Physics 201 Homework 6
Feb 13, 2013
1. A volleyball is spiked so that its incoming velocity of +4.0 m/s is changed
to an outgoing velocity of -21 m/s. The mass of the volleyball is 0.35 kg. What
impulse does the player apply to the ball?
-8.8 N-s
Solution
The impulse is equal to the change in momentum of the volleyball. The momentum before the spike is:
pi = mvi = (0.35)(+4.0) = +1.40
And after:
pf = mvf = (0.35)(−21.0) = −7.35
The total change is -8.75, which is equal to the impulse.
2. A 50.0-kg skater is traveling due east at a speed of 3.00 m/s. A 70.0-kg skater
is moving due south at a speed of 7.00 m/s. They collide and hold on to each
other after the collision, managing to move off at an angle θ south of east, with
a speed of vf . Find (a) the angle θ and (b) the speed vf , assuming that friction
can be ignored.
Solution
The important thing to remember is that when we say momentum is conserved,
this is a vector statement. In other words, if momentum is conserved, it is
conserved along any component as well. In this problem the system includes
both skaters. Therefore the momentum of the system has a eastward component
of
p = mv = (50.0)(3.00) = 150
It also has a southward component of
p = mv = (70.0)(7.00) = 490
The system momentum will have these same components after the collision.
Now, I personally don’t like to use non-standard directions. It’s just too easy to
get confused. According to a map, east is the positive x direction and north is
the positive y direction. This means that the momentum of this system is
px = 150
py = −490
Therefore,
mag ~(p) =
q
(px )2 + (py )2
=
p
(150)2 + (−490)2
= 512.45
and
ang ~(p) = tan−1 (py /px )
= tan−1 ((−490)/(150))
= −72.98◦
(a) The direction of the velocity is the same as the momentum. This is from the
positive x directon. But we should revert back to the convention in the question.
In other words, the answer should be 73.0◦ south of east.
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(a) 73.0◦ south of east
(b) vf = 4.27 m/s
(b) Since the total mass of two skaters is 120.0 kg, the velocity of the pair is given
by
p = mv
=⇒ (512.45) = (120.0)(v)
=⇒ vf = 4.27
3. Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating
rink they stand close together on skates, at rest and facing each other, with a
compressed spring between them. The spring is kept from pushing them apart
because they are holding each other. When they release their arms, Al moves
off in one direction at a speed of 0.90 m/s, while Jo moves off in the opposite
direction at a speed of 1.2 m/s. Assuming that friction is negligible, find Al’s
mass.
m = 96 kg
Solution
Initially, the momentum of the pair is zero. In order to calculate Jo’s momentum
we need to know her mass, but we are only given the combined mass of the pair.
Since we are interested in Al’s mass we should give it the label m. This means
that Jo’s mass is 168 − m. This means that Jo’s momentum is
p = mv = (168 − m)(1.2) = 201.6 − 1.2m
And Al’s momentum is
p = mv = (m)(−0.90) = −0.90m
Notice that I have made sure that the signs for the two velocities are opposite.
Since the total momentum is conserved when we add these two quantities we can
set it equal to zero and solve for m:
(201.6 − 1.2m) + (−0.90m) = 0 =⇒ m = 96kg
4. A two-stage rocket moves in space at a constant velocity of 4900 m/s. The
two stages are then separated by a small explosive charge placed between them.
Immediately after the explosion the velocity of the 1200-kilogram upper stage is
5700 m/s in the same direction as before the explosion. What is the velocity
(magnitude and direction) of the 2400-kilogram lower stage after the explosion?
4500 m/s in the direction of the original
motion
Solution
The total mass of the system is 3600 kilograms. It moves with a velocity of 4900
m/s. So the total inital momentum is
p = mv = (3600)(4900) = 1.7640 × 107
Momentum is conserved (the explosive charge is an internal force), so this is also
the final momentum. The momentum that the upper stage carries away is
p = mv = (1200)(5700) = 6.8400 × 106
This is less than the total momentum. This means that the lower stage will need
to carry the rest of the momentum: 1.080 × 107 kg-m/s. Notice this is still in the
direction of the original motion (both stages are moving in the same direction).
Since we know the mass of the lower stage, we can calculate its velocity:
(1.0800 × 107 ) = (2400)(v) =⇒ v = 4500
5. A space probe is traveling in outer space with a momentum that has a
magnitude of 7.5 × 107 kg-m/s. A retrorocket is fired to slow down the probe.
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5.1 × 107 kg-m/s
It applies a force to the probe that has a magnitude of 2.0 × 106 newtons and
a direction opposite to the probe’s motion. It fires for a period of 12 seconds.
Determine the momentum of the probe after the retrorocket ceases to fire.
Solution
Impulse is what drives the momentum to change (impulse-momentum theorem).
The impulse due to this force is given by J = F ∆t, or
J = (−2.0 × 106 )(12) = −2.4000 × 107
Notice the negative sign. Because impulse and momentum are vector quantities
we need to remember to pay attention to direction. Since the retrorockets are
opposing the motion, the force associated with them is negative (assuming the
original motion is positive).
By the impulse-momentum theorem we have J = ∆p = pf − pi . Since we know
the initial momentum we can now calculate the final momentum:
(−2.4000 × 107 ) = (pf ) − (7.5 × 107 ) =⇒ pf = 5.1000 × 107
The motion is still in the same direction, only slower.
6. A cannon of mass 5800 kilograms is rigidly bolted to the earth so it can recoil
only by a negligible amount. The cannon fires an 85.0-kilogram shell horizontally
with an initial velocity of 551 m/s. Suppose the cannon is then unbolted from
the earth, and no external force hinders its recoil. What would be the velocity of
a shell fired by this loose cannon? (Hint: In both cases assume that the burning
gunpowder imparts the same kinetic energy to the system.)
547 m/s. (The cannon recoils at -8 m/s.)
Solution
In the first situation, the gunpowder creates kinetic energy:
KE = 12 mv 2 = 12 (85.0)(551)2 = 1.2903 × 107
In the second situation, we are told the gunpowder still creates the same kinetic
energy, but now this energy is split (unequally) between the cannon and the shell.
Let the shell be m1 and the cannon m2 . Thus,
(1.2903 × 107 ) = 12 m1 v12 + 12 m2 v22
In addition, the momentum starts at zero so it must end with zero:
0 = m1 v1 + m2 v2
Using this last equation, we can get a relationship between v1 and v2 :
0 = (85.0)(v1 ) + (5800)(v2 ) =⇒ v1 = (−68.235)(v2 )
And we can plug this into the kinetic energy equation:
(1.2903 × 107 ) = 12 (85.0)(68.235)2 (v2 )2 + 21 (5800)(v2 )2
=⇒ (1.2903 × 107 ) = (200780)(v2 )2
=⇒ v2 = −8.0165
This is the recoil of the cannon (we choose the negative root, because we know
that the recoil is backward). The speed of the shell is
v1 = (−68.235)(−8.0165) = 547.01
7. A projectile (mass = 0.20 kilograms) is fired at and embeds itself in a target
(mass = 2.50 kilograms). The target (with the projectile in it) flies off after being
3
7.4%
struck. What percentage of the projectile’s incident kinetic energy does the target
(with the projectile in it) carry off after being struck?
Solution
This is a classic inelastic collision situation. The formula is
m1
u = v1
m1 + m2
The target is at rest, so it is m2 . Plugging in our data yields:
u = v1
0.20
=⇒ u = (0.074074)(v1 )
2.50 + 0.20
This is close to what we need. We are asked for a ratio of kinetic energies. The
initial kinetic energy of the projectile is
KEi = 21 mv 2 = 12 (0.20)(v1 )2 = (0.10)(v1 )2
And the final kinetic energy of the combination is
KEf = 21 mv 2 = 12 (2.70)(u)2 = (1.35)(u)2
The ratio of these two is
KEf
(1.35)(u)2
=
KEi
(0.10)(v1 )2
(1.35) (u)2
=
(0.10) (v1 )2
2
u
= (13.500)
v1
We just calculated this ratio: u/v1 = 0.074074, so
KEf
= (13.50)(0.074074)2 = 0.074074
KEi
8. Two identical balls are traveling toward each other with velocities of -4.0 and
+7.0 m/s, and they experience an elastic head-on collision. Obtain the velocities
(magnitude and direction) of each ball after the collision.
Solution
Since the collision is elastic, both momentum and kinetic energy are conserved.
But since both balls are moving, none of the collision formulas that we have are
applicable. We must go back to first principles. We know the initial (and therefore
total) momentum of the system is
p = m1 v1 + m2 v2
= (m)(−4.0) + (m)(7.0)
= (3.0)(m)
And the inital (and therefore total) kinetic energy of the system is
KE = 12 m1 v12 + 12 m2 v22
= 21 (m)(−4.0)2 + 12 (m)(7.0)2
= (32.5)(m)
We will use the collision convention that u represents the final velocities. The
final momentum of the system is
p = m1 u1 + m2 u2
=⇒ (3.0)(m) = (m)(u1 ) + (m)(u2 )
=⇒ 3.0 = u1 + u2
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They exchange velocities
And the final kinetic energy is
KE = 21 m1 u21 + 12 m2 u22
=⇒ (32.5)(m) = 12 (m)(u1 )2 + 12 (m)(u2 )2
=⇒ 65 = u21 + u22
So, we have two equations with two unknowns. We can take the first to find an
expression for u2 :
u2 = 3.0 − u1
And plug this into the kinetic energy equation:
65 = u21 + (3.0 − u1 )2
=⇒ 65 = u21 + 9.0 − (6.0)(u1 ) + u21
=⇒ 0 = (2)u21 − (6.0)u1 − 56.0
=⇒ 0 = u21 − (3.0)u1 − 28.0
Which is a quadratic equation. We can use the quadratic equation, but miraculously this equation will factor:
0 = u21 − (3.0)u1 − 28.0
=⇒ 0 = (u1 − 7.0)(u1 + 4.0)
=⇒ u1 = 7.0 or − 4.0
The latter solution is just recovering the initial situation, so our answer is
u1 = 7.0
Thus,
u2 = 3.0 − (7.0) = −4.0
The two balls have simply exchanged velocities. This always happens in elastic
collisions with objects of equal mass.
9. John’s mass is 86 kilograms, and Barbara’s is 55 kilograms. He is standing
on the x axis at x1 = 9.0 meters, while she is standing on the x axis at x2 = 2.0
meters. They switch positions. How far and in which direction does their center
of mass move as a result of the switch?
Solution
Be aware that if John and Barbara were an isolated system (two astronauts or
something), then their center of mass would not move at all. Forces internal
to the system cannot change the motion of the center of mass. But presumably John and Barbara are standing in a field or something, so that doesn’t
apply.footnote{Supposing this to be true, the forces provided by pushing off of
the ground are what change the position of the center of mass.}
The formula for the center of mass is
xcm =
m1 x1 + m2 x2
m1 + m2
So before the switch their center of mass is at
xcm =
(86)(9.0) + (55)(2.0)
= 6.2695
(86) + (55)
And after the switch we have
xcm =
(86)(2.0) + (55)(9.0)
= 4.7305
(86) + (55)
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-1.5 meters
The center of mass moves from 6.2695 to 4.7305. This is a change of
∆xcm = 4.7305 − 6.2695 = −1.5390
The center of mass is weighted toward John because he has more mass and follows
him when he moves.
10. A 2.50-gram bullet, traveling at a speed of 425 m/s, strikes the wooden block
of a ballistic pendulum, such as that in Figure 1. The block has a mass of 215
grams. (a) Find the speed of the bullet/block combination immediately after the
collision (b) How high does the combination rise above its initial position?
Figure 1: Problem 7.55
Solution
(a) This problem is split into two parts which correspond to the two phases of
the motion of this system. The first phase is the inelastic collision—energy is not
conserved in this phase. The second phase is the pendulum-like motion in which
energy is conserved. For the inelastic collision, the formula to use is
u = v1
m1
m1 + m2
Plugging in the data, we have:
u = (425)
(0.00250)
= 4.8851
(0.00250) + (0.215)
(b) As mentioned above, some of the original kinetic energy of the bullet is lost
to heat in the inelastic collision. But after the collision is complete, mechanical
energy is conserved. Initially, this is in the form of the kinetic energy of the
bullet/block combination. Thus,
KE = 12 (m1 + m2 )(u)2 = 12 (0.2175)(4.8851)2 = 2.5952
This energy is completely transformed into potential energy when the bullet/block
combination reaches its maximum height. Using P E = mgh, we can calculate
this height. Thus,
(2.5952) = (0.2175)(9.80)(h) =⇒ h = 1.2175
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(a) 4.89 m/s
(b) 1.22 meters