the spring.
|Us = ½kx2 = ½(500 N/m)(0.2 m)2 = 10 J |
b. Determine the distance that the block travels.
|Us = Wf = Ffd = μmgd |
|10 J = (0.25)(5 kg)(10 m/s2)d ∴ d = 0.8 m |
C. Solving Collision Problems
Collision Lab
Observe an elastic collision between a swinging 200-g weight and a golf ball, and compare the actual post-collision velocities with the theoretical velocities.
a. Collect the following data.
|200-g Weight (A) |Golf Ball (B) |
|string length LA | |mass mB | |
|initial angle θA |25o |table height dyB | |
|final angle θA' | |distance dxB' | |
| | | |
|dyA |dyA = LA(1 – cosθA) | |
|vA |vA2 = 2gdyA | |
|dyA'|dyA' = LA(1 – cosθA') | |
|vA' |vA'2 = 2gdyA' | |
|tB' |dyB = ½gtB'2 | |
|vB' |vB' = dxB'/tB' | |
c. Calculate the following theoretical values.
| |Formula |Calculation |
|vA' |vA = -(vA' – vB') | |
| | | |
| |mAvA = mAvA' + mBvB' | |
|vB' |vB' = vA + vA' | |
d. Calculate the percent differences.
| |Formula |Calculation |
|vA' |% = 100|Δv|/vtheory | |
|vB' | | |
Questions 53-59 Briefly explain your answer.
53. A small car and a large truck collide head-on and stick together. Which one has larger momentum change?
(A) car (B) truck (C) tie
|C—impulses are equal (Newton's third law) ∴ change in momentums are |
|equal |
54. A small beanbag and a bouncy rubber ball are dropped from the same height above the floor. The both have the same mass. Which would hurt more if it hit you on the head?
(A) beanbag (B) rubber ball (C) doesn't matter
|B—the rubber ball would bounce back which would generate a greater Δp |
|and impulse force |
55. A box slides with initial velocity 10 m/s on a frictionless surface and collides inelastically with a stationary identical box. What is the final velocity of the combined boxes?
(A) 0 m/s (B) 5 m/s (C) 10 m/s (D) 20 m/s
|B—mAvA + mBvB = (mA + mB)v' |
|mA(10 m/s) + 0 = 2mAv' ∴ v' = 5 m/s |
Questions 56-57 A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light.
56. Which fragment has the greater momentum?
(A) heavier one (B) lighter one (C) tie
|C—initial momentum = final momentum = zero ∴ equal momenta in opposite |
|directions |
57. Which fragment has the greater speed?
(A) heavier one (B) lighter one (C) tie
|B—momentums are equal, p = mv, but with less mass, the lighter one has |
|greater speed |
58. Alice (50 kg) and Bill (75 kg) are standing on slippery ice and push off of each other. If Alice slides at 6 m/s, what speed does Bill have?
(A) 2 m/s (B) 3 m/s (C) 4 m/s (D) 6 m/s
|C—0 = (mAvA' + mBvB' |
|0 = (50 kg)(6 m/s) + (75 kg )vB' ∴ vB' = -4 m/s |
59. A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannon ball is fired at a speed of 50 m/s, what is the recoil speed of the flatcar?
(A) 0 m/s (B) 0.5 m/s (C) 10 m/s (D) 50 m/s
|B—0 = (mA + mB)v' |
|0 = (10)(50 m/s) + (1000)vB' ∴ vB' = -0.5 m/s |
60. A 25-kg child in a stationary 55-kg boat with a 5-kg package throws the package out horizontally at 8 m/s. What is the boat and child's resultant velocity?
|0 = mAvA' + mBvB' |
|0 = (25 kg + 55 kg)v1' + (5 kg)(8 m/s)v = -0.5 m/s |
61. An 85-kg safety running at 5 m/s tackles a 95-kg fullback traveling at 4 m/s from behind. What is their mutual speed just after the tackle?
|mAvA + mBvB = (mA + mB)v' |
|(85)(5) + (95)(4) = (85+ 95)v' ∴ v' = 4.47 m/s |
62. A 0.45-kg ice puck, moving east with a speed of 3.0 m/s, has a head-on elastic collision with a 0.9-kg puck initially at rest. What are the resulting speeds and directions?
|vA + vA' = vB + vB' |
|3.0 + vA' = 0 + vB' ∴ vB' = vA' + 3.0 |
| |
|mAvA + mBvB = mAvA' + mBvB' |
|(.45 kg)(3.0 m/s) = (45 kg)(vA') + (.9 kg)(vA' + 3.0) |
|vA' = -1 m/s (west) |
| |
|vB' = vA' + 3.0 = -1 m/s + 3 = 2 m/s (east) |
63. A 1-kg block traveling at 5 m/s in the direction of 30o south of east collides and sticks with a 2-kg block traveling north at 3 m/s. Determine
a. The x-component of the resulting velocity, vx'.
|px: mAvAcosθA + mBvBcosθB = (mA + mA)vx' |
|(1)(5)cos-30 + 0 = (1 + 2)vx' ∴ vx' = 1.4 m/s |
b. The y-component of the resulting velocity, vy'.
|py: mAvAsinθA + mBvBsinθB = (mA + mA)vy' |
|(1.0)(5)sin-30 + (2)(3)sin90 = (1 + 2)vy' ∴ vy' = 1.2 m/s |
c. The resultant speed.
|v' = (vx'2 + vy'2)½ |
|v' = [(1.4 m/s)2 + (1.2 m/s)2]½ = 1.9 m/s |
d. The resultant direction.
|tanθ = vy'/vx' |
|tanθ = 1.2 m/s/1.4 m/s = 0.81 ∴ θ = 54o |
64. A 0.50-kg softball is traveling at 40 m/s. A bat makes contact with the ball for 0.025 s, after which, the ball's velocity is 35 m/s in the opposite direction (Δv = -75 m/s).
a. Determine the change in the ball's momentum.
|Δp = mΔv = (0.5 kg)(-75 m/s) |
|Δp = -37.5 kg•m/s |
b. Determine the average force exerted by the bat.
|FΔt = mΔv |
|F(0.025 s) = -37.5 kg•m/s ∴ F = -1500 N |
65. Determine the velocity after the following inelastic collisions between mA (1 kg) and mB (3 kg).
|vA = + 2 m/s |θA = 0o |vB = -1 m/s |θB = 0o |
|mAvA + mBvB = (mA + mB)v' |
|(1 kg)(2 m/s) + (3 kg)(-1 m/s) = (1 kg + 3 kg)v' |
|v' = -0.25 m/s (west) |
|vA = + 5 m/s |θA = 53o |vB = 3 m/s |θB = 90o |
|px: mAvAcosθA + mBvBcosθB = (mA + mB)vx' |
|(1)(5cos53) + (3)(3cos90) = (1 + 3)vx' ∴ vx' = 0.75 m/s |
| |
|py: mAvAsinθA + mBvBsinθB = (mA + mB)vy' |
|(1)(5sin53) + (3)(3sin90) = (1 + 3)vy' ∴ vy' = 3.25 m/s |
| |
|v' = (vx'2 + vy'2)½ = [(.75 m/s)2 + (3.25 m/s)2]½ |
|v' = 3.33 m/s |
| |
|tanθ = vy'/vx' = 3.25 m/s/0.75 m/s = 4.33 ∴ θ = 77o |
66. Stationary mA (1 kg) and mB (3 kg) are separated by a compressed spring, which is then released. What is the resulting velocity of mB, when vA is -12 m/s?
|(mA + mB)v = mAvA' + mBvB' |
|0 = (1 kg)(-12 m/s) + (3 kg)vB' |
|vB' = 4 m/s (east) |
67. Determine the velocities after the following elastic collisions between mA (1 kg) and mB (3 kg).
|vA = + 2 m/s |θA = 0o |vB = 0 m/s |θB = 0o |
|vA + vA' = vB + vB' |
|2 + vA' = 0 + vB' ∴ vB' = vA' + 2 |
| |
|mAvA + mBvB = mAvA' + mBvB' |
|(1 kg)(2 m/s) = (1 kg)(vA') + (3 kg)(vA' + 2) |
|vA' = -1 m/s (west) |
| |
|vB' = vA' + 2 = -1 m/s + 2 = 1 m/s (east) |
68. A 15-g bullet penetrates a 1.1-kg block of wood. As a result, the block slides along a surface (μ = 0.85) for 9.5 m.
a. How much work is done by friction?
|W = F||d = Ffd = μmgd |
|W = (0.85)(1.1 kg)(10 m/s2)(9.5) = 89 J |
b. What is the velocity of the system just after the impact?
|W = K = ½mv2 |
|89 J = ½(1.1 kg)v2 ∴ v = 13 m/s |
c. What is the velocity of the bullet just before the impact?
|mAvA + mBvB = (mA + mB)v' |
|(0.015 kg)vA + 0 = (1.1 kg)(13 m/s) ∴ vA = 950 m/s |
69. An 18-g bullet traveling at 230 m/s buries itself in a 3.6-kg pendulum hanging on a 2.8-m long string.
a. Determine for the bullet/pendulum just after impact.
(1) velocity
|mAvA + mBvB = (mA + mB)v' |
|(0.018 kg)(230 m/s) + 0 = (3.6 kg)v' ∴ v' = 1.1 m/s |
(2) kinetic energy
|K = ½mv2 = ½(3.6 kg)(1.1 m/s)2 = 2.2 J |
b. How high does the pendulum rise?
|K = Ug = mgh |
|2.2 J = (3.6 kg)(9.8 m/s2)h ∴ h = 0.062 m |
70. Andre hits a 0.06-kg tennis ball straight up into the air with a 300-N force. The ball remains on the racket for 0.25 m.
a. Using dynamics and kinematics, determine
(1) The acceleration.
|Fnet = ma |
|300 N = (0.06 kg)a ∴ a = 5000 m/s2 |
(2) The initial velocity.
|v2 = vo2 + 2ad |
|v2 = vo2 + 2ad = 2(5000 m/s2)(0.25 m) ∴ v = 50 m/s |
(3) The maximum height reached by the tennis ball.
|v2 = vo2 + 2ad |
|0 = (50 m/s)2 + 2(-10 m/s2)d ∴ d = 125 m |
b. Use energy to determine the maximum height reached by the tennis ball.
|F||d = mgh |
|(300 N)(0.25 m) = (0.06 kg)(10 m/s2)h ∴ h = 125 m |
71. How much power is used to lift 100 kg a distance of 2 m in 4 s?
|P = W/t = mgh/t |
|P = (100 kg)(10 m/s2)(2 m)/(4 s) = 500 W |
72. A 1,000-kg car maintains a constant speed of 30 m/s against a combined friction and air resistance force of 550 N.
a. How much power is needed to cruise at 30 m/s?
|Pf = Ffvav |
|Pf = (550 N)(30 m/s) = 16,500 W |
b. How steep an incline can the car climb if the engine can generate 50,000 W of power?
|P = Pclimb + Pf = Fgv + Pf = mgsinθ + Pf |
|50,000 W = (10,000sinθ N)(30 m/s) + 16,500 W ∴ θ = 6o |
73. A 1-kg block is pushed down against a spring (k = 500 N/m), which is compressed 0.1 m. The block is released and propelled vertically.
a. Determine the potential energy of the spring.
|U = ½kx2 |
|U = ½(500 N/m)(0.1 m)2 = 2.5 J |
b. Determine the maximum height reached by the block.
|Us = Ug = mgh |
|2.5 J = (1 kg)(10 m/s2)h ∴ h = 0.25 m |
74. A 10-kg box is initially at the top of a 5-m long ramp set at 30o. The box slides down to the bottom of the ramp. The force of friction is 26 N. Determine the
a. potential energy at the top of the ramp.
|Ug = mgh = mgLsinθ |
|Ug = (10 kg)(10 m/s2)(5 m)sin30 = 250 J |
b. work done by friction during the slide.
|Wf = Ffd |
|W = (26 N)(5 m) = 130 J |
c. velocity of the box at the bottom of the ramp.
|Ug – Wf = K' → Ug – Wf = ½mv2 |
|250 J – 130 J = ½(10 kg)v2 ∴ v = 4.9 m/s |
75. A 0.050-kg bullet traveling at 1,000 m/s penetrates a 10-kg block of wood.
a. What is the velocity of the block after impact?
|mAvA + mBvB = (mA + mB)v' |
|(0.050 kg)(1,000 m/s) + 0 = (10.05 kg)v' ∴ v' = 5 m/s |
b. How far does the block travel along a rough surface
(μ = 0.25) before stopping?
|K = W = F||d = Ffrd ∴ ½mv2 = μmgd |
|½(5 m/s)2 = (0.25)(10 m/s2)d ∴ d = 5 m |
c. How high does the block rise if it were suspended from a long string?
|K = Ug ∴ ½mv2 = mgh |
|½(5 m/s)2 = (10 m/s2)h ∴ h = 1.25 m |
Ballistic Pendulum Lab
Measure the velocity of a projectile using pendulum data, kinematic data and the velocimeter, and comparing the three values.
a. Collect the following data.
|Pendulum |Kinematics |
|bearing mass, mA | |height, dy | |
|pendulum mass, mB | |distance, dx | |
|pendulum angle, θ | | | | | |
| | | | |Velocimeter |
|lowest height, ho | | | | | |
|height at θ, hθ | |
b. Calculate the following from the data.
| |Formula |Calculation |
|Pendulum |
|h |h = hθ – ho | |
|Ug |Ug = (mA + mB)gh | |
|K |K = Ug | |
|v' |K = ½(mA + mB)v'2 | |
|vA |mAvA + 0 = (mA + mB)v' | |
|Kinematics |
|t |dy = ½gt2 | |
|vA |vA = dx/t | |
|Velocimeter |
|vA |vA = (vkm/hr)/3.6 | |
c. Calculate the percent differences with the velocimeter.
| |Formula |Calculation |
|P |% = 100|Δv|/vvelocimeter | |
|K | | |
Practice Multiple Choice (No calculator)
Briefly explain why the answer is correct in the space provided.
|1 |
2. A 3-kg block, initially at rest, is pulled along a frictionless, horizontal surface with a force shown as a function of time t by the graph.
The speed of the block at t = 3 s is
(A) 3 m/s (B) 4 m/s (C) 6 m/s (D) 8 m/s
|J = FΔt = area under the graph = ½(6 N)(3 s) = 9 N•s |
|J = mΔv → 9 N•s = (3 kg)v ∴ v = 3 m/s |
3. Two pucks, where mI = 3mII, are attached by a stretched spring and are initially held at rest on a frictionless surface.
[pic]
The pucks are released simultaneously. Which is the same for both pucks as they move toward each other?
(A) Speed (B) Velocity
(C) Acceleration (D) Magnitude of momentum
|mIvI = mIIvII and mI = 3mII, then vII = 3 mI ∴ speed, velocity and |
|acceleration will be different. |
4. A 2,000-kg railroad car rolls to the right at 10 m/s and collides and stick to a 3,000-kg car that is rolling to the left at 5 m/s. What is their speed after the collision?
(A) 1 m/s (B) 2.5 m/s (C) 5 m/s (D) 7 m/s
|mAvA + mBvB = (mA + mB)v' |
|(2,000)(10) + (3,000)(-5) = (5,000)v' ∴ v' = 1 m/s |
5. A 5-kg block with momentum = 30 kg•m/s, sliding east across a horizontal, frictionless surface, strikes an obstacle. The obstacle exerts all impulse of 10 N•s to the west on the block. The speed of the block after the collision is
(A) 4 m/s (B) 8 m/s (C) 10 m/s (D) 20 m/s
|po + J = pt → 30 kg•m/s + (- 10 N•s) = 20 kg•m/s |
|pt = mv ∴ v = pt/m = (20 kg•m/s)/(5 kg) = 4 m/s |
6. In the diagram, a block of mass M initially at rest on a frictionless horizontal surface is struck by a bullet of mass m moving with horizontal velocity v.
[pic]
What is the velocity of the bullet-block system after the bullet embeds itself in the block?
(A) (M + v)m/M (B) (m + v)m/M
(C) (m + M)v/M (D) mv/(m + M)
|mAvA + mBvB = (mA + mB)v' |
|mv + 0 = (m + M)v' ∴ v' = mv/(m + M) |
7. A disc of mass m is moving to the right with speed v when it collides and sticks to a second disc of mass 2m. The second disc was moving to the right with speed v/2.
The speed of the composite body after the collision is
(A) v/3 (B) v/2 (C) 2v/3 (D) 3v/2
|mAvA + mBvB = (mA + mB)v' |
|mv + (2m)(v/2) = (m + 2m)v' → v + v = 3v' ∴ v' = 2v/3 |
8. An object of mass m is moving with speed vo to the right on a horizontal frictionless surface when it explodes into two pieces. Subsequently, one piece of mass 2/5m moves with a speed ½vo to the left. The speed of the other piece of the object is
(A) vo/2 (B) vo/3 (C) 7vo/5 (D) 2 vo
|mv = m1v1' + m2v2' → mvo = (2/5m)(-1/2vo) + (3/5m)v2' |
|vo = -2/10vo + 3/5v2' ∴ v2' = 2vo |
9. Two objects of mass 0.2 kg and 0.1 kg, respectively, move parallel to the x-axis. The 0.2 kg object overtakes and collides with the 0.1 kg object. Immediately after the collision, the y-component of the velocity of the 0.2 kg object is 1 m/s upward.
What is the y-component of the velocity of the 0.1 kg object immediately after the collision?
(A) 2 m/s downward (B) 0.5 m/s downward
(C) 0 m/s (D) 0.5 m/s upward
|mAvAy + mBvBy = mAvAy' + mBvBy' |
|0 = (0.2 kg)(1 m/s) + (0.1 kg)vBy' ∴ vBy' = -2 m/s |
10. Two particles of equal mass mo
moving with equal speeds vo
along paths inclined at 60° to the
x-axis, collide and stick together.
Their velocity after the collision
has magnitude
(A) vo/4 (B) vo/2
(C) √vo/2 (D) √3vo/2
|vy' = 0, movocos60 + movocos60 = (mo + mo)vx' |
|movo(½) + movo(½) = 2movx' ∴ vx' = vo/2 |
11. Student A lifts a 50-N box to a height of 0.4 m in 2.0 s. Student B lifts a 40-N box to a height of 0.50 m in 1.0 s. Compared to student A, student B does
(A) the same work but develops more power
(B) the same work but develops less power
(C) more work but develops less power
(D) less work but develops more power
|WA = F||d = (50 N)(.4 m) = 20 J = WB = (40 N)(.50 m) = 20 J |
|PA = W/t = 20 J/2.0 s = 10 W < PB = 20 J/1.0 s = 20 W ∴ |
12. What is the change in gravitational potential energy for a 50-kg snowboarder raised a vertical distance of 400 m?
(A) 50 J (B) 200 J (C) 20,000 J (D) 200,000 J
|Ug = mgh = (50 kg)(10 m/s2)(400 m) = 200,000 J |
13. How high is a 50-N object moved if 250 J of work is done against the force of gravity?
(A) 2.5 m (B) 10 m (C) 5 m (D) 25 m
|W = F||d |
|250 J = (50 N)d ∴ d = 5 m |
14. What is the spring potential energy when a spring (k = 80 N/m) is stretched 0.3 m from its equilibrium length?
(A) 3.6 J (B) 12 J (C) 7.2 J (D) 24 J
|Us = ½kx2 = ½(80 N/m)(0.3 m)2 = 3.6 J |
15. What is the kinetic energy of a 5-kg block that slides down an incline at 6 m/s?
(A) 20 J (B) 90 J (C) 120 J (D) 240 J
|K = ½mv2 = ½(5 kg)(6 m/s)2 = 90 J |
Questions 16-17 A weight lifter lifts a mass m at constant speed to a height h in time t.
16. How much work is done by the weight lifter?
(A) mg (B) mh (C) mgh (D) mght
|W = F||d = mgh |
17. What is the average power output of the weight lifter?
(A) mg (B) mh (C) mgh (D) mgh/t
|P = W/t = mgh/t |
18. The graphs show the position d versus time t of three objects that move along a straight, level path.
[pic]
Which has no change in kinetic energy?
(A) II only (B) III only (C) I and II (D) I and III
|ΔK = 0 ∴ constant velocity (no acceleration). In I velocity is constant|
|and II is stationary. |
19. Which is a scalar quantity that is always positive or zero?
(A) Power (B) Work
(C) Kinetic energy (D) Potential Energy
|All energy/power values are scalar, kinetic energy can only be positive |
|(v2). |
Questions 20-21 A constant force of 900 N pushes a 100 kg mass up the inclined plane at a uniform speed of 4 m/s.
[pic]
20. The power developed by the 900-N force is
(A) 400 W (B) 800 W (C) 900 W (D) 3600 W
|P = Fv = (900 N)(4 m/s) = 3600 W |
21. The gain in potential energy when the mass goes from the bottom of the ramp to the top.
(A) 100 J (B) 500 J (C) 1000 J (D) 2000 J
|Ug = mgh = (100 kg)(10 m/s2)(1 m) = 1000 J |
22. What is the maximum height that a 0.1-kg stone rises if 40 J of work is used to shoot it straight up in the air?
(A) 0.4 m (B) 4 m (C) 40 m (D) 400 m
|W = Ug = mgh |
|40 J = (0.1 kg)(10 m/s2)h ∴ h = 40 m |
23. A 40-N block is released from rest on an incline 8 m above the horizontal.
What is the kinetic energy of the block at the bottom of the incline if 50 J of energy is lost due to friction?
(A) 50 J (B) 270 J (C) 320 J (D) 3100 J
|Initial energy, Ug = mgh = (40 N)(8 m) = 320 J |
|Ug – Wf = K → 320 J – 50 J = 270 J |
24. A 50-kg diver falls freely from a diving platform that is 3 m above the surface of the water. What is her kinetic energy at 1 m above the water?
(A) 0 (B) 500 J (C) 750 J (D) 1000 J
|mgh3 = K + mgh1 |
|(50)(10)(3) = K + (50)(10)(1) ∴ K = 1000 J |
25. A 1000 W electric motor lifts a 100 kg safe at constant velocity. The vertical distance through which the motor can raise the safe in 10 s is most nearly
(A) 1 m (B) 3 m (C) 10 m (D) 32 m
|P = W/t = mgh/t |
|1000 W = (100 kg)(10 m/s2)h/10 s ∴ h = 10 m |
Questions 26-28 The vertical height versus gravitational potential energy for an object near earth's surface is graphed below.
Ug (J)
|80 | | | | | |
26. What is Ug when the object is 2.25 m above the surface?
(A) 50 J (B) 45 J (C) 60 J (D) 55 J
|Ug at 2.25 m is 45 J |
27. What is the mass of the object?
(A) 1.5 kg (B) 2.0 kg (C) 2.5 kg (D) 3.0 kg
|data point (1.0, 20) and Ug = mgh. |
|m = Ug/gh = (20 J)/(10 m/s2)( 1.0 m) = 2 kg |
28. What does the slope of the graph represent?
(A) mass of the object
(B) gravitational force on the object
(C) kinetic energy of the object
(D) potential energy of the object
|slope = Δy/Δx = Ug/h |
|Ug = mgh ∴ Ug/h = mg, which is gravitational force |
29. If an object with greater mass was graphed instead of the object graphed above, how would the slope of the graph differ from the above graph?
(A) more positive (B) less positive
(C) equal but negative (D) be the same
|Greater mass = greater gravitational force and slope = gravitational |
|force ∴ the slope would be greater. |
30. Which is the graph of the spring potential energy of a spring versus elongation from equilibrium?
(A) (B) (C) (D)
|Us = ½kx2 ∴ Us increases with the square of the elongation, which forms |
|a curve that slopes upward. |
31. Which is the graph of the gravitational potential energy of an object versus height? (Assume height ................
................
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