Holt Physics Problem 6D - Hays High School

NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6D
CONSERVATION OF MOMENTUM
PROBLEM
SOLUTION
1. DEFINE
2. PLAN
A 20.0 kg cannonball is fired from a 2.40 × 10 3 kg. If the cannon
recoils with a velocity of 3.5 m/s backwards, what is the velocity of the
cannonball?
Given:
m 1 = mass of cannonball = 20.0 kg
m 2 = mass of cannon = 2.40 × 10 3 kg
v 1,i = initial velocity of cannonball = 0 m/s
v 2, i = initial velocity of cannon = 0 m/s
v 2,f = final velocity of cannon = 3.5 m/s backwards =−3.5 m/s
Unknown: v 1, f = final mass of cannonball = ?
Choose the equation(s) or situation: Because the momentum of the cannoncannonball
system is conserved and therefore remains constant, the total initial
momentum of the cannon and cannonball will equal the total final momentum
of the cannon and cannonball.
m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f
Because the cannon and cannonball are initially at rest, the initial momentum for
each is zero. From momentum conservation it follows that the total final momentum
is also zero.
m 1 v 1,f + m 2 v 2,f = 0
Rearrange the equation(s) and isolate the unknown(s):
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
v 1,f = ⎯ −m 2v ⎯ 2,f
m1
Substitute the values into the equation(s) and solve:
v 1,f =
v 1,f =
−(2.40 × 10 3 kg)(−3.5 m/s)
⎯⎯⎯
20.0 kg
420 m/s forward
=420 m/s
The velocity is positive, indicating the forward direction. The cannonball’s mass
is less than one-hundredth the mass of the cannon, so its speed should be over a
hundred times greater than the recoil speed of the cannon.
1. A student stumbles backward off a dock and lands in a small boat. The
student isn’t hurt, but the boat drifts away from the dock with a
velocity of 0.85 m/s to the west. If the boat and student each have a
mass of 68 kg, what is the student’s initial horizontal velocity?
Problem 6D Ch. 6-7

NAME ______________________________________ DATE _______________ CLASS ____________________
2. A coal barge with a mass of 1.36 × 10 4 kg drifts along a river. When it
passes under a coal hopper, it is loaded with 8.4 × 10 3 kg of coal. What
is the speed of the unloaded barge if the barge after loading has a speed
of 1.3 m/s?
3. A child jumps from a moving sled with a speed of 2.2 m/s and in the
direction opposite the sled’s motion. The sled continues to move in the
forward direction, but with a new speed of 5.5 m/s. If the child has a
mass of 38 kg and the sled has a mass 68 kg, what is the initial velocity
of the sled?
4. A swimmer with a mass of 58 kg and a velocity of 1.6 m/s to the north
climbs onto a 142 kg raft. The combined velocity of the swimmer and
raft is 0.32 m/s to the north. What is the raft’s velocity before the swimmer
reaches it?
5. A 50.0 g shell fired from a 3.00 kg rifle has a speed of 400.0 m/s. With
what speed does the rifle recoil in the opposite direction?
6. Momentum conservation often assumes that the mass of an object remains
constant throughout a process or event. However, a change in
momentum can also occur when mass changes. Consider an automobile
with a full tank of gasoline traveling at a velocity of 88.0 km/h to
the east. The mass of the car when the fuel tank is full is 1292 kg. Suppose
that the car travels along a highway that extends eastward for
600 km. By the time the car has traveled this distance, its mass is
1255 kg. What is the car’s velocity at the end of the journey?
7. In 1976, Comet West was observed to break into four smaller parts as it
orbited near the sun. Suppose a comet with a mass of 5.0 × 10 14 kg and
moving with a speed of 74.0 km/s breaks into two equal parts. One
part moves 15.0° above the original orbit with a speed of 105 km/s,
while a second fragment moves 30.0° below the original orbit. What is
the velocity of the second comet fragment?
8. A twig floating in a small pond is initially at rest. On the twig is a snail,
which begins moving along the length of the twig with a speed of
1.2 cm/s. The twig moves in the opposite direction with a speed of
0.40 cm/s. If the snail’s mass is 2.5 g, what is the mass of twig?
9. A toy that is initially at rest consists of three parts that are held together
by spring-loaded clips. At a given instant, the toy “explodes.” Two of the
pieces, which each have a mass of 25.0 g, travel with velocities of
7.0 cm/s to the south and 7.0 cm/s to the west, respectively. The third
piece has a velocity of 3.3 cm/s at an angle of 45° north of east. What is
the mass of the third piece?
10. An ice skater at rest catches a bag of sand moving to the north with a
speed of 5.4 m/s. This causes both the skater and the bag to move to
the north at a speed of 1.5 m/s. If the skater’s mass is 63 kg, what is the
mass of the bag of sand?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Ch. 6-8
Holt Physics Problem Bank

Additional Practice 6D
Givens
1. m 1 = 68 kg
m 2 = 68 kg
v 2, i = 0 m/s
v 1, f = 0.85 m/s to the west
=−0.85 m/s
v 2, f = 0.85 m/s to the west
=−0.85 m/s
Solutions
m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f
m
v 1, i = 1 v 1, f + m 2 v 2, f − m 2 v
⎯⎯⎯ 2, i
=
mi
v 1, i =−0.85 m/s + (−0.85 m/s) =−1.7 m/s
v 1, i = 1.7 m/s to the west
(68 kg)(−0.85 m/s) + (68 kg)(−0.85 m/s) − (68 kg)(0 m/s)
⎯⎯⎯⎯⎯⎯
68 kg
2. m i = 1.36 × 10 4 kg
m 2 = 8.4 × 10 3 kg
v 2, i = 0 m/s
v 1, f = v 2, f = 1.3 m/s
m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f
v 1, i =
v 1, i =
m 1 v 1, f + m 2 v 2, f − m 2 v 2, i
⎯⎯⎯
m1
(1.36 × 10 4 kg)(1.3 m/s) + (8.4 × 10 3 kg)(1.3 m/s) − (8.4 × 10 3 kg)(0 m/s)
⎯⎯⎯⎯⎯⎯⎯⎯
1.36 × 10 4 kg
1.8 × 10 4 kg•m/s + 1.1 × 10 4 kg•m/s
v 1, i = ⎯⎯⎯⎯ =
1.36 × 10 4 kg
2.9 × 10 4 kg•m/s
⎯⎯
1.36 × 10 4 kg
v 1, i = 2.1 m/s
3. v 1, f = 2.2 m/s backwards
= −2.2 m/s
v 2, f = 5.5 m/s forward
=+5.5 m/s
m 1 = 38 kg
m 2 = 68 kg
m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f
v 1, i = v 2, i , so
v 1, i = ⎯ m 1v1,
f + m2v 2, f (38 kg)(−2.2 m/s) + (68 kg)(5.5 m/s)
⎯ = ⎯⎯⎯⎯
m1
+ m2
38 kg + 68 kg
−84 kg•m/s + 370 kg•m/s 290 kg•m/s
v 1, i = ⎯⎯⎯ = ⎯⎯ = 2.7 m/s
106 kg 106 kg
v 1, i = 2.7 m/s forward
V
4. m 1 = 38 kg
v 1, i = 1.6 m/s to the north
m 2 = 142 kg
v 1, f = 0.32 m/s to the north
v 2, f = 0.32 m/s to the north
5. m 1 = 50.0 g
v 1, i = 0 m/s
v 1, f = 400.0 m/s forward
m 2 = 3.00 kg
v 2, i = 0 m/s
m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f
v 2, i =
v 2, i =
m 1 v 1, f + m 2 v 2, f − m 1 v 1, i
⎯⎯⎯
m2
(38 kg)(0.32 m/s) + (142 kg)(0.32 m/s) − (38 kg)(1.6 m/s)
⎯⎯⎯⎯⎯⎯
142 kg
12 kg•m/s + 45 kg•m/s − 61 kg•m/s −4.0 kg•m/s
v 2, i = ⎯⎯⎯⎯ = ⎯⎯ =−2.8 × 10 −2 m/s
142 kg
142 kg
v 2, i = 2.8 × 10 −2 m/s to the south
Because the initial velocities for both rifle and projectile are zero, the momentum conservation
equation takes the following form:
m 1 v 1, f + m 2 v 2, f = 0
−m −(50.0 × 10 −3 1 v kg)(400.0 m/s)
v 2, f = ⎯ 1, f
= ⎯⎯⎯ = −6.67 m/s
m2
3.00 kg
v 2, f = 6.67 m/s backward
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V Ch. 6–6
Holt Physics Solution Manual

Givens
6. m i = 1292 kg
v i = 88.0 km/h to the east
m f = 1255 kg
Solutions
m i v i = m f v f
v f = ⎯ m ivi
(1292 kg)(88.0 km/h)
⎯ = ⎯⎯⎯
mf
1255 kg
v f = 90.6 km/h to the east
7. m = 5.0 × 10 14 kg
v i = 74.0 km forward
m i = m 2 = ⎯ 1 2 ⎯ m
v 1, f = 105 km/s at 15.0°
above forward
v 2, f is at an angle of −30.0°
to the forward direction
mv i = m 1 v 1, f + m 2 v 2, f
To solve for velocity in two dimensions, the momentum conservation equation must
be written as two equations, one for both the x and y directions.
In the x-direction:
mv i = m 1 v 1, f (cos q 1 ) + m 2 v 2, f (cos q 2 )
v i = ⎯ 1 2 ⎯ v 1, f (cos q 1 ) + ⎯ 1 2 ⎯ v 2, f (cos q 2 )
v 2, f =
v 2, f =
2v i − v 1, f (cos q 1 )
⎯⎯
cos q2
(2)(74.0 km/s) − (105 km/s)(cos 15.0°)
⎯⎯⎯⎯
cos(−30.0°)
148 km/s − 101 km/s
v 2, f = ⎯⎯⎯ =
cos(−30.0°)
v 2, f = 54 km/s
In the y-direction (check):
0 = m 1 v 1, f (sin q 1 ) + m 2 v 2, f (sin q 2 )
v 1, f (sin q 1 ) =−v 2, f (sin q 2 )
47 km/s
⎯⎯
cos(−30.0°)
(105 km/s)(sin 15.0°) =−(54 km/s)[sin(−30.0°)]27.2 km/s = 27 km/s
The slight difference arises from differences in the number of significant figures and
from rounding.
v 2,f = 54 km/s at 30.0° below the initial forward direction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. v 1, i = 0 cm/s
v 1, f = 1.2 cm/s forward
=+1.2 cm/s
v 2, i = 0 cm/s
v 2, f = 0.40 cm/s backward
=−0.40 cm/s
m 1 = 2.5 g
m 1 v 1, f + m 2 v 2, f = 0
−m
m 2 = 1 v −(2.5 g)(1.2 cm/s)
⎯ 1, f
= ⎯⎯
v2, f −0.40 cm/s
m 2 = 7.5 g
V
Section Five—Problem Bank V Ch. 6–7

Givens
9. v i = 0 cm/s
m 1 = 25.0 g
m 2 = 25.0 g
v 1 = 7.0 cm/s to the south
= 7.0 cm/s at −90° from
east
v 2 = 7.0 cm/s to the west
= 7.0 cm/s at 180° from
east
v 3 = 3.3 m/s at 45° north of
east
Solutions
m 1 v 1, f + m 2 v 2, f + m 3 v 3, f = 0
In the x-direction:
m 1 v 1, f (cos q 1 ) + m 2 v 2, f (cos q 2 ) + m 3 v 3, f (cos q 3 ) = 0
m 3 =
m 3 =
m 1 v 1, f (cos q 1 ) + m 2 v 2, f (cos q 2 )
⎯⎯⎯⎯
−v 3, f (cos q 3 )
− (25.0 g)(7.0 cm/s)(cos −90°) + (25.0 g)(7.0 cm/s)(cos 180°)
⎯⎯⎯⎯⎯⎯⎯
−(3.3 cm/s)(cos 45°)
(25.0 g)(7.0 cm/s)
m 3 = ⎯⎯ =
(3.3 cm/s)(cos 45°)
In the y-direction (check):
75 g
m 1 v 1, f (sin q 1 ) + m 2 v 2, f (sin q 2 ) + m 3 v 3, f (sin q 3 ) = 0
(25.0 g)(7.0 cm/s)[sin(−90°)] + (25.0 g)(7.0 cm/s)(sin 180°)
+ (75 g)(3.3 cm/s)(sin 45°) = 0
−180 g•cm/s + 0 g•cm/s + 180•g•cm/s = 0
10. v 1, i = 0 m/s
v 2, i = 5.4 m/s to the north
v 1, f = 1.5 m/s to the north
v 2, f = 1.5 m/s to the north
m 1 = 63 kg
m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f
m 1 v 1, f − m 1 v (63 kg)(1.5 m/s) − (63 kg)(0 m/s)
m 2 = ⎯⎯ 1, i
= ⎯⎯⎯⎯ =
v2, i − v 2, f
5.4 m/s − 1.5 m/s
m 2 = 24 kg
(63 kg)(1.5 m/s)
⎯⎯
3.9 m/s
Additional Practice 6E
1. m 1 = 1550 kg
m 2 = 770 kg
v 2, i = 0 m/s
v f = 9.44 m/s forward
2. m 1 = 0.17 kg
m 2 = 0.75 kg
v 2, i = 0.50 m/s to the left
=−0.50 m/s
v f = 4.2 m/s to the right
=+4.2 m/s
v 1, i =
(m 1 + m 2 )v f − m 2 v 2, i
⎯⎯⎯
m1
(1550 kg + 770 kg)(9.44 m/s) − (770 kg)(0 m/s)
v 1, i = ⎯⎯⎯⎯⎯ =
1550 kg
v 1, i = 14.0 m/s forward
v 1, i =
v 1, i =
v 1, i =
(m 1 + m 2 )v f − m 2 v 2, i
⎯⎯⎯
m1
(0.17 kg + 0.75 kg)(4.2 m/s) − (0.75 kg)(−0.50 m/s)
⎯⎯⎯⎯⎯⎯
0.17 kg
(0.92 kg)(4.2 m/s) + 0.38 kg•m/s
⎯⎯⎯⎯
0.17 kg
3.9 kg•m/s + 0.38 kg•m/s
v 1, i = ⎯⎯⎯ =
0.17 kg
v 1, i = 25 m/s to the right
4.3 kg•m/s
⎯⎯ 0.17 kg
(2320 kg)(9.44 m/s)
⎯⎯
1550 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
V
V Ch. 6–8
Holt Physics Solution Manual