Chapter 8

Question 1

A - Calculate the momentum of a 2000 kg elephant charging a hunter at a speed of 7.50 m/s

Momentum (p)=mv
p=( 2103 kg )( 7.5 m/s )
p=( 2103 kg )( 7.5 m/s )=1.5104 kgm/s

B - Compare the elephant’s momentum with the momentum of a 0.040 kg tranquilizer dart fired at a speed of 600 m/s

p=( 4102 kg )( 6.0102 m/s )=24 kgm/s
1.5104 kgm/s24 kgm/s=625

C - What is the momentum of the 90.0\ kg hunter running at 7.40 m/s after missing the elephant?

p=( 90.0 kg )( 7.4 m/s )=666 kgm/s

Question 3

A - At what speed would a 2.00104 kg airplane have to fly at to have a momentum of 1.60109 kg m/s ( the same as the ship's momentum in the problem above ) ?

1.60109 kg m/s=( 2.00104 kg )( X m/s )
X m/s=( 1.60109 kg m/s )( 2.00104 kg )=0.8105=8104 m/s

B - What is the plane’s momentum when it is taking off at a speed of 60.0 m/s

p=( 2.00104 kg )( 60.0 m/s )=1.2106 kgm/s

C - If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

0=p1+p2 + ...pn
0=( m1v1 )airplane+( m2v2 )ship
( m2v2 )ship=( m1v1 )airplane
Recoil Velocity ( v2 )=1pairplanemass of ship
m=1.6109 kgm/s( 48.0 km/h )1000 meters1 second1 hour3600 seconds=1.2108 kg
Recoil Velocity ( v2 )=11.2106 kgm/s1.2108 kg=0.01 m/s

Question 15 - A cruise ship with a mass of 1.00107 kg strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. ( Hint: First calculate the time it took to bring the ship to rest )

Δx=v¯Δt
( 6.0 m0 m )=( 0.750 m/s+0 m/s2 )Δt
Δt=6.0 m0.375 m/s=16.0 seconds
Δp=FnetΔt
 Fnet=pfinalpinitialΔt
pinitial=( 1.0107 kg )( 0.750 m/s )=7.5106 kg m/s
Fnet=( 0 kgm/s )( 7.5106 kg m/s )16.0 s=4.6875105 kgms2=4.6875105 Newtons

Question 23 - Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s , and the second having a mass of 110,000 kg and a velocity of 0.120 m/s. ( The minus indicates direction of motion ) What is their final velocity?

( m1v1 )+( m2v2 )=( m1v1 )+( m2v2 )
( m1v1 )+( m2v2 )=( m1v )+( m2v )
( m1v1 )+( m2v2 )=v( m1+m2 )
v=( m1v1 )+( m2v2 )( m1+m2 )
v=( ( 1.5105 kg )( 3101 m/s ) )+( ( 1.1105 kg )( 1.2101 m/s ) )( ( 1.5105 kg )+( 1.1105 kg ) )=1.22101 m/s

Question 25 - Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is . Would the answer to this question be different if the car with the passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer.

Fnet=ΔpΔt
pfinal=( 70.0 kg )( 10 m/s )=7102 kg m/s
Fnet=ΔpΔt=( 7102 kg m/s )( 0 kg m/s )2.6101 s=2.6923076923076924103 kgms2=2.6923103 Newtons

Question 31 - A 0.240 kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s.

A - Calculate the average force exerted on the ball by the bumper.

Fnet=ΔpΔt=mΔvΔt=m( vfvi )Δt
Fnet=( 2.4101 kg )( ( 2.4 m/s )( 3.0 m/s ) )1.5102 s=86.4 kgms2=86.4 Newtons

B - How much kinetic energy in joules is lost during the collision?

ΔKE=( mvfinal22 )( mvinitial22 )
ΔKE=( m2 )( vfinal2vinitial2 )
ΔKE=( 2.4101 kg2 )( ( 2.4 m/s )2( 3.0 m/s )2 )
ΔKE=( 1.2101 kg )( ( 5.76 m2/s2 )( 9.0 m2/s2 ) )=0.3888¯ kgm2s2=0.3888¯ Joules

C - What percent of the original energy is left?

KEinitial=( 2.4101 kg )( 3.0 m/s )22=1.08 kgm2s2=1.08 Joules
% KE Left=1.08 J0.389 J1.08 J100%=63.98148148148148%

Question 44

A - During an ice skating performance, an initially motionless 80.0 kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s , what is the mass of the barbell?

0=m1v1+m2v2
0=( 80.0 kg0.5 m/s )+( X kg10.0 m/s )
1( X kg10.0 m/s )=80.0 kg0.5 m/s
X kg=80.0 kg0.5 m/s110.0 m/s=4.0 kg

B - How much kinetic energy is gained by this maneuver?

ΔKE=KEfinalKEinitial
KEfinal=( m1v122 )+( m2v222 )
KEfinal=( 80.0 kg( 0.5 m/s )22 )+( 4.0 kg( 10.0 m/s )22 )=210.0 kgm2s2=210.0 Joules
KEinitial=( m1v122 )+( m2v222 )
KEinitial=( 80.0 kg( 0.0 m/s )22 )+( 4.0 kg( 0.0 m/s )22 )=0.0 Joules
ΔKE=210.0 Joules0.0 Joules=210.0 Joules

C - Where does the kinetic energy come from?

Question 45 - Two identical pucks collide on an air hockey table. One puck was originally at rest.

A - If the incoming puck has a speed of 6.0 m/s and scatters to an angle of 30.0, what is the velocity ( magnitude and direction ) of the second puck? ( You may use the result that θ1θ2=90.0 for elastic collisions of objects that have identical masses. )

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