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Unformatted text preview: impulsemomentum theorem
(Equation 7.4) to find the velocity of the first ball just before the collision.
SOLUTION According to the impulsemomentum theorem, F ∆t = mvf − mv0 , and setting
v0 = 0 m/s and solving for vf , we find that the velocity of the first ball after it is struck by
the pool stick and just before it hits the second ball is vf = F ∆t +1.50 N ⋅ s
=
= 9 .09 m / s
m
0.165 kg Substituting values into Equation 7.8b (with v01 = 9.09 m/s), we have 2m1 2m vf2 = v= v = v = +9.09 m/s m + m 01 m + m 01 01 1
2 38. REASONING
a. Since air resistance is negligible as the ball swings downward, the work done by
nonconservative forces is zero, Wnc = 0 J. The force due to the tension in the wire is
perpendicular to the motion and, therefore, does no work. Thus, the total mechanical energy,
which is the sum of the kinetic and potential energies, is conserved (see Section 6.5). The Chapter 7 Problems 369 conservation of mechanical energy will be used to find the speed of the ball just before it
collides with the block.
b. When the ball collides with the stationary block, the collision is elastic. This means that,
during the collision, the total kinetic energy of the system is conserved. The horizontal or
xcomponent of the total momentum is conserved, because the horizontal surface is
frictionless, and so the net average external force acting on the ballblock system in the
horizontal direction is zero. The total kinetic energy is conserved, since the collision is
known to be elastic.
SOLUTION
a. As the ball falls, the total mechanical energy is conserved. Thus, the total mechanical
energy at the top of the swing is equal to that at the bottom:
m1 gL
{
Total mechanical
energy at top of
swing, all
potential energy = 2
1
2 m1v
123 Total mechanical
energy at bottom
of swing, all
kinetic energy In this expression L is the length of the wire, m1 is the mass of the ball, g is the magnitude of
the acceleration due to gravity, and v is the speed of the ball just before the collision. We
have chosen the zerolevel for the potential energy to be at ground level, so the initial
potential energy of the ball is m1gL. Solving for the speed v of the ball gives
v = 2 gL = 2 ( 9.80 m/s 2 ) (1.20 m ) = 4.85 m/s At the bottom of the swing the ball is moving horizontally and to the right, which we take to
be the +x direction (see the drawing in the text). Thus the velocity of the ball just before
impact is vx = +4.85 m/s .
b. The conservation of linear momentum and the conservation of the total kinetic energy can
be used to describe the behavior of the system during the elastic collision. This situation is
identical to that in Example 7 in Section 7.3, so Equation 7.8a applies: m − m2 vf = 1
v
m1 + m2 x where vf is the final velocity of the ball after the collision, m1 and m2 are, respectively, the
masses of the ball and block, and vx is the velocity of the ball just before the collision. Since
vx = +4.85 m/s, we find that 370 IMPULSE AND MOMENTUM 1.60 kg − 2.40 kg vf = ( +4.85 m/s ) = −0.97 m/s 1.60 kg + 2.40 kg The minus sign indicates that the ball rebounds to the left after the collision. 39. REASONING AND SOLUTION The total linear momentum of the twocar system is
conserved because no net external force acts on the system during the collision. We are
ignoring friction during the collision, and the weights of the cars are balanced by the normal
forces exerted by the ground. Momentum conservation gives
(m1 + m2 )vf = m1v01 + m2v02
14 3
24
14244
4
3
Total momentum
after collision Total momentum
after collision where v02 = 0 m/s since the 1900kg car is stationary before the collision.
a. Solving for vf , we find that the velocity of the two cars just after the collision is
vf = m1v01 + m2 v02
m1 + m2 = ( 2100 kg ) ( +17 m/s ) + (1900 kg ) ( 0 m/s ) =
2100 kg + 1900 kg +8.9 m/s The plus sign indicates that the velocity of the two cars just after the collision is in the same
direction as the direction of the velocity of the 2100kg car before the collision.
b. According to the impulsemomentum theorem, Equation 7.4, we have
F ∆t
+
+
{ = (m1 4m2 )vfinal − (m1 4m2 )vafter
14 244
3
14 244
3
Impulse
due to
friction Final momentum
when cars come
to a halt Total momentum
just after collision where vfinal = 0 m/s since the cars come to a halt, and vafter = vf = +8.9 m/s . Therefore, we
have F ∆t = ( 2100 kg + 1900 kg ) ( 0 m/s ) – ( 2100 kg + 1900 kg ) ( 8.9 m/s ) = – 3.6 × 104 N ⋅ s
The minus sign indicates that the impulse due to friction acts opposite to the direction of
motion of the locked, twocar system. This is reasonable since the velocity of the cars is
decreasing in magnitude as the cars skid to a halt.
c. Using the same notation as in part (b) above, we have from the equations of kinematics
(Equation 2.9) that
2
2
vfinal = vafter + 2ax Chapter 7 Problems 371 where vfinal = 0 m/s and vafter = vf = +8.9 m/s . From Newton's second law we have that a = – f k / ( m1 + m2 ) , where f is the force of kinetic friction that acts on the cars as they skid
k
to a halt. Therefore, – fk 0 = v +2 x
m +m 1
2
2
f or x= ( m1 + m2 ) vf2
2 fk According to Equation 4.8, f...
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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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