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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999
Solutions for Assignment # 6
by Dru Renner [email protected]
November 1, 1999 10:14 am
Problem 6.1 (Ohanian, page 291, problem 13)
The masses of the two automobiles are m 1 = 540 kg and m 2 = 1400 kg andtheir velocities, measuredalong the original direction of m 1 , are v 1 = 80 km/h ≈ 22 .22 m/s and v 2 = −80 km/h ≈ − 22 .22 m/s.
(a) We are toldthat the two automobiles remain lockedtogether after the collision.
After the collision both automobiles move as one, hence they both must have the same velocity which must also be the velocity of the center of mass (vCM ). There are no outside forces, so the velocity of the center of mass remains unchanged during the collision. Therefore
vCM =
m 1 v 1 + m 2 v 2 m 1 + m 2
· 80 km/h ≈ −35 km/h ≈ − 9 .9 m/s
The velocity of the wreck is given by vCM. Note, it was unnecessary to convert from the non-metric unit “km/h” if we wantedthe value of vCM in units of “km/h”.
(b) The total kinetic energy before the collision is
Kbef ore =
m 1 v 12 2
m 2 v^22 2
(540)(22.22)^2 +
(1400)(22.22)^2 ≈ 4. 8 × 105 J
The total kinetic energy after the collision is
Kaf ter =
(m 1 + m 2 )v^2 CM 2
≈ 8. 2 × 104 J
Note, it was necessary to convert from “km/h” to “m/s” to achieve an answer in units of “J”.
(c) This calculation is best done in the center of mass frame. Since there are no
outside forces, this frame is inertial; hence the acceleration in this frame is the same as the acceleration relative to the ground. We will make use of the kinematic relation
a =
v^2 − v^20 2(x − x 0 )
For the automobile of mass m 1 , its speedin the center of mass frame is
v 1 ′ = v 1 − vCM = 22. 2 − (− 9 .9) = 32.1 m/s
The acceleration the passenger compartment experiences is given by
a 1 =
v′ 12 2 d
≈ 8. 6 × 102 m/s^2
For the automobile of mass m 2 , its speedin the center of mass frame is
v′ 2 = v 2 − vCM = − 22. 2 − (− 9 .9) = − 12 .3 m/s
The acceleration the passenger compartment experiences is given by
a 2 =
v′ 22 2 d
≈ 1. 3 × 102 m/s^2
Problem 6.2 (Ohanian, page 292, problem 23)
This problem is illustratedin Figure 11.12 on page 292. Let m be the mass of each steel ball. The first steel ball swings andgains kinetic energy until it collides with the secondsteel ball. Initially the first steel ball is at rest at an angle θ relative to vertical, so its height relative to the bottom of the swing is
h = l(1 − cos θ)
Its mechanical energy is
E = mgh = mgl(1 − cos θ)
At the bottom of its swing, just before the collision, its mechanical energy is
E =
mv^2 2
Conservation of mechanical energy gives
mgl(1 − cos θ) =
mv^2 2
=⇒ v =
√ 2 gl(1 − cos θ)
Also in terms of just h, conservation of mechanical energy gives
mgh =
mv^2 2
=⇒ h =
v^2 2 g
conserved: John and Nancy can interchange internal energy with kinetic energy by sliding andcatching the block.
(a) Let vJ 1 be the velocity of John’s sledand vB 1 be the velocity of the block after John
releases the block. The speedof the block relative to John’s sledis saidto be 3 m/s, thus
vB 1 = vJ 1 + 3
Momentum conservation gives
0 = mJ vJ 1 + mB vB 1 = mJ vJ 1 + mB (vJ 1 + 3) =⇒
vJ 1 = −
3 mB mJ + mB
≈ − 0 .27 m/s =⇒
vB 1 = vJ 1 + 3 =
3 mJ mJ + mB
≈ 2 .7 m/s
(b) Let vN 1 be the velocity of Nancy’s sledafter she catches the block. Momentum
conservation gives
mB vB 1 = (mB + mN )vN 1 =⇒
vN 1 =
mB mB + mN
· vB 1 =
mB mB + mN
3 mJ mJ + mB
≈ 0 .30 m/s
(c) Let vN 2 be the velocity of Nancy’s sledand vB 2 be the velocity of the block after
Nancy slides the block to John. The speedof the block relative to Nancy’s sledis saidto be 3 m/s, thus
vB 2 = vN 2 − 3
(Note that here it is “−3” andabove it is “+3”. This is due to our choice for positive velocity.) Momentum conservation gives
(mB + mN )vN 1 = mB vB 2 + mN vN 2 = mB (vN 2 − 3) + mN vN 2 =⇒
vN 2 =
(mB + mN )vN 1 + 3mB mB + mN
≈ 0 .63 m/s =⇒
vB 2 = vN 2 − 3 ≈ − 2 .4 m/s
(d) Let vJ 2 be the velocity of John’s sledafter he catches the block. Momentum
conservation gives
mJ vJ 1 + mB vB 2 = (mJ + mB )vJ 2 =⇒ vJ 2 =
mJ vJ 1 + mB vB 2 (mJ + mB )
≈ − 0 .46 m/s
(e) The total kinetic energy of the block andsledafter John releases the block is
K =
mJ v J^21 2
mB v B^21 2
≈ 41 J
Initially there was no kinetic energy, but John converts chemical energy (storedin his muscles) to kinetic energy by sliding the block to Nancy.
(f ) The total kinetic energy of the two sleds and the block just after Nancy catches the
block is
K =
mJ v J^21 2
(mB + mN )v N^2 2
≈ 7 .9 J
There are internal forces involvedwhen Nancy catches the block. These forces convert kinetic energy into other forms of energy.
(g) The total kinetic energy of the two sleds and the block just after Nancy releases
the block is
K =
mJ v^2 J 1 2
mB v^2 B 2 2
mN v^2 N 2 2
≈ 48 J
Nancy converts chemical energy (storedin her muscles) to kinetic energy by sliding the block back to John.
(h) The total kinetic energy of the sleds after John catches the block is
K =
(mB + mJ )v^2 J 2 2
mN v^2 N 2 2
≈ 28 J
There are internal forces involvedwhen John catches the block. These forces again convert kinetic energy into other forms of energy.
Problem 6.
The center of mass frame is defined as the frame in which the center of mass is at rest, so the two pucks must travel in opposite directions in that frame. If v 1 and v 2 are the velocities respectively of masses m 1 and m 2 , then
m 1 v 1 + m 2 v 2 = 0 =⇒ v 1 = −
m 2 m 1
· v 2
Clearly the total momentum is zero, but the kinetic energy is
(c) The total kinetic energy in the center of mass frame before the collision was given
above as
K =
m 2 v 22 2
( m 2 m 1
)
(d) We assumedthat the collision was elastic, so kinetic energy is conserved. Therefore
the total kinetic energy after the collision is given as
K =
m 2 |v′ 2 |^2 2
( m 2 m 1
)
m 2 v 22 2
( m 2 m 1
)
Problem 6.
The tennis ball and basketball do not collide until after the basketball strikes the ground. Both balls fall the same distance, call it h; hence both have the same velocity, v =
2 gh just before the basketball strikes the ground. The mass of the basketball is insignificant comparedto the mass of the planet; thus the basketball will bounce back up with the same speed, v. (We assume that the collision is elastic, i.e. kinetic energy is conserved.) Now the tennis ball andbasketball collide. This collision is best viewedfrom the frame moving with the basketball. In this frame the tennis ball travels at a speed2 v and strikes the basketball at rest. The basketball is much heavier than the tennis ball, thus the tennis ball will bounce back up with the same speed, 2v. (Again, we assume that the collision is elastic.) In the original frame, the tennis ball has the speed2 v + v = 3v. Therefore, the height it will reach is h′^ = (3v)
2 2 g = 9
v^2 g = 9h.
Problem 6.
Let m = 6 kg, v = 350 m/s, m 1 = 2 kg, v 1 = 250 m/s, m 2 = 4 kg, and v 2 = 400 m/s.
(a) The total momentum before the collision is
Pbef ore = mv = 2. 1 × 103 kg · m/s
The total momentum after the collision is
Paf ter = m 1 v 1 + m 2 v 2 = 2. 1 × 103 kg · m/s
Therefore, Pbef ore = Paf ter, andmomentum is conserved.
(b) The total kinetic energy before the collision is
Kbef ore =
mv^2 2
≈ 3. 68 × 105 J
The total kinetic energy after the collision is
Kaf ter =
m 1 v^21 2
m 2 v 22 2
≈ 3. 83 × 105 J
Therefore, Kbef ore = Kaf ter, andkinetic energy is not conserved.
(c) The center of mass velocity before the collision is
Vbef ore =
mv m
= v = 350 m/s
The center of mass velocity after the collision is
Vaf ter =
m 1 v 1 + m 2 v 2 m 1 + m 2
= 350 m/s
Therefore, Vbef ore = Vaf ter, andcenter of mass velocity does not change.
(d) First we must transform the velocities v, v 1 , and v 2 into the appropriate center of
mass. Let v′, v 1 ′, and v′ 2 denote the corresponding velocities in the appropriate center of mass.
v′^ = v − Vbef ore = v − v = 0
v′ 1 = v 1 − Vaf ter = 250 − 350 = −100 m/s
v 2 ′ = v 2 − Vaf ter = 400 − 350 = 50 m/s
The total momentum before the collision in the center of mass frame is
Pbef ore = mv′^ = 0
The total momentum after the collision in the center of mass frame is
Paf ter = m 1 v′ 1 + m 2 v′ 2 = 0
Therefore, Pbef ore = Paf ter, andmomentum is conservedin the center of mass frame. The total kinetic energy before the collision in the center of mass frame is
Kbef ore =
mv′^2 2
The total kinetic energy after the collision in the center of mass frame is
Kaf ter =
m 1 v 1 ′^2 2
m 2 v 2 ′^2 2
= 1. 5 × 104 J
Therefore, Kbef ore = Kaf ter, andkinetic energy is not conservedin the center of mass frame. None of the previous conclusions change.
0 = mAxA + mB xB = mxA + 3mxB =⇒ xB = −
xA
Then the force acting on A is
mx¨A = F = −k(xA + xB − l 0 ) = −k(xA +
xA − l 0 ) = −
4 k 3
· (xA −
l 0 )
Thus the equation of motion is
x¨A = −
4 k 3 m
· (xA −
l 0 )
This indicates that the motion is simple harmonic with frequency
ωA =
√ 4 k 3 m
A similar argument for B yields
3 mx¨B = F = −k(xA + xB − l 0 ) = −k(3xB + xB − l 0 ) = − 4 k · (xA − l 0 ) =⇒
ωB =
√ 4 k ·
3 m
√ 4 k 3 m
Thus the motion is also simple harmonic in xB with the same frequency ω given by
ω =
√ 4 k 3 m
Problem 6.8 (Ohanian, page 294, problem 38)
Collisions in two dimensions are discussed in Section 11.3 beginning on page 281. The relevant results, using conservation of momentum andkinetic energy, are
m 1 v 1 = m 1 v 1 ′ cos θ 1 ′ + m 2 v′ 2 cos θ′ 2
0 = m 1 v′ 1 sin θ 1 ′ − m 2 v 2 ′ sin θ′ 2
m 1 v^21 2
m 1 v′ 12 2
m 2 v′ 22 2
where the various quantities are illustratedin Figure 11.6 on page 281. For the problem at hand, m 1 = m 2 = m where m = 1. 673 × 10 −^27 kg is the mass of the proton; this allows us to cancel the masses on each side of the equation. The equations above become
v 1 = v′ 1 cos θ′ 1 + v′ 2 cos θ 2 ′ (2)
0 = v′ 1 sin θ 1 ′ − v 2 ′ sin θ′ 2 (3)
v^21 = v 1 ′^2 + v′ 22 (4)
We are given the energy for the initial proton, which will give us v 1 , andwe can measure θ′ 1 and θ′ 2 from the picture; thus the above equations will suffice to determine v 1 ′ and v′ 2 , which will then determine the energy of each outgoing proton. To solve the above equations, multiply Equation (??) by sin θ′ 2 , andmultiply Equation ( ??) by cos θ′ 2. The resulting equations are
sin θ′ 2 v 1 = v′ 1 sin θ 2 ′ cos θ′ 1 + v′ 2 sin θ 2 ′ cos θ 2 ′ (5)
0 = v′ 1 sin θ′ 1 cos θ 2 ′ − v 2 ′ sin θ′ 2 cos θ′ 2 (6)
If we add Equation (??) andEquation ( ??) then we get
sin θ′ 2 v 1 = v 1 ′(sin θ 2 ′ cos θ 1 ′ + sin θ 1 ′ cos θ′ 2 )
This gives us an equation for v 1 ′
v′ 1 =
sin θ′ 2 sin θ 2 ′ cos θ 1 ′ + sin θ 1 ′ cos θ′ 2
· v 1
We can measure the angles from the picture: θ′ 1 ≈ 45 ◦^ is the angle of P B relative to AP and θ 2 ′ ≈ 44 ◦^ is the angle of P C relative to AP. This gives
v 1 ′ ≈ 0. 695 · v 1
Using Equation (??) gives
v 12 = v′ 12 + v 2 ′^2 =⇒ v′ 2 =
√ v 12 − v′ 12 ≈ 0. 719 · v 1
The energy for the proton initially traveling along AP is
E 1 =
mv 12 2
= 8. 0 × 10 −^13 J
uex =
Fth R
34 × 106
13. 8 × 103
≈ 2. 5 × 103 m/s
(b) The engines will burn until there is no more fuel. If the initial mass of the
rocket andfuel is Mi = 2. 85 × 106 kg andthe final mass of the rocket (with no fuel) is Mf = 0. 77 × 106 kg then the mass of fuel MF is
MF = Mi − Mf = 2. 85 × 106 − 0. 77 × 106 = 2. 08 × 106 kg
If the burn rate is constant, then the burn time T is given by
R =
MF
T
=⇒ T =
MF
R
2. 08 × 106
13. 8 × 103
≈ 151 s
(c) The force equation for the rocket is
M (t) · a = Fth − M (t) · g =⇒ a =
Fth M (t)
− g
where M (t) is the mass of the rocket andfuel inside the rocket at time t. The initial mass is M (0) = Mi, so the initial acceleration is
a =
Fth Mi
− g =
34 × 106
2. 85 × 106
− 10 ≈ 1 .9 m/s^2
(d) The final mass is M (T ) = Mf , so the acceleration just before the engines stop is
a =
Fth Mf
− g =
34 × 106
0. 77 × 106
− g ≈ 34 m/s^2
(e) The solution to the above differential equation is
vf = −u ln
( M
f Mi
) − gt
At t = T = M RF , RT = MF and Mi − MF = Mf , so the speed is
vf = − 2. 5 × 103 · ln
(
- 77 × 106
- 85 × 106
) − 10 · 151 ≈ 1. 8 × 103 m/s
Problem 6.11 (Ohanian, page 271, problem 54)
We are told that the reaction products have a mass of 4.4 kg and that these products have a kinetic energy of 4.2 × 107 J. If the velocity of this mass is u then
- 4 · u^2 2
= 4. 2 × 107 =⇒ u = 4. 4 × 103 m/s
