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Unformatted text preview: 0kg ball define the positive direction.
Substituting the values given in the text, these equations give F kg – 7.50 kg I(2.00 m / s) =
5.00
G kg + 7.50 kg J
5.00
H
K
F 2(5.00 kg) I(2.00 m / s) =
=G
5.00
H kg + 7.50 kg J
K 5.00  kg ball v f1 = –0.400 m / s 7.50  kg ball v f2 +1.60 m / s The signs indicate that, after the collision, the 5.00kg ball reverses its direction of motion,
while the 7.50kg ball moves in the direction in which the 5.00kg ball was initially moving.
b. When the collision is completely inelastic, the balls stick together, giving a composite
body of mass m1 + m2 that moves with a velocity vf . The statement of conservation of linear
momentum then becomes
( m1 + m2 ) v f = m1 v 01 + 0
14 24 3 1 24
44
43
Total momentum
after collision Total momentum
before collision The final velocity of the two balls after the collision is, therefore,
vf = m1 v 01
m1 + m2 = (5.00 kg)(2.00 m / s)
= +0.800 m / s
5.00 kg + 7.50 kg 366 IMPULSE AND MOMENTUM 34. REASONING The net external force acting on the twopuck system is zero (the weight of
each ball is balanced by an upward normal force, and we are ignoring friction due to the
layer of air on the hockey table). Therefore, the two pucks constitute an isolated system, and
the principle of conservation of linear momentum applies.
SOLUTION Conservation of linear momentum requires that the total momentum is the
same before and after the collision. Since linear momentum is a vector, the x and y
components must be conserved separately. Using the drawing in the text, momentum
conservation in the x direction yields
mA v0A = mA vfA ( cos 65° ) + mBvfB ( cos 37° ) (1) while momentum conservation in the y direction yields
0 = mA vfA ( sin 65° ) − mBvfB ( sin 37° ) (2) Solving equation (2) for vfB, we find that
vfB = mA vfA ( sin 65° )
mB ( sin 37° ) (3) Substituting equation (3) into Equation (1) leads to m v ( sin 65° ) mA v0A = mA vfA ( cos 65° ) + A fA ( cos 37° )
sin 37° a. Solving for vfA gives
vfA = v0 A
+5.5 m/s
=
= 3.4 m/s sin 65° sin 65° cos 65° + cos 65° + tan 37° tan 37° b. From equation (3), we find that
vfB = ( 0.025 kg ) ( 3.4 m/s ) ( sin 65° ) =
( 0.050 kg )( sin 37° ) 2.6 m/s 35. SSM REASONING Batman and the boat with the criminal constitute the system.
Gravity acts on this system as an external force; however, gravity acts vertically, and we are
concerned only with the horizontal motion of the system. If we neglect air resistance and
friction, there are no external forces that act horizontally; therefore, the total linear Chapter 7 Problems 367 momentum in the horizontal direction is conserved. When Batman collides with the boat,
the horizontal component of his velocity is zero, so the statement of conservation of linear
momentum in the horizontal direction can be written as (m1 + m2 )vf
14 244
4
3 = Total horizontal momentum
after collision m1v01 + 0
14243
Total horizontal momentum
before collision Here, m1 is the mass of the boat, and m2 is the mass of Batman. This expression can be
solved for vf , the velocity of the boat after Batman lands in it.
SOLUTION Solving for vf gives
vf = m1v01 m1 + m2 = ( 510 kg ) ( +11 m/s ) =
510 kg + 91 kg +9.3 m/s The plus sign indicates that the boat continues to move in its initial direction of motion. 36. REASONING According to the momentumconservation principle, the final total
momentum is the same as the initial total momentum. The initial total momentum is the
vector sum of the two initial momentum vectors of the objects. One of the vectors (p0A)
points due east and one (p0B) due north, so that they are perpendicular (see the drawing on
the left).
North North p0B p0A East Pf θ
p0A East p0B Based on the conservation principle, the direction of the final total momentum Pf must be
the same as the direction of the initial total momentum. From the drawing on the right, we
can see that the initial total momentum has a component p0A pointing due east and a
component p0B pointing due north. The final total momentum Pf has these same
components and, therefore, must point north of east at an angle θ. 368 IMPULSE AND MOMENTUM SOLUTION Based on the conservation of linear momentum, we know that the magnitude
of the final total momentum is the same as the magnitude of the initial total momentum.
Using the Pythagorean theorem with the initial momentum vectors of the two objects, we
find that the magnitude of the final total momentum is ( mAv0A )2 + ( mBv0B )2 Pf = 2
2
p0A + p0B = Pf = (17.0 kg )2 (8.00 m/s )2 + ( 29.0 kg )2 ( 5.00 m/s )2 = 199 kg ⋅ m/s The direction is given by p0B −1 mBv0B −1 ( 29.0 kg ) ( 5.00 m/s ) = tan = tan = 46.8° north of east (17.0 kg )(8.00 m/s ) p0A mAv0A θ = tan −1 37. REASONING The velocity of the second ball just after the collision can be found from
Equation 7.8b (see Example 7). In order to use Equation 7.8b, however, we must know the
velocity of the first ball just before it strikes the second ball. Since we know the impulse
delivered to the first ball by the pool stick, we can use the...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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