Unformatted text preview: tion 7.8b: 2m1 2 ( 715 kg ) vf 2 = v01 = ( +2.25 m/s ) = +1.82 m/s
m +m 715 kg + 1055 kg 2
1 30. REASONING Since momentum is conserved, the total momentum of the twoobject
system after the collision must be the same as it was before the collision. Momentum is
mass times velocity. Since one of the objects is at rest initially, the total initial momentum
comes only from the moving object.
Let m1 and v01 be, respectively, the mass and initial velocity of the moving object before the
collision. In addition, m and vf are the total mass and final velocity of the two objects (which
stick together) after the collision. The conservation of linear momentum can be written as = mvf
{
Total momentum
after the collision m1v01
123
Total momentum
before the collision Solving this equation for vf, the final velocity of the twoobject system gives
vf = m1v01
m SOLUTION
Largemass object (8.0 kg) moving initially: Assume that, before the collision, the object is
moving in the + direction so that v01 = +25 m/s. Then, vf = m1v01 (8.0 kg ) ( +25 m/s )
=
= +18 m/s
m
3.0 kg + 8.0 kg Chapter 7 Problems 363 The final speed is 18 m/s .
Smallmass object (3.0 kg) moving initially: vf = m1v01 ( 3.0 kg ) ( +25 m/s )
=
= +6.8 m/s
m
3.0 kg + 8.0 kg The final speed is 6.8 m/s . 31. SSM REASONING We obtain the desired percentage in the usual way, as the kinetic
energy of the target (with the projectile in it) divided by the projectile’s incident kinetic
energy, multiplied by a factor of 100. Each kinetic energy is given by Equation 6.2 as 1
mv 2 ,
2 where m and v are mass and speed, respectively. Data for the masses are given, but the
speeds are not provided. However, information about the speeds can be obtained by using
the principle of conservation of linear momentum. SOLUTION We define the following quantities:
KETP = kinetic energy of the target with the projectile in it
KE0P = kinetic energy of the incident projectile
mP = mass of incident projectile = 0.20 kg
mT = mass of target = 2.50 kg
vf = speed at which the target with the projectile in it flies off after being struck
v0P = speed of incident projectile
The desired percentage is KE TP
Percentage =
×100 =
KE 0P 1
2 ( mT + mP ) vf2
1
m v2
2 P 0P ×100 % (1) According to the momentumconservation principle, we have
mT
vf
(14+ mP )3
4
244
Total momentum of target and
projectile after target is struck = 0 + mP v0P
14 244
4
3
Total momentum of target and
projectile before target is struck Note that the target is stationary before being struck and, hence, has zero initial momentum.
Solving for the ratio vf /v0P, we find that 364 IMPULSE AND MOMENTUM vf
mP
=
v0P mT + mP
Substituting this result into Equation (1) gives
Percentage = = 1
2 ( mT + mP ) vf2
1
m v2
2 P 0P ×100 % = 1
2 ( mT + mP ) 1
m
2P 2 × 100 % m +m P
T
mP mP
0.20 kg
× 100 % =
×100 % = 7.4 %
mT + mP
2.50 kg + 0.20 kg 32. REASONING The weight of each vehicle is balanced by the normal force exerted by the
road. Assuming that friction and other resistive forces can be ignored, we will treat the twovehicle system as an isolated system and apply the principle of conservation of linear
momentum. SOLUTION Using v0, car and v0, SUV to denote the velocities of the vehicles before the
collision and applying the principle of conservation of linear momentum, we have 0
1 24
43 = mcar v0, car + mSUV v0, SUV
1444 24444
4
3
Total momentum
before collision Total momentum
after collision Note that the total momentum of both vehicles after the collision is zero, because the
collision brings each vehicle to a halt. Solving this result for v0, SUV and taking the direction
in which the car moves as the positive direction gives
v0, SUV = − mcar v0, car
mSUV = − (1100 kg ) ( 32 m/s )
2500 kg = −14 m/s This result is negative, since the velocity of the sport utility vehicle is opposite to that of the
car, which has been chosen to be positive. The speed of the sport utility vehicle is the
magnitude of v0, SUV or 14 m/s . 33. SSM WWW REASONING The system consists of the two balls. The total linear
momentum of the twoball system is conserved because the net external force acting on it is
zero. The principle of conservation of linear momentum applies whether or not the collision
is elastic. Chapter 7 Problems 365 m v f1 + m2 v f2 = m1 v 01 + 0
11 4 44
42 3
14 4
23
Total momentum
after collision Total momentum
before collision When the collision is elastic, the kinetic energy is also conserved during the collision
2
2
1
1
m v 2 + 1 m v f2 = 2 m1 v 01 + 0
2 1 f1
144 22 2 3 14243
4 444
Total kinetic energy
after collision Total kinetic energy
before collision SOLUTION
a. The final velocities for an elastic collision are determined by simultaneously solving the
above equations for the final velocities. The procedure is discussed in Example 7 in the text,
and leads to Equations 7.8a and 7.8b. According to Equation 7.8: v f1 = F − m Iv
m
G +m J
m
HK
1 2 01 1 and v f2 = 2 F 2 m Iv
G +m J
m
HK
1 01 1 2 Let the initial direction of motion of the 5.0...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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