Download Physics 100 Group Session for Chapters 1 – 3

Answers to Physics 101 Team Session for Chapter 6&7
1) In the figure, 1.10 x 103 J of work is done by the force F (magnitude = 30 N), when moving the case a
distance of 50 m. At what angle Θ is the force oriented with respect to the ground?
SOLUTION Solving Equation 6.1 for the angle  , we obtain
W 
1.10 103 J 
1 

cos


  42.8
Fs
 (30.0 N)(50.0 m) 
  cos 1 
2) An astronaut using a jet propulsion unit flies straight upwards from rest. At a height of 16 m, his velocity is
5 m/s. His mass, including the propulsion unit and flight suit is 136 kg. What is the work done by the generated
thrust force?
SOLUTION From Equation 6.7b, we have
Wnc  KE + PE 

1
2

mvf2  12 mv02   mghf  mgh0 
Since rocket man starts from rest, v0 = 0 m/s. If we take h0 = 0 m on the ground, we have
1
2
Wnc  mvf2  mghf  m

1 v2
2 f
 ghf

2
2
4
 (136 kg)  12 (5.0 m/s)  (9.80 m/s )(16 m)   2.3  10 J


3) A 0.149-kg baseball approaches a bat horizontally at a speed of 40.4 m/s and is hit straight back at a speed of
45.6 m/s. If the ball is in contact with the bat for 1.10 x 10-3 s, what is the average force exerted on the ball by
the bat? (Note: neglect any effects from air resistance or gravity.)
REASONING During the collision, the bat exerts an impulse on the ball. The impulse is the product of the
average force that the bat exerts and the time of contact. According to the impulse-momentum theorem, the
impulse is also equal to the change in the momentum of the ball. We will use these two relations to
determine the average force exerted by the bat on the ball.

SOLUTION The impulse J is given by Equation 7.1 as J = Ft , where F is the average force that the bat
exerts on the ball and t is the time of contact. According to the impulse-momentum theorem, Equation 7.4,
the net average impulse  F  t is equal to the change in the ball’s momentum;  F  t  mvf  mv0 .
Since we are ignoring the weight of the ball, the bat’s force is the net force, so F  F . Substituting this
value for the net average force into the impulse-momentum equation and solving for the average force
gives
F
mvf  mv 0
t

 0.149 kg  45.6 m/s    0.149 kg  40.2 m/s  
1.10 103 s
11 600 N
where the positive direction for the velocity has been chosen as the direction of the incoming ball.
4) In comic strips, Superman is known for hurling big asteroids without himself moving an inch. In reality, if
Superman threw an asteroid that was 1000 times more massive than he at a speed of 800 m/s is he able to stay at
rest? If not, what would his recoil velocity be?
SOLUTION According to the conservation of momentum, the total momentum of the superman and
asteroid before the throwing, has to equal the total momentum after throwing. Also, if we set superman’s
mass equal to M, then the asteroid’s mass is 1000M. Using these two relationships, we can solve for
superman’s velocity.
velocity of Superman & asteroid before throw, when Superman is holding the asteroid = vs+a
mass of superman = M
velocity of Superman after throwing = vs
mass of asteroid = 1000M
velocity of asteroid after throwing = va
total momentum before hurling asteroid = total momentum after hurling asteroid
momentum of Superman + momentum of asteroid = momentum of Superman + momentum of asteroid
(M + 1000M)vs+a = (M x vs)+ (1000M x va)
0 kg m/s = (M x vs) + 1000M x 800 m/s
-(M x vs) = 800 000M m/s
vs = - 800 000 m/s
(where the negative sign shows that Superman’s velocity is in the opposite direction of the asteroid’s)