Preparation problems
11-16. A 340-g ball moving at 7.40 m/s collides with a
230-g ball initially at rest. After the collision the first ball is
moving at 1.60 m/s and the second at 8.57 m/s, both in the initial
direction of the first ball. (a) Calculate the momenta before
and after the collision and show that, to three significant figures,
momentum is conserved. (b) Is the collision elastic, totally
inelastic, or somewhere in between? Justify your answer.
Solution:
(a) We have a straightforward problem to calculate the momentum
using our definition p = mv. So first calculate the initial momentum of
the two balls, before the collision.
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(0.340 kg) (7.40 m/s) + (0.230 kg)(0 m/s) |
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Now we can calculate the momentum of the two balls after the collision:
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(0.340 kg) (1.60 m/s) + (0.230 kg)(8.57 m/s) |
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And we see that momentum is indeed conserved in the collision.
(b) We can immediately rule out a totally inelastic
collision, as the two balls do not stick to each other. In order to
determine whether the collision is elastic or not, we need to determine
whether energy is conserved. First the initial kinetic energy:
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1
2
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m1 v1i2 + |
1
2
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m2 v2i2 |
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1
2
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(0.340 kg) (7.40 m/s)2 + |
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2
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(0.230 kg)(0 m/s)2 |
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And now the final kinetic energy:
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1
2
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m1 v1f2 + |
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2
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m2 v2f2 |
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1
2
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(0.340 kg) (1.60 m/s)2 + |
1
2
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(0.230 kg)(8.57 m/s)2 |
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Since the kinetic energies are not equal, the collision must be
partially inelastic.
11-17. In a railroad switchyard, a 45-ton freight car
is sent at 8.0 mi/h toward a 28-ton car that is moving in the same
direction at 3.4 mi/h. (a) What is the speed of the pair after
they couple together? (b) What fraction of the initial kinetic
energy was lost in the collision?
Solution:
This problem involves a collision between two objects in which
they couple together-in other words, a totally inelastic collision. As
such, we know that momentum is conserved during the collision, and there
is maximal energy loss.
(a) To solve the first part of the problem, we first
convert all our known values to SI units:
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| Big Car | Little Car
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| Mass | m1 = 1.98 ×105 kg | m2 = 1.23 ×105 kg
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| Init. Velocities | v1i = 3.57 m/s | v2i = 1.60 m/s
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| Final Mass |
mf = m1 + m2 = 3.21×105 kg
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| Final Velocities | vf = unknown |
Now we write down our equation for the conservation of
momentum:
(b) Now we need to calculate the kinetic energy for
both the initial and final convigurations. First the intial
configuration:
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1
2
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m1 v1i2 + |
1
2
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m2 v2i2 |
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1.26 ×106 J + 1.57 ×105 J |
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Now the final configuration:
And we find the fraction of energy lost using the formula
Follow-up problems
10-23. A 780-g wood block is at rest on a frictionless
table when a 30-g bullet is fired into it. If the block with the
embedded bullet moves off at 17 m/s, what was the original speed of the
bullet?
Solution:
Since there are no external forces acting on the block and the
bullet, we can apply the principle of the conservation of momentum.
This gives us the equation
We can write the intial momentum as
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pi = mb vbi + mB vBi = mb vbi |
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And the final momentum as
Putting this all together, we see
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(0.03 kg + 0.78 kg)
0.03 kg
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(17 m/s) |
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10-39. If a rocket's exhaust speed is 200 m/s relative
to the rocket, what fraction of its initial mass must be ejected to
increase the rocket's speed by 50 m/s?
Solution:
We are looking for the fraction of mass lost, or the quantity
We can use the rocket equation from our constants and equation sheet to
start the problem.
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vf = vi + vex ln |
é
ê
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mi
mf
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ù
ú
û
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We can solve this equation for the [(mi)/( mf)], and then calculate
the quantity we desire.
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vi + vex ln |
é
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mi
mf
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ù
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û
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1 - e- [(vf - vi)/( vex)] |
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1 - e- [(50 m/s)/( 200 m/s)] |
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Solutions translated from TEX by TTH, version 1.57.
