Unformatted text preview: e plate breaks,
we have,
−m1 ( 3.00 m/s ) sin 25.0° + m2 (1.79 m/s ) cos 45.0° = 0
(1)
m1 ( 3.00 m/s ) cos 25.0° + m2 (1.79 m/s ) sin 45.0° − (1.30 kg ) ( 3.07 m/s ) = 0
Subtracting (2) from (1), and noting that cos 45.0º = sin 45.0º, gives (2) m1 = 1.00 kg . Substituting this value into either (1) or (2) then yields m2 = 1.00 kg . 24. REASONING We will divide the problem into two parts: (a) the motion of the freely
falling block after it is dropped from the building and before it collides with the bullet, and
(b) the collision of the block with the bullet.
During the falling phase we will use an equation of kinematics that describes the velocity of
the block as a function of time (which is unknown). During the collision with the bullet, the
external force of gravity acts on the system. This force changes the momentum of the
system by a negligibly small amount since the collision occurs over an extremely short time
interval. Thus, to a good approximation, the sum of the external forces acting on the system
during the collision is negligible, so the linear momentum of the system is conserved. The
principle of conservation of linear momentum can be used to provide a relation between the
momenta of the system before and after the collision. This relation will enable us to find a
value for the time it takes for the bullet/block to reach the top of the building.
SOLUTION Falling from rest (v0, block = 0 m/s), the block attains a final velocity vblock just
before colliding with the bullet. This velocity is given by Equation 2.4 as
vblock
{
Final velocity of
block just before
bullet hits it = v0, block + a t
13
2
Initial velocity of
block at top of
building where a is the acceleration due to gravity (a = −9.8 m/s2) and t is the time of fall. The
upward direction is assumed to be positive. Therefore, the final velocity of the falling block
is
vblock = a t
(1) 358 IMPULSE AND MOMENTUM During the collision with the bullet, the total linear momentum of the bullet/block system is
conserved, so we have that
m
v
(144+ m )3
4
2444
bullet block f = mbullet vbullet + mblock vblock
144424444
4
3 (2) Total linear momentum
before collision Total linear momentum
after collision Here vf is the final velocity of the bullet/block system after the collision, and vbullet and
vblock are the initial velocities of the bullet and block just before the collision. We note that
the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of
the building. This means that vf , the final velocity of the bullet/block system after the
collision must have the same magnitude as vblock, the velocity of the falling block just
before the bullet hits it. Since the two velocities have opposite directions, it follows that
vf = −vblock. Substituting this relation and Equation (1) into Equation (2) gives (m bullet + mblock ) ( −a t ) = mbullet vbullet + mblock ( a t ) Solving for the time, we find that
t= − mbullet vbullet a ( mbullet + 2mblock ) = − ( 0.015 kg ) ( +810 m/s ) ( −9.80 m/s ) ( 0.015 kg ) + 2 (1.8 kg ) 2 = 0.34 s 25. REASONING During the time that the skaters are pushing against each other, the sum of
the external forces acting on the twoskater system is zero, because the weight of each skater
is balanced by a corresponding normal force and friction is negligible. The skaters constitute
an isolated system, so the principle of conservation of linear momentum applies. We will
use this principle to find an expression for the ratio of the skater’s masses in terms of their
recoil velocities. We will then obtain expressions for the recoil velocities by noting that each
skater, after pushing off, comes to rest in a certain distance. The recoil velocity,
acceleration, and distance are related by Equation 2.9 of the equations of kinematics. SOLUTION While the skaters are pushing against each other, the total linear momentum of
the twoskater system is conserved: m1vf 1 + m2vf 2 = {
0
14 244 Total momentum
4
3
Total momentum
after pushing before pushing Solving this expression for the ratio of the masses gives Chapter 7 Problems m1
m2 =− vf 2 359 (1) vf 1 For each skater the (initial) recoil velocity vf , final velocity v, acceleration a, and
displacement x are related by Equation 2.9 of the equations of kinematics: v 2 = vf2 + 2a x .
Solving for the recoil velocity gives vf = ± v 2 − 2ax . If we assume that skater 1 recoils in
the positive direction and skater 2 recoils in the negative direction, the recoil velocities are
Skater 1 vf 1 = + v12 − 2a1 x1 Skater 2 2
vf 2 = − v2 − 2a2 x2 Substituting these expressions into Equation (1) gives
2
− v2 − 2a2 x2
m1
=−
=
m2
v12 − 2a1 x1 2
v2 − 2a2 x2 (2) v − 2a1 x1
2
1 Since the skaters come to rest, their final velocities are zero, so v1 = v2 = 0 m/s. We also
know that their accelerations have the same magnitudes. This means that a2 = −a1, where
the minus sign denotes that the acceleration of skater 2 is opposite that of skater 1, since
they are moving in oppo...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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