MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 4: WORK AND ENERGY
WORKSHEET SOLUTIONS
1. A ball hits a wall and bounces back at 2/5 the original speed. What part of the original kinetic energy
of the ball did it lose in the collision?
1A.
(1) KI = ½ mv2
(2) KF = ½ m(2/ 5 v)2
(3) KF= ½ m (4/ 25 v2)
(4) KF = 4/ 50 mv2
(5) KI = ½ mv2 = 25/ 50 mv2
(6) KF / KI = 4/ 50 ~ 25/ 50
(7) KF / KI = 4/ 25
(8) The ball lost 21/ 25 of its initial kinetic energy.
2. A 4 kg object starting at rest falls from a height of 12 m to the ground. In this instance, the force of
the air is not negligible so that the magnitude of work done by this frictional force is 25 J. What is the
object’s kinetic energy prior to hitting the ground?
2A.
(1) Ug = K
(2) mgh = K
(3) (4 kg)(9.8 m/s2)(12 m)
(4) 470 J = Ug (possible)
(5) K = Ug - Wf
(6) K = 470 J - 25 J
(7) K = 445 J
3. A 50 N crate is pulled 7 m up an inclined plane at a constant velocity. If the plane is inclined at an
angle of 22 degrees to the horizontal, what is the magnitude of the work done on the crate by the force of
gravity?
Question 3. Diagram
(w = 40 N
Rx = Rcos
Ry = h = Rsin)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3A.
Ramp height:
(1) Ry = Rsin = h = (7 m)(sin 22o) = 2.62 m
Gravity potential energy:
(2) Ug = mgh = wh
(3) Ug = (50 N)(2.62 m)
(4) Ug = 131 J
(5) Wg = - Ug
(6) Wg = - 131 J
4. A worker pushes a 250 N wheelbarrow up a ramp 6.00 m in length and inclined at 20 degrees with the
horizontal. What potential energy change does the wheelbarrow experience?
4A.
Ramp height:
(1) (6 m)(sin 20o) = 2.05 m
Gravity potential energy:
(2) Ug = mgh = wh
(3) Ug = (250 N)(2.05 m)
(4) Ug = 513 J
5. A baseball catcher puts on an exhibition by catching a 0.12 kg ball dropped from a helicopter at a
height of 71 m. If the catcher “gives” with the ball for a distance of 0.62 m while catching it, what
average force is exerted on the mitt by the ball?
5A.
(1) Ug = mgh
(2) Ug = (0.12kg)(9.8 m/s2)(71 m)
(3) Ug = 83.5 J
(4) Ug = - Wg
(5) - Wg = - Fr
(6) Ug = - Fr (the negative sign means downward direction)
(7) F = Ug / r
(8) F = (83.5 J) / (0.62 m)
(9) F = 135 N
6. A 1500 kg car travels along a highway at a speed of 20 m/s. What is its kinetic energy?
6A.
(1) K = ½ mv2
(2) K = ½ (1500 kg)(20 m/s)2
(3) K = 300,000 J
(4) K = 300 kJ
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7. A pulley-cable system on a crate hoists a bucket of cement with a total weight of 23,500 N to a height
of 47 m. If this is accomplished in 2 minutes, what is the power output by the pulley-cable system?
7A.
(1) P = W / t
(2) W = Fr
(3) P = (Fr) / t
(4) P = (23,500 N)(47 m) / 120 s
(5) P = 9,204 W
(6) P = 9.2 kW
8. A horizontal force of 200 N is applied to move a 55 kg cart across a 10 m level surface. If the cart
accelerates at 2.00 m/s2, starting from rest, then what kinetic energy does it gain while moving the 10 m
distance?
8A.
(1) v2 = vo2 + 2ad
(2) v2 = 0 + 2ad
(3) v = (2ad)1/2
(4) v = [(2)(2.00 m/s2)(10 m)]1/2
(5) v = 6.32 m/s
(6) K = ½ mv2
(7) K = ½ (55 kg)(6.32 m/s)2
(8) K = 1098 J
Questions 9 - 14. A box weighing 120 N is pulled across a horizontal floor by a 100 N force directed
30 degrees above the horizontal. The box is pulled 5 m across the floor. Friction between box and floor
results in a 30 N force of friction on the box.
Questions 9 - 14.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. Draw a free-body diagram.
9A.
(Fx: Fcos – f
Fy: Fsin + n – w)
10. How much work is done by F?
10A.
(1) W = Fr
(2) W = (100 N)(cos 30o)(5 m)
(3) W = 433 J
11. What is the work done by the normal force?
11A. Zero. The normal force is perpendicular to the displacement.
12. What is the work done by gravity?
12A. Zero. Gravity acts perpendicular to the displacement.
13. How much work is done by friction?
13A.
(1) Wf = Fr
(2) Wf = - f d
(3) Wf = - (30 N)(5 m)
(4) Wf = - 150 J
14. Calculate the net work done in this displacement.
14A. W = 433 J – 150 J = 283 J
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Questions 15 - 17. A crane lifts a 1500 kg car from rest on the ground to a height of 8 meters where its
velocity is 4 m/s, as shown below. T is the tension in the cable.
Questions 15 - 17. Diagram
15. What is the work done on the car by gravity in the 8 m displacement (in kJ)?
15A.
(1) Ug = mgh
(2) Wg = - Ug
(3) Wg = - mgh
(4) Wg = - (1500 kg)(9.8 m/s2)(8 m)
(5) Wg = - 117,600 J
(6) Wg = - 117.6 kJ
16. What is the change in the car’s kinetic energy in the lifting process (in kJ)?
16A.
(1) K = ½ mv2
(2) K = ½ (1500 kg)(4 m/s)2
(3) K = 12,000 J
(4) K = 12 kJ
17. How much work is done by T in the lift?
17A. The crane creates gravitational potential energy and provides kinetic energy to the car. The cable
must support both types of energy.
(1) E = K + Ug
(2) E = 12 kJ + 117.6 kJ
(3) E = 129.6 kJ
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 5: MOMENTUM
WORKSHEET SOLUTIONS
1. A 0.5 kg tennis ball, initially moving at a speed of 17 m/s, is struck by a racket causing it to rebound
in the opposite direction at a speed of 23 m/s. A high speed movie film determines that the racket and
ball are in contact for 0.06 s. What is the average net force exerted on the ball by the racket?
1A.
(1) pI = mv
(2) pI = (0.5 kg)(17 m/s) = 8.5 kg.m/s
(3) pF = m(- v)
(4) pF = - (0.5 kg)(23 m/s) = - 11.5 kg.m/s
(5) p = Ft
(6) F = p / t
(7) p = pF – pI
(8) F = (pF – pI) / t
(9) F = (- 11.5 kg.m/s – 8.5 kg.m/s) / 0.06 s
(10) F = (- 20 kg.m/s) / 0.06 s
(11) F = - 333 N (to the left)
2. During a snowball fight, two balls, with masses of 0.3 kg and 0.5 kg respectively, are thrown in such
a manner that they meet head-on and combine to form a single mass. The magnitude of initial velocity
for each is 16 m/s. The final kinetic energy of the system just after the collision is what percentage of
kinetic energy just before the collision?
2A.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) V = (m1v1 + m2v2) / (m1 + m2)
(3) V =[(0.3kg)(16m/s) + (0.5kg)(-16m/s)] / (0.3kg + 0.5kg)
(4) V = - 4 m/s
(5) KI = ½ m1(v1)2 + ½ m2(v2)2
(6) KI = ½ (0.3 kg)(16 m/s)2 + ½ (0.5 kg) (-16 m/s)2
(7) KI = 102.4 J
(8) KF = ½ (m1 + m2)V
(9) KF = ½ (0.3 kg + 0.5 kg)(- 4 m/s)2
(10) KF = 6.4 J
(11) % K = (KF / KI) x 100
(12) % K = (6.4 J / 102.4 J) x 100
(13) % K = 6.3%
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3. A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2 m/s. It
collides and couples with a 50,000 kg second car, initially at rest and with brakes released. What is the
speed of the two cars after coupling?
3A.
Let the moving car be m1, and let the stationary car be m2.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) m1v1 + 0 = (m1 + m2)V
(3) V = (m1v1) / (m1 + m2)
(4) V = (15,000 kg)(2 m/s) / (15,000 kg + 50,000 kg)
(5) V = 0.46 m/s
4. A miniature, spring-loaded, radio-controlled gun is mounted on an air puck. The gun’s bullet has a
mass of 0.005 kg and the gun and puck have a combined mass of 0.120 kg. With the system initially at
rest, the radio-controlled trigger releases the bullet causing the puck and empty gun to move with a
speed of 0.5 m/s. After release, what is the speed of the center of mass of the gun-puck-bullet system?
Question 4. Diagram
After Release
(Gun)
Before Release
(Gun-Puck)
After Release
(Puck)
4A. Zero. The center of mass is an imaginary point in space where we apply Newton’s laws to an
object. Before the puck is released, the center of mass is in the middle of the gun-puck system. After
the puck is released, momentum is conserved. The gun’s momentum (m1v1) must equal the puck’s
momentum (m2v2). The center of mass has not moved.
5. A fire hose directs a steady stream of 15 kg/sec of water with velocity 28 m/s against a flat plate.
What force is required to hold the plate in place?
5A.
(1) F = ma
(2) F = m (v/t)
(3) F = (15 kg) [(28 m/s) / (1 sec)]
(4) F = 420 N
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
6. Two skaters, both of mass 50 kg, are at rest on skates on a frictionless ice pond. One skater throws a
0.2 kg Frisbee at 5 m/s to her friend, who catches it and throws it back at 5 m/s. When the first skater
has caught the returned Frisbee, what is the velocity of each of the two skaters?
6A.
(1) Define the x-axis: Motion to the right is positive, motion to the left is negative.
(2) First Skater’s Throw : Frisbee moves 5 m/s to the right.
(3) m1( - v1) = m2v2
(m1 = mass of skater, m2 = mass of Frisbee)
(4) - v1 = m2v2 / m1
(5) - v1 = (0.2 kg)(5 m/s) / 50 kg
(6) v1 = - 0.02 m/s = 0.02 m/s to the left
(7) Second Skater’s Catch: 0.02 m/s to the right
(8) Second Skater’s Throw: Frisbee moves 5 m/s to the left, skater moves an additional 0.02 m/s to the
right
(9) First Skater’s Catch: skater moves 0.02 m/s to the left
(10) Both skaters are moving away from center of mass at 0.04 m/s
7. A 0.03 kg bullet is fired into, and becomes embedded in, a 1.8 kg wooden ballistic pendulum. If the
pendulum rises a vertical distance of 0.09 m, what is the initial speed of the bullet?
7A.
Let m1 be the mass of the bullet, let m2 be the mass of the wooden pendulum.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) m1v1 + 0 = (m1 + m2) V
(3) m1v1 = (m1 + m2) V
(4) v1 = (m1 + m2)(V) / m1
Determine value of V, using Conservation of Energy:
(5) Ug = K
(6) (m1 + m2)gh = ½ (m1 + m2)(V)2
(7) gh = ½ V2
(8) 2gh = V2
(9) V = (2gh)1/2
Substitute (2gh)1/2 for V:
(10) v1 = (m1 + m2)(2gh)1/2 / (m1) (from step 4)
(11) v1 = (1.83 kg)[(2)(9.8 m/s2)(0.09 m)]1/2 / (0.03 kg)
(12) v1 = 81 m/s
8. A railroad cart of 1250 kg is rolling with a speed of 4.0 m/s when 2000 kg of coal is dropped from a
height of 1.5 m into the cart. What is the final speed of the cart after the coal is in the cart?
8A.
(1) pI = pF
(2) m1v1 = m1’v1’
(3) v1’ = (m1v1) / (m1’)
(4) v1’ = [(1250 kg)(4 m/s)] / (3250 kg)
(5) v1’ = 1.54 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. A 1.47 kg duck decoy is flying overhead at 1.03 m/s when a hunter fires straight up. The 0.02 kg
bullet is moving 352 m/s when it hits the decoy and stays lodged in the decoy’s body. What is the
momentum of the decoy and bullet immediately after the hit?
9A.
Let m1 be the mass of the decoy, let m2 be the mass of the bullet.
(1) px: m1v1 + m2v2 = (m1 + m2)Vx
(2) m1v1 + 0 = (m1 + m2)Vx
(3) m1v1 = (m1 + m2)Vx
(4) Vx = (m1v1) / (m1 + m2)
(5) Vx = [(1.47 kg)(1.74 m/s)] / (1.49 kg)
(6) Vx = 1.72 m/s
(7) py: m1v1 + m2v2 = (m1 + m2)Vy
(8) 0 + m2v2 = (m1 + m2)Vy
(9) m2v2 = (m1 + m2)Vy
(10) Vy = (m2v2) / (m1 + m2)
(11) Vy = [(0.02 kg)(352 m/s)] / (1.49 kg)
(12) Vy = 4.72 m/s
(13) V2 = Vx2 + Vy2
(14) V = (Vx2 + Vy2)1/2
(15) V = [(1.72 m/s)2 + (4.72 m/s)2]1/2
(16) V = 5.03 m/s
(17) p = (m1 + m2)V
(18) p = (1.49 kg)(5.03 m/s)
(19) p = 7.49 kg.m/s
Questions 10 - 13. Two carts used in a classroom demonstration make a head-on collision. Before
the collision, cart #2 is at rest and #1 is moving towards it at 60 cm/s. The mass ratio is 2:1, that is
m1 / m2 = 2.
Questions 10 – 13.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Suppose that velcro makes the carts stick together on contact.
10. What kind of collision is this and how is it analyzed?
10A. This is an inelastic collision.
11. What is the final velocity of the system?
11A.
(1) m1v1 + m2v2 = (m1 + m2)V
(2) V = (m1v1 + m2v2) / (m1 + m2)
(3) V = (m1v1 + 0) / (m1 + m2)
(4) V = (m1v1) / (m1 + m2)
(5) Substitute 2m2 for m1:
V = (2m2v1)/ (2m2 + m2)
(6) V = (2m2v1) / (3m2)
(7) V = [2 (60 cm/s)] / 3
(8) V = 40 cm/s
Now suppose magnets in the carts cause the carts to repel one another, so they never make physical
contact. After collision the velocities are v1f and v2f.
12. What kind of collision is this and how is it analyzed?
12A. This is an elastic collision.
13. Given the 2:1 mass ratio in the collision, what relation involving final velocities is obtained from
momentum conservation (for velocities in units of cm/s)?
(A) v1f + v2f = 60
(B) v1f + 2v2f = 120
(C) v1f + 3v2f = 120
(D) v1f + 2v2f = 60
(E) 2v1f + v2f = 120
13A.
(1) m1v1 + m2v2 = m1v1’ + m2v2’
(2) m1v1 + 0 = m1v1’ + m2v2’
(3) m1v1 = m1v1’ + m2v2’
(4) Substitute 2m2 for m1:
(2m2)v1 = (2m2)vl’ + m2v2’
’
’
(5) 2m2v1 = 2m2vl + m2v2
(6) 2v1 = 2vl’ + v2’
(7) 2vl’ + v2’ = 2vl
(8) 2vl’ + v2’ = 120
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Questions 14 - 15. Two particles have a head-on collision. Assume no external forces are present. The
before and after pictures are shown below.
Questions 14 – 15 Diagrams
BEFORE
Questions 14 – 15 Diagrams
AFTER
14. Is linear momentum conserved?
14A. Yes, but some energy is lost due to friction.
15. Gravity is an external force that will change the total linear momentum of the two particles. Why is
it that gravity can be ignored in almost all collisions as they occur in practice?
15A.
(1) Gravity is negligible compared to internal collision forces.
(2) The collision takes place in a brief time interval.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Questions 16 - 19. A ball of mass 0.10 kg traveling at 30 m/s in the +x direction hits a racket which
returns the ball in the –x direction at the same speed. The collision with the racket lasts 0.10 second.
16. What is the ball’s momentum before the collision?
16A.
(1) p = mv
(2) p = (0.10 kg)(30 m/s)
(3) p = 3 kg.m/s
17. What is the ball’s momentum after the collision?
17A.
(1) p’ = mv’
(2) p’ = (0.10 kg)(- 30 m/s)
(3) p’ = - 3 kg.m/s
18. What is the change in the ball’s momentum?
18A.
(1) p = p’ – p
(2) p = - 3 kg.m/s – 3 kg.m/s
(3) p = - 6 kg.m/s
(4) p = 6 kg.m/s to the left
19. Use the Impulse and Momentum theorem to calculate the x-component of the average force on the
ball.
9A.
(1) J = Ft = p
(2) F = p / t
(3) F = - 6 kg.m/s / 0.10 s
(4) F = - 60 N (to the left)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTERS 4 - 5: WORK, ENERGY, and MOMENTUM
QUIZ SOLUTIONS
1. A ball hits a wall and bounces back at 2/5 the original speed. What part of the original kinetic energy
of the ball did it lose in the collision?
1A.
(1) KI = ½ mv2
(2) KF = ½ m(2/ 5 v)2
(3) KF= ½ m (4/ 25 v2)
(4) KF = 4/ 50 mv2
(5) KI = ½ mv2 = 25/ 50 mv2
(6) KF / KI = 4/ 50 ~ 25/ 50
(7) KF / KI = 4/ 25
(8) The ball lost 21/ 25 of its initial kinetic energy.
2. A horizontal force of 100 N is applied to move a 35 kg cart across a 7 m level surface. If the cart
accelerates at 2.00 m/s2, starting from rest, then what kinetic energy does it gain while moving the 7 m
distance?
2A.
(1) Kinematic equation: V2 = Vo2 + 2ad
(2) V2 = 0 + 2ad
(3) V = (2ad)1/2
(4) V = [(2)(2.00 m/s2)(7 m)]1/2
(5) V = 5.29 m/s
(6) Kinetic Energy equation: K = ½ mv2
(7) K = ½ (35 kg)(5.29 m/s)2
(8) K = 490 J
3. A 7 kg object starting at rest falls from a height of 12 m to the ground. In this instance, the force of
the air is not negligible so that the magnitude of work done by this frictional force is 20 J. What is the
object’s kinetic energy prior to hitting the ground?
3A.
(1) Kinematic equation: V2 = Vo2 + 2ad
(2) V2 = 0 + 2ad
(3) V2 = 2ad
(4) V = (2gd)1/2
(5) V = [(2)(9.8 m/s2)(12 m)]1/2
(6) V = 15.33 m/s
(7) Kinetic Energy equation: K = ½ mv2
(8) K = ½ (7 kg)(15.33 m/s)2
(9) K (possible) = 822.5 J
(10) 822.5 J – 20 J = 802.5 J
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4. A 1725 kg car travels along a highway at a speed of 57.6 kilometers per hour. What is its kinetic
energy?
4A.
(1) (57.6 km / 1 hour)(1000 m / 1 km)(1 hour / 3600 s)
(2) v = 16 m/s
(3) K = ½ mv2
(4) K = ½ (1725 kg)(16 m/s)2
(5) K = 220.8 kJ
5. A 55 N crate is pulled 7 m up an inclined plane at a constant velocity. If the plane is inclined at an
angle of 42 degrees to the horizontal, what is the work done on the crate by the force of gravity?
5A.
(1) Ramp height: (7 m)(sin 42o) = 4.68 m
(2) Ug = mgh = wh
(3) Ug = (55 N)(4.68 m)
(4) Ug = 257.6 J
(5) Wg = - Ug
(6) Wg = - 257.6 J
6. A baseball catcher puts on an exhibition by catching a 0.12 kg ball dropped from a helicopter at a
height of 81 m. If the catcher “gives” with the ball for a distance of 0.95 m while catching it, what
average force is exerted on the mitt by the ball?
6A.
(1) Ug = mgh
(2) Ug = (0.12 kg)(9.8 m/s2)(81 m)
(3) Ug = 95.3 J
(4) Ug = - Wg
(5) - Wg = - Fr
(6) Ug = - Fr
(7) F = Ug / r
(8) F = (95.3 J) / (0.95 m)
(9) F = 100.3 N
7. A worker pushes a 250 N weight wheelbarrow up a ramp 4.50 m in length and inclined at 17 degrees
with the horizontal. What potential energy change does the wheelbarrow experience?
7A.
(1) Ramp height: (4.50 m)(sin 17o) = 1.32 m
(2) Ug = mgh = mw
(3) Ug = (250 N)( 1.32 m)
(4) Ug = 328.9 J
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
8. A simple pendulum, 1.00 m in length, is released from rest when the support string is at an angle of
25 degrees from the vertical. What is the speed of the suspended mass at the bottom of the swing?
(Ignore air resistance, g = 9.8 m/s2)
8A.
(1) V2 = Vo2 + 2ad
(2) V2 = 0 + 2ad
(3) V = (2ad)1/2
(4) V = (2gh)1/2
(5) h = (1.00 m)(sin 25o) = 0.42 m
(6) V = [2(9.8 m/s2)(0.42 m)]1/2
(7) V = 2.88 m/s
9. A 0.2 kg tennis ball, initially moving at a speed of 19 m/s, is struck by a racket causing it to rebound
in the opposite direction at a speed of 23 m/s. A high speed movie film determines that the racket and
ball are in contact for 0.05 s. What is the average net force exerted on the ball by the racket?
9A.
(1) pI = mv
(2) pI = (0.2 kg)(19 m/s) = 3.8 kg.m/s
(3) pF = m(- v)
(4) pF = - (0.2 kg)(23 m/s) = - 4.6 kg.m/s
(5) p = Ft
(6) F = p / t
(7) p = pF – pI
(8) F = (pF – pI) / t
(9) F = (- 4.6 kg.m/s – 3.8 kg.m/s) / 0.05 s
(10) F = (- 8.4 kg.m/s) / 0.05 s
(11) F = 168 N to the left
10. A railroad car of 1115 kg is rolling with a speed of 2.35 m/s when 1750 kg of coal is dropped from a
height of 1.0 m into the car. What is the final speed of the cart after the coal is in the cart?
Question 10. Diagram
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
10A.
(1) pI = pF
(2) m1v1 = m1’v1’
(3) v1’ = (m1v1) / (m1’)
(4) v1’ = [(1115 kg)(2.35 m/s)] / (2865 kg)
(5) v1’ = 0.92 m/s
11. A railroad freight car, mass 16,253 kg, is allowed to coast along a level track at a speed of 2 m/s. It
collides and couples with a 42,350 kg second car, initially at rest and with brakes released. What is the
speed of the two cars after coupling?
11A.
Let m1 be the moving car, and let m2 be the stationary car.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) V = (m1v1 + m2v2 ) / (m1 + m2)
(3) V = [(16,253 kg)(2 m/s) + 0] / (16,253 kg + 42,350 kg)
(4) V = 0.56 m/s
12. A 1.22 kg duck decoy is flying overhead at 1.74 m/s when a hunter fires straight up. The 0.010 kg
bullet is moving 425 m/s when it hits the decoy and stays lodged in the decoy’s body. What is the
momentum of the decoy and bullet immediately after the hit?
12A.
Let m1 be the decoy, and let m2 be the bullet.
(1) px: m1v1 + m2v2 = (m1 + m2)Vx
(2) m1v1 + 0 = (m1 + m2)Vx
(3) m1v1 = (m1 + m2)Vx
(4) Vx = (m1v1) / (m1 + m2)
(5) Vx = [(1.22 kg)(1.74 m/s)] / (1.23 kg)
(6) Vx = 1.73 m/s
(7) py: m1v1 + m2v2 = (m1 + m2)Vy
(8) 0 + m2v2 = (m1 + m2)Vy
(9) m2v2 = (m1 + m2)Vy
(10) Vy = (m2v2) / (m1 + m2)
(11) Vy = [(0.01 kg)(425 m/s)] / (1.23 kg)
(12) Vy = 3.46 m/s
(13) V2 = Vx2 + Vy2
(14) V = (Vx2 + Vy2)1/2
(15) V = [(1.73 m/s)2 + (3.46 m/s)2]1/2
(16) V = 3.86 m/s
(17) p = (m1 + m2)V
(18) p = (1.23 kg)(3.86 m/s)
(19) p = 4.75 kg.m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
13. During a snowball fight, two balls, with masses of 0.35 and 0.55 kg respectively, are thrown in such
a manner that they meet head-on and combine to form a single mass of 0.90 kg. The magnitude of
initial velocity for each is 12 m/s. The final kinetic energy of the system just after the collision is what
percentage of kinetic energy just before the collision?
13A.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) V = (m1v1 + m2v2 ) / (m1 + m2)
(3) V =[(0.35 kg)(12 m/s) + (0.55 kg)(- 12 m/s)] / (0.35 kg + 0.55 kg)
(4) V = - 2.67 m/s
(5) KI = ½ m1(v1)2 + ½ m2(v2)2
(6) KI = ½ (0.35 kg)(12 m/s)2 + ½ (0.55 kg) (12 m/s)2
(7) KI = 64.8 J
(8) KF = ½ (m1+m2)V2
(9) KF = ½ (0.35 kg + 0.55 kg)(- 2.67 m/s)2
(10) KF = 3.2 J
(11) KF = 3.2 J / 64.8 J = 4.9 %
Questions 14 - 15. During a collision, an impulse of 400 N.s is experienced by a car passenger.
14. Luckily, air bags are deployed, and the passenger comes to a complete stop in 5 seconds. Calculate
the average force experienced by the passenger.
14A.
(1) J = Ft
(2) F = J / t
(3) F = 400 N.s / 5 s
(4) F = 80 N (18 lbs)
15. Unfortunately, the air bags fail to deploy, and the seatbelts malfunction. The passenger collides
against the dash board in 0.15 seconds. Calculate the new average force experienced by the passenger.
15A.
(1) J = Ft
(2) F = J / t
(3) F = 400 N.s / 0.15 s
(4) F = 2667 N (599 lbs)
16. A 0.07 kg bullet is fired into, and becomes embedded in, a 1.9 kg wooden ballistic pendulum. If the
pendulum rises a vertical distance of 0.18 m, what is the initial speed of the bullet?
16A.
Let m1 be the bullet, let m2 be the pendulum.
(1) m1v1 + m2v2 = (m1 + m2) V
(2) m1v1 + 0 = (m1 + m2) V
(3) m1v1 = (m1 + m2) V
(4) v1 = (m1 + m2)(V) / m1
(5) Determine value of V: Ug = K
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
16A. (continued…)
(6) (m1 + m2)gh = ½ (m1 + m2)(V)2
(7) gh = ½ V2
(8) 2gh = V2
(9) (2gh)1/2 = V
(10) v1 = (m1 + m2)(2gh)1/2 / (m1)
(11) v1 = (1.97 kg)[(2)(9.8 m/s2)(0.18 m)]1/2 / (0.07 kg)
(12) v1 = 52.9 m/s
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PHYSICS