PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 1
A 2.0 kg fish moving with a velocity of 6.0 m/s to the East is swallowed by a 6.0 kg fish moving with a
velocity of 1.0 m/s to the West.
a) Which of the fish has the larger magnitude (size) of momentum?
The smaller fish has a third of the mass and 6 the speed of the large fish. Momentum is the
product of the mass and speed so the smaller fish has
1
6  2
3
twice the momentum of the larger fish.
Just by looking at the relative sizes and speeds of the fish we can estimate which has the
greater momentum. The momentum is conserved so we can expect the combination of the
two fish to have a momentum in the direction of the 2.0 kg fish. We can then use this to
check whether the final answer is reasonable.
b)
Use conservation of momentum to find the final velocity of the fish. Assume that the final
velocity of both fish is the same.
terms:
m1  2.0 kg, m2  6.0 kg, v1i  6.0 m/s E, v 2i  1.0 m/s W  1.0 m/s E,
v1 f  v 2 f  v  ?
The initial momentum of the system is
pi  p1i  p 2i
 m1v1i  m2 v 2i
  2.0 kg  6.0 m/s E    6.0 kg  1.0 m/s E 
 6.0 kg  m/s E
The final momentum of the system is
p f  p1 f  p 2 f
 m1v1 f  m2 v 2 f
  2.0 kg  v   6.0 kg  v
  8.0 kg  v
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
Use the law of conservation of momentum
pi  p f
6.0 kg  m/s E   8.0 kg  v
Rearrange to find the velocity
v
6.0 kg  m/s E
8.0 kg
 0.75 m/s E
The final velocity of the ice bowler is 0.75 m/s to the East.
When we are using conservation of momentum, we are comparing the momenta just before and just
after the collision. There is no time for other forces to affect the total momentum.
c)
Compare the initial and final mechanical (kinetic) energy of the two fish system.
Initial kinetic energy
Ki  K1i  K 2i
 m1v12i  m2v22 f
  2.0 kg  6.0 m/s    6.0 kg  1.0 m/s 
2
2
 78 J
Final kinetic energy
K f   m1  m2  v 2
  8.0 kg  0.75 m/s 
2
 4.5 J
The kinetic energy has changed. Although this appears to violate the conservation of energy rule,
remember that we just changed two velocities. In this case, there may be elastic membranes inside the
fish that get stretched or compressed. There is also friction and, possibly some noise as the fish collide.
The energy is not lost; it is just converted into different forms of energy.
The kinetic energy is conserved in special cases known as elastic collisions.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 2
Anne, Bob and Carla are sliding towards each other on a frictionless ice surface. If Anne has a mass of 55
kg and initial velocity of 4.5 m/s North, Bob has a mass of 85 kg and initial velocity of 3.5 m/s East, and
Carla has a mass of 62 kg and initial velocity of 3.0 m/s West, what is the speed of Anne and Bob if they
cling to each other while Carla travels off at 2.0 m/s North after their collision?
terms:
mA  55 kg, mB  85 kg, mC  62 kg, v Ai  4.5 m/s E, v Bi  3.5 m/s N,
vCi  3.0 m/s W  3.0 m/s E, vCf  2.0 m/s N, v Af  v Bf  v  ?
Before the collision
The initial total momentum is
pi  p Ai  p Bi  pCi
 mA v Ai  mB v Bi  mC vCi
  55 kg  4.5 m/s E    85 kg  3.5 m/s N    62 kg  3.0 m/s E 
 61.5 kg  m/s E  297.5 kg  m/s N
After the collision
The final total momentum is
p f  p ABf  pCf
  mA  mB  v  mC vCf
  55 kg  85 kg  v   62 kg  2.0 m/s N 
 140 kg  v  124 kg  m/s N
Using the law of conservation of momentum
pi  p f
61.5 kg  m/s E  297.5 kg  m/s N  140 kg  v  124 kg  m/s N
Rearrange to find the velocity
v
61.5 kg  m/s E  297.5 kg  m/s N  124 kg  m/s N
140 kg
 0.439 m/s E  1.239 m/s N
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
Converting into magnitude-direction form (practice)
v  vE2  vN2

 0.439 m/s 2  1.239 m/s 2
 1.31 m/s
vN
vE
  arctan
 arctan
1.239
0.439
 70.48
The final velocity of Anne and Bob is 1.3 m/s 70.° North of East.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 3
Bob sits in the middle of a round, frictionless ice rink. He needs to get to one side as fast as possible.
Bob’s mass is 85 kg. Bob has two rocks, each having a mass of 12 kg. Bob can throw the rocks, either
individually or together at a velocity of 15 m/s with respect to Bob (the difference between Bob’s
velocity and the rock(s)’s velocity is 15 m/s). Is it better to
a) launch both rocks at once,
b) or launch the rocks one-at-a-time?
We will assume that all of the motion is in 1D, so we can use scalars. The motion of the rocks is in the
positive direction.
terms:
a)
mB  85 kg, mR  12kg, vR  vB  15 m/s , v Bf  ?
First note that Bob and the rocks are motionless so the initial momentum of the system is 0.
pi  0
Bob throws both rocks at once so the final momentum of the system is
p f  pBf  pRf
 mB vBf  2mR vRf
  85 kg  vBf   24 kg  vRf
We know that the rocks are moving at 15 m/s with respect to Bob so
vRf  vBf  15 m/s
Substituting this into the final momentum gives

p f   85 kg  vBf   24 kg  vBf  15 m/s

 109 kg  vBf  360 kg  m / s
From conservation of momentum
pi  p f
0  109 kg  vBf  360 kg  m / s
Isolating Bob’s speed
vBf 
360 kg  m / s
 3.30 m/s
109 kg
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
Bob’s final speed is 3.30 m/s (ignore the direction).
b) Once again, the initial momentum of the system is 0.
pi  0
Bob throws the first rock so the first final momentum of the system is
p f 1  pBRf  pRf
  mB  mR  vBRf  mR vRf
  85 kg  12 kg  vBRf  12 kg  vRf
We know that the first rock is moving at 15 m/s with respect to Bob and the second rock so
vRf  vBRf  15 m/s
Substituting this into the final momentum gives

p f 1   97 kg  vBRf  12 kg  vBRf  15 m/s

 109 kg  vBRf  180 kg  m / s
From conservation of momentum
pi  p f 1
0  109 kg  vBRf  180 kg  m / s
Isolating Bob’s speed
vBRf 
180 kg  m / s
 1.65 m/s
109 kg
Bob and the second rock have a speed of 1.86 m/s.
Bob then throws the second rock. We will focus on the system with Bob and the second rock sot the
initial momentum is not zero.
pi 2  pB 2  pR 2
 mB vBRf  mR vRRf
  85 kg  1.65 m/s   12 kg  1.65 m/s 
 160.2 kg  m/s
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
The final momentum of the system is
p f 2  pBf  pRf
 mB vBf  mR vRf
  85 kg  vBf  12 kg  vRf
We know that the second rock is moving at 15 m/s with respect to Bob so
vRf  vBf  15 m/s
Substituting this into the final momentum gives

p f 2   85 kg  vBf  12 kg  vBf  15 m/s

  97 kg  vBf  180 kg  m / s
From conservation of momentum
pi 2  p f 2
160.2 kg  m/s   97 kg  vBf  180 kg  m / s
Isolating Bob’s speed
vBf 
160.2 kg  m / s  180 kg  m / s
 3.51 m/s
97 kg
Bob final speed after throwing the second rock is 3.5 m/s.
This is an improvement over the final speed when Bob threw both rocks at the same time. Bob should
throw the rocks separately if he is in a hurry to get to shore.
We should also assume that there is no time separation between the two rock tosses so Bob spends
very little time at 1.65 m/s. The more time spent at 1.65 m/s, the lower the average speed. You can
check this out by comparing the average speeds to travel 100 m when Bob spends 1 s at 1.65 m/s and 10
s at 1.65 m/s. You should get v  3.44 m/s for 1 s at 1.65 m/s and v  2.95 m/s for 10 s at 1.65 m/s.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 4
A 45 kg crate slides along a horizontal, frictionless surface. What is the constant force needed to change
the velocity of the crate from 24 m/s East to 18 m/s North over a span of 5.0 s? Express the final answer
in component form.
We can use the velocities and time to calculate the acceleration, and then find the force by multiplying
the acceleration by the mass. Instead, we will find the impulse and then divide the impulse the time to
find the force.
terms:
mB  45 kg, vi  24 m/s E, v f  18 m/s N , F  ?
The impulse is the change in momentum.
I  p f  pi
 mv f  mv i
  45 kg 18 m/s N    45 kg  24 m/s E 
 810 kg  m/s N  1080 kg  m/s E
Recall that
I  Ft
or
F

I
t
 810 kg  m/s N  1080 kg  m/s E
5.0 s
 162 N N  216 N E
The constant force is 160 N North and -220 N East (or 160 N North and 220 N West).
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 5
Two identical pool balls move towards each other with velocities of v1i = 15 m/s and v2i = –25 m/s. Find
the velocities of the balls after the head-on elastic collision.
1.
2.
1D coordinates are not really needed here
sketch
v1i = 15 m/s
v2i = –25 m/s
known terms are
defined in the sketch
m1 = m
m2 = m
We are not told the masses of the two balls, just that the balls are identical. In this case we
can just leave the mass as “m”, or pick an arbitrary value (1 kg works well). Be careful if
you use an arbitrary value. There are cases where the mass is buried within the question or
is some specific value, like the mass of a proton. It is best to use the mass m and solve the
problem algebraically. This reduces round off error and avoids cases where the arbitrary
value is wrong.
3.
Conservation of Momentum (C of M)
Working with the conservation of momentum equation gives us
p f  pi
p1 f  p2 f  p1i  p2i
mv1 f  mv2 f  mv1i  mv2i


m  v1 f  v2 f    10 m/s  m
m v1 f  v2 f  m 15 m/s   m  25 m/s 
v1 f  v2 f  10 m/s
4.
•
1
Conservation of Energy (C of E)
The question states that this is an elastic collision so kinetic energy is conserved.
Conservation of kinetic energy gives us
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
K f  Ki
K1 f  K 2 f  K1i  K 2i
1
2
m v12f 
1
2
m v22 f 
1
2
m v12i 
1
2
m v22i
v12f  v22 f  15 m/s    25 m/s 
2
2
v12f  v22 f  850 m 2 /s 2
 2
Note that mass is not a factor in either of equations 1 or 2.
5.
Solve system of equations
We can isolate the final speed of the second ball using equation 1.
•
1'
v2 f  v1 f  10 m/s
We can now substitute this into equation 2.
Leave out the units, they get in the way, and everything is in m/s so we expect an answer in m/s.

v12f  v1 f  10

2
 850
v12f  v12f  20v1 f  100  850
2v12f  20v1 f  750  0
v12f  10v1 f  375  0
Here we have a quadratic equation in the first ball velocity.
*** Big Trick: The two roots of this equation are the initial and final velocity. We already
know one of the roots of this equation, the initial velocity, so factoring is easy.
***
v12f  10v1 f  375  0
 v1 f  15 v1 f  25  0
The first binomial is from the initial velocity.
The two roots are v1 f  15 m/s, which is the initial velocity, and v1 f  25 m/s which is the actual
final velocity.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
We can now sub the first ball final velocity into equation 1’ to get.
v2 f  v1 f  10 m/s
   25 m/s   10 m/s
=15 m/s
The final velocities are -25 m/s and 15 m/s for balls 1 and 2, respectively.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 6 HINT
A block of mass m1 = 2.40 kg, initially moving to the right at v1i = 4.00 m/s on a frictionless horizontal
track, collides with a spring attached to a second block of mass m2 = 3.20 kg and moving to the left with
a velocity of v2i = –3.00 m/s. The spring has a spring constant of k = 8.00 × 102 N/m.
a) Determine the velocity of block 2 at the instant when block 1 is moving to the right with a
velocity of v1f = 3.00 m/s.
Momentum is always conserved when there are no unbalanced external forces acting on the system for
any length of time. The spring forces are internal to the system so there are no unbalanced external
forces.
b)
Determine the spring compression at the time of part a).
Assume the spring is perfect so no energy is lost. Kinetic energy is converted to spring energy and then
back to kinetic energy.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 6
A block of mass m1 = 2.40 kg, initially moving to the right at v1i = 4.00 m/s on a frictionless horizontal
track, collides with a spring attached to a second block of mass m2 = 3.20 kg and moving to the left with
a velocity of v2i = –3.00 m/s. The spring has a spring constant of k = 8.00 × 102 N/m.
In this case, the momentum transfer isn’t instantaneous. As the blocks come together, the spring is
compressed, storing some of the energy. As the energy transfer continues, the spring reaches its
maximum compression, and then expands, pushing the blocks apart.
The compression of the spring determines the force between the two blocks. The forces on the two
blocks change their velocities. We will assume the spring is perfect, so no energy is lost in the
compression and expansion of the spring. Both C of M and C of E are valid throughout the contact
between blocks and spring. The elastic potential energy of the spring must be included in the total
energy.
a)
Determine the velocity of block 2 at the instant when block 1 is moving to the right with a
velocity of v1 = 3.00 m/s..
Assume the spring may be compressed when v1 = 3.00 m/s.
Sketch
before
v1i = 4.00 m/s
m1 = 2.40 kg
v1f = 3.00 m/s
during
v2i = –3.00 m/s
m2 = 3.20 kg
v2f = ?
We have everything in the conservation of momentum equation except v2f, so we can use C of M to find
v2f .
p  pi
p1  p2  p1i  p2i
m1v1  m2v2  m1v1i  m2v2i
m2v2  m1v1i  m2v2i  m1v1
v2 
m1v1i  m2 v2i  m1v1
m2
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
Substituting values
v2 

m1v1i  m2v2i  m1v1
m2
 2.40 kg  4.00 m/s    3.20 kg  3.00 m/s    2.40 kg 3.00 m/s 
3.20 kg
 2.25 m/s
The velocity of block 2 is –2.25 m/s when the velocity of block 1 is 3.00 m/s.
b)
Determine the spring compression at the time of part a).
Here is where we use C of E.
The energy of the system before contact is
Ei  K1i  K 2i
 12 m1v12i  12 m2v22i

 2.40 kg  4.00 m/s 2  12  3.20 kg  3.00 m/s 2
1
2
 33.6 J
The energy of the system when block 1 is traveling at 3.00 m/s includes the kinetic energy of the blocks,
plus the potential energy of the spring. We will let x be the compression of the spring.
E  K1  K 2  U S
 12 m1v12  12 m2 v22  12 kx 2

1
2
 2.40 kg  3.00 m/s 2  12  3.20 kg  2.25 m/s 2  12 8.00  102 N/m  x 2


 18.9 J  4.00  102 N/m x 2
Conservation of energy gives us
E  Ei


18.9 J  4.00  102 N/m x 2  33.6 J
x
33.6 J  18.9 J
4.00  102 N/m
 0.1917 m
The spring is compressed by 0.192 m when block 1 is travelling at 3.00 m/s.
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
PROBLEM 7
In a game of pool, a player wishes to sink the “11” ball in the corner pocket. In order to land in the
corner, the “11” ball must travel at an angle of 55° relative to the direction of the cue ball. If the initial
speed of the cue ball is 5.0 m/s, and the “11” ball is stationary. Find the final speed of the two balls and
direction of the cue ball. Assume the collision is elastic, both balls have a mass of 0.17 kg, and the “11”
lands in the corner pocket.
1.
Coordinates
Choose the positive x-direction as the original direction of the cue ball, v1i  5.0 m/s ˆi and
v 2i  0 .
y
2.
Create sketches.
x
m2 = 0.17 kg, v2i = 0
before
m1 = 0.17 kg, v1i = 5.0 m/s i
m2, v2f
55°
after

m1, v1f
3.
The initial total momentum is
pi  p1i  p 2i
 m1v1i  m2 v 2i


  0.17 kg  5.0 m/s ˆi   0.17 kg  0 
 0.85 kg  m/s ˆi
In component form (leave units out for simplicity)
pix  0.85
and
piy  0
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
The final momentum is
p f  p1 f  p 2 f
 m1v1 f  m2 v 2 f
 0.17 v1 f  0.17 v 2 f
In component form
p fx  0.17v1 f cos   0.17v2 f cos 55
 0.17v1 f cos   0.0975v 2 f
and
p fy  0.17v1 f sin   0.17v2 f sin 55
 0.17v1 f sin   0.1393v2 f
Conservation of Momentum gives us
pix  p fx
0.85  0.17v1 f cos   0.0975v 2 f
divide by 0.17
5.0  v1 f cos   0.5736v 2 f
1
and
piy  p fy
4.
0  0.17v1 f sin   0.1393v2 f
divide by 0.17
0  v1 f sin   0.8192v2 f
 2
The collision is elastic so energy is conserved
Ei  E f
1
2
m1v12i  12 m2v22i  12 m1v12f  12 m2 v22 f
1
2
 0.17  5.0 2  12  0.17  0 2  12  0.17  v12f
25  v12f  v22 f
5.
 12  0.17  v22 f
 3
Solve for v1f , θ, and v2f
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
In any problem solving involving trigonometry, it is usually a good idea to get rid of the angles. We will
make use of sin 2   cos2   1 with equations 1 and 2 to generate an equation without angle.
Equation 1 can be rearranged as
5.0  v1 f cos   0.5736v 2 f
v1 f cos   5.0  0.5736v 2 f
We can now square this
5.0  v1 f cos   0.5736v 2 f
v12f cos2   25.0  5.736v 2 f  0.329v22 f
1'
Equation 2 can be rearranged as
v1 f sin   0.8192v2 f
We can now square this
 2'
v12f sin 2   0.671v22 f
Adding the squared equations gives us
v12f sin 2   v12f cos 2   0.671v22 f  25.0  5.736v 2 f  0.329v22 f


v12f sin 2   cos 2   v22 f  5.736v 2 f  25.0
v12f  v22 f  5.736v 2 f  25.0
 4
We can now substitute equation 4 into equation 3.
 3
25  v12f  v22 f
25  v22 f  5.736v2 f  25  v22 f
0  2v22 f  5.736v2 f

0  2v2 f v2 f  2.868

This has solutions v2 f  0 and v2 f  2.868 .
Like problem 5, we have two solutions. One is the final solution and one is the initial solution. We know
that the zero velocity is the initial solution, so the final solution must be v2 f  2.9 m/s .
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PHYSICS SEMESTER ONE
UNIT 6: ANSWERS TO PROBLEMS
Substituting this into equation 3 gives us
 3
25  v12f  v22 f
25  v12f   2.868 
2
v1 f  25  8.225
 4.096
We can now use equation 2 to find the angle for the cue ball.
0  v1 f sin   0.8192v2 f
sin  
0.8192v2 f
v1 f
  arcsin
0.8192(2.868)
4.096
 35
Note that I chose to use the equation sin because I expected an angle between -90° and 0° ( v1 fy
would have to be negative to balance the positive v2 fy ). The inverse sine function produces angles
between -90° and 90°. The inverse cosine function produces angles between 0° and 180°. I could also
have converted to component form to find v1 fx and v1 fy .
The final cue ball velocity is 4.1 m/s at an angle of -35° from its original direction. The 11 ball final
velocity is 2.9 m/s at an angel of 55° from the original cue ball direction.
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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18