Answers  Center of Mass, Momentum,
and Collisions


1.

Impulse J = p_{f}  p_{i } = mv_{f}
 mv_{i}. Taking the return direction as positive,
J = 0.060 kg{(40  (50)}m/s = 5.4 kgm/s.


2.

X_{CM }= (m_{1}x_{1 }+ m_{2}x_{2}
+ m_{3}x_{3} + m_{4}x_{4})/(
m_{1} + m_{2} + m_{3} + m_{4})
=
{(1.0 kg)(0) + (4.0 kg)(2.0 m) + (2.0 kg)(0) + (3.0 kg)(2.0
m)}/(10.0 kg) = 1.4 m.
Y_{CM} = (m_{1}y_{1 }+ m_{2}y_{2}
+ m_{3}y_{3} + m_{4}y_{4})/(
m_{1} + m_{2} + m_{3} + m_{4})
=
{(1.0 kg)(0) + (4.0 kg)(0) + (2.0 kg)(2.0 m) + (3.0 kg)(2.0
m)}/(10.0 kg) 1.0 m.


3.

 Taking east in the +i direction and north in the
+j direction,
J = p_{f}  p_{i }=
(1500 kg)(8.0 m/s)(j  i) = 12 x 10^{3}
kgm/s(j  i).
 J = F_{av }t
F_{av} = J/t = {12 x 10^{3}
kgm/s(j  i)}/3.0 s = 4 x 10^{3}
N(j  i).
 By Newton's third law of motion, the average force of
the car on the road = 4 x 10^{3} N(j  i).


4.

Divide the system into three pairs of particles, consisting
of 1 + 2, 3 + 4, and 5 + 6. The midway for 1 + 2 is at a and
the midpoint for 5 + 6 is c. We can treat 1 + 2 as a point particle
of mass of 2m at a and 5 + 6 as a point particle of mass 2m
at c. We can combine these two point particles to be another
point particle with mass 4m at b. 3 + 4 have their midpoint
at b so we can replace them with a point particle of mass 2m
at b. Finally the whole works acts like a point particle of
mass 6m at b.


5.

 p_{A } = (1.0 kg)(4.0 m/s) = 4.0 kgm/s.
p_{B } = (2.0 kg)(2.0 m/s) = 4.0 kgm/s.
 K_{A} = 1/2(1.0 kg)(4.0 m/s)^{2 }= 8.0
J.
K_{B} = 1/2(2.0 kg)(2.0 m/s)^{2} = 4.0 J.
While the momenta are equal, the kinetic energies are not.


6.

Take the distance to the edge of the
rink as x. Then the velocity v_{1 }after the collision
for the skater of mass m_{1 }equals +x/12 s and the
velocity v_{2} of the other skater equals x/18 s. From
conservation of momentum,
p_{i }= p_{f}
0 = (50 kg)(x/12 s) + (m_{2})(x/18 s) or
(m_{2})(x/18 s) = (50 kg)(x/12 s)
m_{2} = (50 kg)((18/12) = 75 kg


7.

Take to the right to be positive. From
conservation of momentum,
p_{i
}= p_{f}
0.050 kg(0.20 m/s) + (0.10 kg)((0.04 m/s) = (0.05 + 0.10)kg
v_{f}
(0.010
 0.004)kgm/s = (0.15 kg)v_{f
} v_{f}
= 0.040 m/s to the right


8.

For Fig. 3a:
(a) The area of the triangle is 1/2 (base)(altitude) = ba/2.
(b) The mass per unit area σ
= mass/area = M/(ba/2) = 2M/ba.
(c) The area of the element dA = (y)dx.
(d) Since σ = mass/area,
the mass of any element equals σ
times the area of the element. dm = (σ)dA
= (2M/ba)ydx.
(e) dm x = (2M/ba)ydx (x)
(f) In general, the equation of a line is y = mx + b, where
m is the slope of the line and b is the intercept on the yaxis.
In Fig. 3, we see b = 0 and m = a/b, so y = (a/b)x.
(g) From the equation of the line found in (f) we substitute
y into dm x =
(2M/ba) y dx x, or dm x = (2M/ba) (a/b)x dx x = 2M/b^{2}
x^{2} dx, and
(h)
For Fig. 3b:
(a) and (b) The solutions to both (a) and (b) are the same
for both Fig. 3a and Fig. 3b.
(c ) dA = (b  x)dy
(d) dm = (2M/ba) (b  x)dy
(e) and (f) The solutions to both (e) and (f) are the same
for both Fig. 3a and Fig. 3b.
(g) from the equation of the line y = (a/b) x, x =
by/a
(h) y dm = y (2M/ba)(b  by/a)dy = (2M/ba) (b/a)(a  y)
y dy =
(2M/a^{2})(a  y) y dy
Y_{CM } = ∫ dm y/M = ∫(2M/a^{2})
(ay  y^{2}) dy /M = 2/a^{2} (ay^{2}/2
 y^{3}/3)
taking the integral from 0 to a,
Y_{CM} = (2/a^{2}) (a^{3}/2  a^{3}/3)
= (2/a^{2}) (3a^{3}/6  2a^{3}/6)
= a/3


9.

The center of mass of the four particles at the base of the
pyramid is the center of the base. The center of mass of the
fiveparticle system lies on the line perpendicular to the
base and passes through the peak of the pyramid. This line
passes through the center of mass of the four particles at
the base and gives the xcoordinate of the center of mass
at the center of the base.
Y_{CM } = {(m)(h) + (4m)(0)}/5m = h/5.


10.

In the figure above, b represents the center of mass of the
barge and Δb the distance
the barge moves to the right as the man moves to the left. No
external forces act on the bargeman system and the center of
mass CM of this system remains at rest.
Initially, X_{CM }= {(75 kg)(10 m) + (225 kg)(b)}/(75
+ 225)kg (Equation
1)
Finally, X_{CM }= {(75 kg)x + (225 kg)(b + Δb)}/(75
+ 225)kg (Equation
2)
Since the CM of the system remains at rest, equate Eq. 1 and
Eq. 2 and cancel denominators,
(75 kg)(10 m) + (225 kg)(b) = (75 kg)x + (225 kg)(b + Δb)
(Equation 3)
Distance moved by man relative to shore = (10 m  x) = 2.0 m
 Δb
or Δb
= x  8.0 m.
Substituting this expression for Δb
into Eq. 3:
750 N + (225 kg)(b) = (75 kg)x + (225 kg)(b + x  8.0 m)
750 N + 1800 N = (300 kg)x and
x = 8.5 m.


11.

Momentum is conserved:
p_{i} = p_{f}
p_{in } = p_{1f} + p_{2f}
Since the final momentum of product 1, p_{1f}, =
 p_{2f}, the negative of the final
momentumof product 2, the initial momentum p_{in}
of the nucleus must equal zero.


12.

(p_{i})_{x}
= (p_{f})_{x}
(3.0 kg)(1.0 m/s) = (5.0 kg)(v_{f})_{x} (v_{f})_{x
}= 0.60 m/s
(p_{i})_{y
}= (p_{f})_{y
}(2.0 kg)(2.0 m/s) = (5.0 kg)(v_{f})_{y } (v_{f})_{y}
= 0.80 m/s
v_{f} = {(0.60)^{2 }+ (0.80)^{2}}^{1/2
}= 1.0 m/s. tan Θ
= 0.80/0.60. Θ = 53^{o}.


13.

p_{i} = p_{f} 
mv_{o} 
= 
mv_{o} 
K_{i} = K_{f} 
1/2 mv_{o}^{2} 
= 
1/2 mv_{o}^{2} 
p_{i} = p_{f} 
mv_{o} 
= 
mv_{o}/2 + mv_{o}/2 
K_{i } ≠
K_{f} 
1/2 mv_{o}^{2} 
≠ 
1/2 m(v_{o}/2)^{2
}_{ }+ 1/2 m(v_{o}/2)^{2} 
Another conservation law is at work here. In an elastic collision,
kinetic energy is also conserved. For an elastic collision,
(b) cannot occur.


14.

For an elastic collision,
p_{i}
= p_{f}
m_{1}v_{1i } = m_{1}v_{1f}
+ m_{2}v_{2f
}(Equation
1)
where I have assumed both final velocities to the right. In
any calculation, if one of them is to the left, you will get
a minus answer in your solution.
K_{i}
= K_{f
}1/2 m_{1}v_{1i}^{2 }= 1/2 m_{1}v_{1f}^{2}
+ 1/2 m_{2}v_{2f}^{2
}(Equation
2)
Solving Eq. 1 and Eq. 2 simultaneously for v_{1f} and
v_{2f} gives:
v_{1f} = {(m_{1}  m_{2})/(m_{1}
+ m_{2})} v_{1i} and
v_{2f} = {2m_{1}/(m_{1} + m_{2})}
v_{1i}
For v_{1i }= 2.0 m/s, m_{1} = 1.0 kg
and
 m_{2 }= 1.0 kg, v_{1f} = 0 and
v_{2f} = 2.0 m/s = v_{1f}
Example: Cue ball colliding with another billiard ball.
 m_{2} = 1000 kg, v_{1f }=  (999/1001)(2.0
m/s) ≈ (1)(2.0 m/s) =  v_{1i}
v_{2f} = (2.0/1001)(2.0 m/s) ≈ 0.004 m/s,
which is small compared to v_{1i}
Example: A basketball colliding with a bowling ball is an
example of a relatively small mass colliding with a large
mass.
 m_{2} = 0.001 kg, v_{1f }= (0.999)(2.0
m/s) ≈ (1)(2.0 m/s) = v_{1i}
v_{2f} = (2.0/1.001)(1.0 m/s) ≈ 2 v_{1i}
Example: A bowling ball colliding with a basketball is an
example of a relatively large mass colliding with a small
mass.


15.

Momentum is conserved, 



p_{i} 
= 
p_{f} 

(4.0
kg)(2.0 m/s) + (1.0 kg)(3.0 m/s) 
= 
(4.0
kg)v_{1f} + (1.0 kg)v_{2f} 

5.0 m/s 
= 
4.0v_{1f
} + 1.0v_{2f} 
or 
v_{2f} 
= 
5.0 m/s
 4.0v_{1f } (Equation
1) 




Kinetic energy is conserved, 



K_{i} 
= 
K_{f} 
1/2(4.0kg)(4.0m^{2}/s^{2})+1/2(1.0kg)(9.0m^{2}/s^{2}) 
= 
1/2(4.0kg)v_{1f}^{2}+1/2(1.0kg)v_{2f}^{2} 

1/2 (25)(m/s)^{2} 
= 
1/2(4.0v_{1f}^{2}+
1.0v_{2f}^{2}) 
or 
(25)(m/s)^{2} 
= 
4.0v_{1f}^{2}
+ 1.0v_{2f}^{2} (Equation
2) 
Substituting Eq. 1 into Eq. 2:
25 (m/s)^{2} =
4.0v_{1f}^{2} + (5.0 m/s  4.0v_{1f})^{2}
25 (m/s)^{2} = 4.0v_{1f}^{2} +_{
}25 (m/s)^{2}  40v_{1f} m/s + 16v_{1f}^{2
} 0 = 20v_{1f}^{2}  40v_{1f}^{
}m/s or
0 = v_{1f}(v_{1f}  2 m/s)
Solutions are v_{1f} = 0 and
2.0 m/s. To decide which to choose, use Eq. 1.
For v_{1f }= 0, v_{2f} = 5.0 m/s.
For v_{1f} = 2.0 m/s, v_{2f} =  3.0 m/s.
The latter is not possible because they would collide. The correct
solution is shown in Fig. for 15a.


16.

The velocity of the center of mass V = (m_{1}v_{1i}
+ m_{2}v_{2i})/(m_{1} + m_{2}) =
{4.0(2.0)+1.0(3.0)}kgm/s/5.0 kg = 1.0 m/s.
In the center of mass coordinate system moving with V = 1.0
m/s to the right,
v_{1i}’ = v_{1i } V = 2.0
m/s  1.0 m/s = 1.0 m/s.
v_{2i} = v_{2i } V = 3.0 m/s  1.0 m/s =
4.0 m/s._{
}v_{1f}^{’}_{ }= v_{1f
} V =  1.0 m/s = 1.0 m/s.
v_{2f’ }= v_{2f } V = 5.0 m/s  1.0
m/s = 4.0 m/s.
Prime velocities are shown in Fig. 16b above.
In center of mass system, p_{i }= (4.0 kg)(1.0 m/s)
+ (1.0 kg)(4.0 m/s) = 0
= p_{f} = (4.0 kg)(1.0 m/s) + (1.0 kg)(4.0 m/s). The
center of mass system is a zero momentum system.


17.

If momentum, a vector quantity, is conserved, its components
are conserved.
(p_{i})_{x} = (p_{f})_{x}
10 kg(24 m/s) = (20 kg)(v_{2f})_{x
} (v_{2f})_{x} = 12 m/s
(p_{i})_{y} = (p_{f})_{y
} 0=10 kg(10 m/s)+20 kg (v_{2f})_{y
} (v_{2f})_{y} = 5.0 m/s
v_{2f} = {(v_{2f})_{x}^{2 }+
(v_{2f})_{y}^{2}}^{1/2
} v_{2f} = (12^{2} + 5.0^{2})^{1/2}
m/s = 13 m/s.
tan Θ= 5/12. Θ
= 22.6^{o}.


18.

(p_{i})_{x} = (p_{f})_{x
} cos 37^{o} = 0.80 and
sin 37^{o} = 0.60
2.0 kg(39 m/s) = 2.0 kg(v_{1f})_{x} + 8.0
kg(10 m/s)(0.80)
(78  64)m/s = 2.0(v_{1f})_{x}
(v_{1f})_{x} = 7.0 m/s
(p_{i})_{y } = (p_{f})_{y}
0 = 2.0 kg(v_{1f})_{y }+ 8.0 kg(10 m/s)(0.6)
 24 m/s = (v_{1f})_{y
}v_{1f }= {(7.0)^{2} + (24)^{2}}^{1/2
}m/s = 25 m/s
tan Θ =  24/7.
Θ =  74^{o}.


19.

Momentum is conserved
(p_{x})_{i} = (p_{x})_{f}
mv_{1i} = m(v_{1f})_{x} + m(v_{2f})_{x
} (Equation
1)
(p_{y})_{i} = (p_{y})_{f}
0 = mv_{1f} sin Θ_{1}
+ mv_{2f} sin Θ_{2}.
Since v_{1f} = v_{2f},
Θ_{1} =  Θ_{2
} and
(v_{1f})_{x} = (v_{2f})_{x } (Equation
2)
From Eq. 1,
v_{1i} = 2 (v_{1f})_{x }= 2v_{1f}
cos Θ_{1} (Equation
3)
From conservation of kinetic energy, K_{i }= K_{f}
1/2 m v_{1i}^{2} = 1/2 m v_{1f}^{2}
+ 1/2 m v_{2f}^{2} = mv_{1f}^{2}
or
v_{1f} = v_{2f} = v_{1i}/(2)^{1/2
}From Eq. (3), v_{1i} = 2 [v_{1i}/(2)^{1/2}]
cos Θ_{1}; cos
Θ_{1} = (2)^{1/2}/2
and
Θ_{1} =45^{o and }Θ_{2}
=  45^{o
} 

20.

 Take the zero of gravitational potential energy at the
bottom of the ramp, U_{af} = 0. At the top
of the ramp, U_{ai }= mgh, where h = 0.40 m.
On the ramp, from conservation of energy
U_{ai
} +
K_{ai} = U_{af }+ K_{af
} m_{A}(9.8 m/s^{2})(0.40 m) +
0 = 0 + 1/2 m_{A}v_{fa}^{2}
v_{fa}
= [19.6(0.4)]^{1/2}m/s = 2.8 m/s
 When A collides with B, momentum is conserved.
p_{bi} = p_{bf}
m_{A}v_{ib }= (m_{A} + m_{B})v_{fb}
v_{fb} = [m_{A}/(m_{A} + m_{B})]v_{ib}
= [2.0/(11.2)]2.8 m/s = 0.50 m/s
 Work done by friction = (U_{cf }+ K_{cf}
)  (U_{ci }+ K_{ci})
fd cos 180^{o} =
(2.8 N)d = (0 + 0)  [0  1/2(11.2 kg)(0.50 m/s)^{2}]
(2.8 N)d = 1.4 Nm and
d = 0.50 m.


21.

 For an elastic collision in one dimension,
v_{1f} = (m_{1}  m_{2})/(m_{1
}+ m_{2}) v_{1i } and
v_{2f} = 2m_{1}/(m_{1 }+ m_{2})
v_{1i}
In this problem, m_{1} = m_{2 }
and v_{1i }= v_{o. }Thus the first
block stops after the collision, v_{1f} =
0, and the second block gains a velocity v_{2f}
= v_{o} (Fig. 6a' above). Now the second block swings
up to height h. Energy is conserved. Taking the middle
of the block as the zero of potential energy at the initial
position i, the potential energy at "final"
position of height h, where the block comes momentarily
to rest is mgh.
U_{i }+ K_{i
} = U_{f}
+ K_{f
} 0 + 1/2 mv_{o}^{2 }= mgh +
0
and
h = v_{o}^{2}/2g
 For the inelastic collision,
p_{i } = p_{f}
mv_{o }= 2mv_{b}
_{ }v_{b} = v_{o}/2
U_{ib }+ K_{ib
} =
U_{fb} + K_{fb
} 0 + 1/2(2m)(v_{o}^{2}/4)^{
}= 2mgh + 0 and
h_{b} = v_{o}^{2}/8g


22.

p_{ia}
= p_{fa
} 0.020 kg(v_{o1}) = (0. 020 + 5.0)kg v_{af}
v_{af}
= (0.02/5.02)v_{o1
} It takes 0.80 N to keep the 5.0 kg moving with constant
velocity. The increase in the normal force due to the weight
of the bullet does not change the frictional force to two significant
figures. Take the gravitational potential at the surface = 0
= U_{fb }= U_{ib}.
Work by friction = (U_{fb }+ K_{fb})  (U_{ib}
+ K_{ib})
(0.80 N)(1.5 m) = (0 + 0) 
{0 + 1/2(0.502 kg)[0.02v_{o1}/5.02]^{2}}
1.2 kg(m/s)^{2} = 3.98 x 10^{6}
v_{o1}^{2 }kg
v_{o1} = 549 m/s


23.

Take the potential energy at the bottom of swing to be zero.
The kinetic energy at the top of the swing is zero since the
velocity of A there v_{bi }= 0.
See the figure for #23 above for parts (a), (b), (c) and (d).
 U_{bf} + K_{bf} = U_{bi }+ K_{bi}
0 + K_{bf }= (2.0kg)(10 m/s^{2})(0.80 m)
+ 0
K_{bf }= 16 J = 1/2(2.0 kg)v_{bf}^{2}
 v_{bf} = 4.0 m/s
 p_{i }= p_{f
} 2.0 kg(4.0 m/s) = (2.0 + 3.0)kg v_{cf}
v_{cf} = 1.6 m/s
 U_{df} + K_{df} = U_{di }+ K_{di}
5.0 kg(10 m/s^{2})h + 0 = 0 + 1/2(5.0 kg)(1.6 m/s)^{2}
h = 0.13 m


24.

 Parts (a) and (b) above have the same approach as #23
above. The figures for (a) and (b) are the same as for #23
above.
 Parts (a) and (b) above have the same approach as #23
above. The figures for (a) and (b) are the same as for #23
above.
 For the elastic collision,
v_{Bf} = 2m_{A}/(m_{A }+ m_{B})(4.0
m/s)= (4/5)(4.0 m/s) = 3.2 m/s.
v_{Af} = (m_{A}  m_{B})/(m_{A
}+ m_{B})(4.0 m/s) = (1.0/5)(4.0 m/s) = 0.8
m/s.
 U_{df} + K_{df} = U_{di }+ K_{di}
For B, (3.0 kg)(10 m/s^{2})h_{B} = 1/2(3.0
kg)(3.2 m/s)^{2}. h_{B }= 0.51 m.
For A, (2.0 kg)(10 m/s^{2})h_{A }= 1/2(2.0
kg)(0.8 m/s)^{2}. h_{A} = 0.032 m.


25.

 Do the problem in steps. First there is the collision
between the bullet of mass m = 5.0 x 10^{3} kg
travelling with a speed of 400 m/s and the block of mass
M = 1.0 kg. After the collision, the bullet has a
velocity v and the block a velocity v_{M}.
From conservation of momentum,
p_{i}
= p_{f}
(5.0 x 10^{3} kg)(400 m/s) = (1.0 kg)v_{M}
+ (5.0 x 10^{3} kg)v (Equation
1)
Just after the collision and just before the block compresses
the spring, the initial potential energy of the blockspring
system is 0 and the kinetic energy of the block is 1/2 (1.0
kg) v_{M}^{2}. When the spring has
been compressed to its maximum, the block momentarily comes
to rest and its "final" kinetic energy is zero,
but the potential energy of the blockspring system is 1/2
kx^{2} = 1/2(900 N/m)((5 x 10^{2}m)^{2}.
From conservation of energy,
U_{i } + K_{i
} =
U_{f}
+
K_{f} _{ }
0 + 1/2(1.0 kg) v_{M}^{2} = 1/2(900 N/m)(5.0
x 10^{2} m)^{2} + 0 or
v_{M} = 1.5 m/s
Substituting this value of v_{M} back into Eq. 1:
(5.0 x 10^{3} kg)(400m/s) = (1.0kg)(1.5m/s) + (5.0
x10^{3} kg)v
or
v = 100 m/s
 For the collision,
K_{i } = 1/2 (5 x 10^{3} kg)(400 m/s)^{2}
= 400 J
K_{f} = 1/2(5.0 x 10^{3} kg)(100 m/s)^{2}
+ 1/2(1.0 kg)(1.5 m/s)^{2}
= (25 + 1.12)J
K lost = (400  26.12)J = 374 J


26.

This problem is a reminder to choose your system carefully.
When you drop an object, its momentum does not stay constant
because the gravitational attraction of the earth, an external
force, acts on it. If you take your system to be the object
and the earth, then momentum is conserved because the force
of the earth on the object and the force of trhe object on
the earth are internal forces. There is no net external force
on the system of the object and the earth and momentum is
conserved. Let the object be a textbook with mass m_{b}
= 2.0 kg and velocity v_{b.} The earth has a mass
M_{e} = 5.98 x 10^{24} kg. Before you release
the book, the momentum of the system is zero.
p_{i} = p_{f}
0 = m_{b}v_{b }+ M_{e}v_{e},
where v_{e} = the speed acquired by the earth.
v_{e} = (m_{b}/M_{e})v_{b}
= (2.0 kg/5.98 x 10^{24} kg)v_{b} = 3.3 x 10^{25}
v_{b}.
So the earth does move up, but we do not detect the motion.

