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Unformatted text preview: all. 7. REASONING The impulsemomentum theorem (Equation 7.4) states that the impulse of an
applied force is equal to the change in the momentum of the object to which the force is
applied. We will use this theorem to determine the final momentum from the given value of
the initial momentum. The impulse is the average force times the time interval during which
the force acts, according to Equation 7.1. The force and the time interval during which it
acts are given, so we can calculate the impulse.
SOLUTION According to the impulsemomentum theorem, the impulse applied by the
retrorocket is
J = mvf − mv0
(7.4)
The impulse is J = F ∆t (Equation 7.1), which can be substituted into Equation 7.4 to give F ∆t = mvf − mv0 or mvf = F ∆t + mv0 where mvf is the final momentum. Taking the direction in which the probe is traveling as
the positive direction, we have that the initial momentum is mv0 = +7.5 × 107 kg⋅m/s and the
force is F = −2.0 × 106 N . The force is negative, because it points opposite to the direction
of the motion. With these data, we find that the final momentum after the retrorocket ceases
to fire is ( ) mvf = F ∆t + mv0 = −2.0 × 106 N (12 s ) + 7.5 × 107 kg ⋅ m/s = +5.1× 107 kg ⋅ m/s 8. REASONING We will apply the impulse momentum theorem as given in Equation 7.4 to
solve this problem. From this theorem we know that, for a given change in momentum,
greater forces are associated with shorter time intervals. Therefore, we expect that the force
in the stifflegged case will be greater than in the kneesbent case.
SOLUTION a. Assuming that upward is the positive direction, we find from the impulsemomentum theorem that ΣF = mvf − mv 0 ∆t = ( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( −6.4 m/s )
2.0 × 10 −3 s = +2.4 × 105 N Chapter 7 Problems 347 b. Again using the impulsemomentum theorem, we find that ΣF = mvf − mv 0 = ∆t ( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( −6.4 m/s )
0.10 s = +4.8 × 103 N c. The net average force acting on the man is ΣF = FGround + W , where FGround is the
average upward force exerted on the man by the ground and W is the downwardacting
weight of the man. It follows, then, that FGround = ΣF − W . Since the weight is W = −mg,
we have
Stiff − legged FGround = ΣF − W ( ) ( ) = +2.4 × 105 N − − ( 75 kg ) 9.80 m/s 2 = +2.4 × 105 N Knees − bent FGround = ΣF − W
= +4.8 × 103 N − − ( 75 kg ) 9.80 m/s 2 = +5.5 × 103 N 9. REASONING AND SOLUTION a. According to Equation 7.4, the impulsemomentum theorem, ( ΣF ) ∆t = mvf − mv0 . Since the only horizontal force exerted on the puck is the force F exerted by the goalie,
ΣF = F . Since the goalie catches the puck, vf = 0 m/s . Solving for the average force
exerted on the puck, we have F= m( v f − v 0 )
∆t = ( 0.17 kg ) ( 0 m/s ) − ( +65 m/s ) 5.0 ×10 −3 s = –2.2 ×103 N By Newton’s third law, the force exerted on the goalie by the puck is equal in magnitude
and opposite in direction to the force exerted on the puck by the goalie. Thus, the average
force exerted on the goalie is +2.2 × 103 N .
b. If, instead of catching the puck, the goalie slaps it with his stick and returns the puck
straight back to the player with a velocity of –65 m/s, then the average force exerted on the
puck by the goalie is
F= m( v f − v 0 )
∆t = 0.17
b kg g(–65 m / s) – (+65 m / s)
5.0 × 10 –3 s = – 4.4 × 10 3 N The average force exerted on the goalie by the puck is thus +4.4 × 103 N . 348 IMPULSE AND MOMENTUM The answer in part (b) is twice that in part (a). This is consistent with the conclusion of
Conceptual Example 3. The change in the momentum of the puck is greater when the puck
rebounds from the stick. Thus, the puck exerts a greater impulse, and hence a greater force,
on the goalie.
10. REASONING During the time interval ∆t, a mass m of water strikes the turbine blade. The
incoming water has a momentum mv0 and that of the outgoing water is mvf. In order to change the momentum of the water, an impulse ( ΣF ) ∆t is applied to it by the stationary turbine blade. Now ( Σ F ) ∆t = F ∆t , since only the force of the blade is assumed to act on
the water in the horizontal direction. These variables are related by the impulsemomentum
theorem, F ∆t = mvf − mv0, which can be solved to find the average force F exerted on the
water by the blade.
SOLUTION Solving the impulsemomentum theorem for the average force gives
F= mv f − mv 0
∆t = m
( v − v0 )
∆t f The ratio m / ( ∆t ) is the mass of water per second that strikes the blade, or 30.0 kg/s, so the
average force is F= m
( v − v 0 ) = ( 30.0 kg/s ) ( −16.0 m/s ) − ( +16.0 m/s ) = −960 N ∆t f The magnitude of the average force is 960 N . 11. REASONING We will divide this problem into two parts, because the forces acting on the
student change abruptly at the instant of impact. In the first part, the student falls freely from
rest, under the sole influence of the conservative gravitational force. Thus, the studen...
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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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