Problem Set 7.3:  1
 2 
3  4  5  6

7  8  9  10
 11  14
 15  16  18
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 by Chris Murray, Sonja Scherer, To Chong 2002
1. What is the magnitude of the momentum of a 22g sparrow flying with a speed of 8.1 m/s?
m = 22 g = .022 kg
p = mv = (.022 kg)(8.1 m/s) = .1782 kgm/s = .18 kgm/s
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2. A child in a boat throws a 5.40kg package out horizontally with a speed of 10.0m/s, See below. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg and that of the boat is 55.0 kg.
Here you are comparing the momentum of the boat, boy and box when they are at rest (no momentum) to just after he throws the package. In the after picture, you have the 55.0 kg boat and the 26.0 kg boy moving (to the left) at some unknown velocity, and the package moving to the right (+) at 10.0 m/s
(Momentum is Zero) 0 = (26.0 kg + 55.0 kg)v + (5.40 kg)(+10.0 m/s) v = .667 m/s
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3. Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1300kg/s with a speed of 40,000 m/s (at the moment of takeoff).
We know that
Impulse = Dp = FDt
and implicitly, Dp = mDv
so ultimately,
FDt = mDv
Good so far, but what do we put into the equation? We know that in 1 second, 1300 kg of combustion products leaves the engine undergoing a change in velocity of 40,000 m/s.
FDt = mDv
F(1 s) = (1300 kg)(40,000 m/s)
F = 52000000 N = 52 MN (M = Mega = 10^{6})
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4. A halfback on an apparent breakaway for a touchdown is tacked from behind. If the halfback has a mass of 95 kg and was moving 4.2 m/s when he was tackled by an 85kg cornerback running at 5.5 m/s in the same direction, what was their mutual speed immediately after the touchdownsaving tackle?
You have in the first frame, the pursuing 85 kg cornerback running to the right (+) at 5.5 m/s, overtaking the 95 kg halfback moving at 4.1 m/s to the right (+) as well.. In the second frame, you have the two of them together 85 kg + 95 kg = 180 kg moving (to the right) at some unknown velocity. I am going to make to the right positive.
(85 kg)(5.5 m/s) + (95 kg)(4.1 m/s) = (180 kg)v v = 4.761 m/s = 4.8 m/s
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5. A 12,500kg railroad car travels alone on a level frictionless track with a constant speed of 18.0m/s. A 5750kg additional load is dropped on a car. What then will be the car's speed?
Assuming the load is not moving when it is dropped in the rail car, we compare the momentum before to the momentum after:
mv = mv
(12,500 kg)(18.0 m/s) = (12,500 kg + 5750 kg)v
v = 12.329 = 12.3 m/s
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6. A 9500kg boxcar traveling at 16 m/s strikes a second car. The two stick together and move off with a speed of 6.0 m/s. What is the mass of the second car?
You have in the first frame, the 9500 kg boxcar moving to the right (+) at 16 m/s, and the second boxcar sitting still (assumed) on the tracks. In the second frame, the two boxcars with a total mass of 9500 kg + M, where M is the unknown mass of the second car, move along to the right (+) at 6.0 m/s. I am going to make to the right positive.
(9500 kg)(16 m/s) + 0 = (9500 kg + M)(6.0 m/s) M = 15833.33 kg = 1.6x10^{4} kg
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7. A gun is fired vertically into a 1.40kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 210m/s, how high will the block rise into the air after the bullet becomes embedded in it?
This is a twoparter, in the first part, where the bullet collides with the block, I will use conservation of momentum. Energy would not be conserved as it is not an elastic collision. In the second part, where the block starts from its original height with some velocity, and goes up to some height, I will use conservation of energy. I could also use simple linear kinematics, but I think the book wants me to use energy.
Part I  the bullet strikes the block. In the first frame you have the .021 kg bullet moving upward (+) at 210 m/s, and the block at rest. In the second frame you see the bullet block combo (1.40 kg + .021 kg = 1.421 kg) moving up at a velocity that we need to know:
(.021 kg)(210 m/s) + 0 = (1.40 kg + .021 kg)v v = 3.103 m/s
Part II  the bulletblock combo starts at zero elevation, and rises some unknown height into the air. Kinetic energy turns into potential energy:
^{1}/_{2}mv^{2} = mgDh ^{1}/_{2}mv^{2} = mgDh
m will cancel:
^{1}/_{2}v^{2} = gDh
Dh = v^{2}/2g
Dh = (3.103 m/s)^{2}/2(9.8 N/kg) = 0.491 m
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8. A 15g bullet strikes and becomes embedded in a 1.10kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.25, and the impact drives the block a distance of 9.5 m and before it comes to rest, what was the muzzle speed of the bullet?
This is also a twoparter, in the first part, where the bullet collides with the block, I will use conservation of momentum. Energy would not be conserved as it is not an elastic collision. In the second part where the block starts out with a kinetic energy and then comes to a halt, having done work against friction, I will use conservation of energy. I could also use simple linear kinematics, but I think the book wants me to use energy.
Here is a cartoon of the whole thing:
Let's start with the last part, where the kinetic energy turns into work against friction:
^{1}/_{2}mv^{2} = Fs ^{1}/_{2}mv^{2} = Fs
Since F = the force of friction, F = F_{fr} = µ_{k}F_{N }
and _{}F_{N } = mg in this case, so ultimately, F = µ_{k}mg
So let's substitute this in for F:
^{1}/_{2}mv^{2} = Fs
^{1}/_{2}mv^{2} = µ_{k}mgs
And so m cancels, and we solve for the velocity:
^{1}/_{2}v^{2} = µ_{k}gs
v^{2} = 2µ_{k}gs
v = Ö{2µ_{k}gs} = Ö{2(.25)(9.80 N/kg)(9.5 m)} = 6.823 m/s
Now we need to do some momentum:
In the first frame, you have the 15 g (.015 kg) bullet moving to the right at an unknown velocity, and the block of wood at rest. In the second frame, you have the bullet block combo moving to the right at 6.823 m/s as we figured out in the first part above.
(.015 kg)v + 0 = (1.10 kg + .015 kg)(6.823 m/s) v = 507.16 m/s = 510 m/s
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9. An atomic nuclear at rest decays radioactively into an alpha particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 3.8 x 10^{5} m/s? Assuming the recoiling nucleus has a mass 57 times greater than that of the alpha particle.
In the first frame, you have the parent nucleus at rest with zero momentum, and in the second frame, you have the daughter nucleus with a mass of 57 m recoiling to the left () and the alpha particle (a helium nucleus) with a mass of m moving to the right (+) at 3.8x10^{5} m/s.
0 = (57m)v + m(3.8x10^{5} m/s) the m divides out of both sides, and you get that the recoil velocity is 57 times smaller and opposite in direction:
v = 6666.666667 (YAAAAAHHHHH!!!) = 6.7x10^{3} m/s opposite direction of the alpha.
An interesting note is that when we solve these problems of alpha decay (in chapter 30) we assume that the alpha particle has all the kinetic energy that the decay releases. This is of course not totally true, as momentum must be conserved, and so the daughter nucleus must carry away an equal but opposite momentum and therefore must have some kinetic energy. But it is almost true, as if we compare the kinetic energies:
KE_{alpha}/KE_{daughter} = ^{1}/_{2}m(3.8x10^{5} m/s)^{2}/ ^{1}/_{2}57m(6666.666667)^{2}
KE_{alpha}/KE_{daughter} = (3.8x10^{5} m/s)^{2}/57(6666.666667)^{2}
KE_{alpha}/KE_{daughter} = 57 times which means that the daughter nucleus has 57 times less the kinetic energy of the alpha particle.
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10. An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?
In the first frame, you have the parent nucleus with a mass of 222u (a u is 1/12 the mass of carbon 12  about the mass of a nucleon {neutrons, protons}) bopping along to the right (+) at 420 m/s, and in the second frame, you have the daughter nucleus with a mass of 218u still moving in the original direction (+) at 350 m/s, and the alpha particle (a helium nucleus) with a mass of 4u moving to the right (+) at some unknown velocity.
(222u)(420 m/s) = (218u)(350) + (4u)v the u divides out of both sides, and you get:
(222)(420 m/s) = (218)(350) + (4)v
v = 4235 m/s = 4.2x10^{3} m/s
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11. a 13g bullet traveling 230 m/s penetrates a 2.0kg block of wood and emerges going 170 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
(.013 kg)(230 m/s) = (2.0 kg)v + (.013 kg)(170 m/s) In the first frame, you have the .013 kg bullet going 230 m/s, and in the second frame, the block of wood (that miraculously has not lost any pieces or changed in mass) going an unknown velocity, and the bullet going 170 m/s.
Solving for v gives .39 m/s
14. A tennis ball may leave the racket of a top player on the serve with a speed of 65.0 m/s. If the ball's mass is 0.0600 kg and it is in contact with the racket for 0.0300 s, what is the average force on the ball? Would this force be large enough to lift a 60kg person?
We know that
Impulse = Dp = FDt
and implicitly, Dp = mDv
so ultimately,
FDt = mDv
We know that in .0300 second, a .0600 kg tennis ball undergoes a change in velocity of 65.0 m/s.
FDt = mDv
F(.0300 s) = (.0600 kg)(65.0 m/s)
F = 130. N, you would need (60 kg)(9.8 N/kg) = 588 N to lift a person.
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15. A 0.145kg baseball pitched at 39.0 m/s is hit on horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is 1.00 x 10^{3} s, calculate the average force between the ball and the bat during contact.
We know that
Impulse = Dp = FDt
and implicitly, Dp = mDv
so ultimately,
FDt = mDv
We know that in 1.00x10^{3} second, a .145 kg baseball goes from moving 39 m/s toward the batter, to 52 m/s toward the outfield. If we make toward the outfield the positive direction, the initial velocity (toward the batter) is 39 m/s, and the change in velocity of the ball is 52.0 m/s  (39.0 m/s) = +91.0 m/s. (Change is always final minus initial)
FDt = mDv
F(1.00x10^{3} s) = (.145 kg)(+91.0 m/s)
F = +13195 N = 1.32x10^{4} N (toward the outfield on the ball)
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16. A golf ball of mass 0.045 kg is hit off the tee at the speed of 45 m/s. The golf club was in contact with the ball for 5.0 x 10^{3} s. Find (a) the impulse imparted to the golf ball, and (b) the average force exerted on the ball by the golf club.
From the HL Data packet:
Impulse = Dp = FDt
and implicitly, Dp = mDv
So the Impulse imparted to the ball by the club is
Dp = mDv = (.045 kg)(45 m/s) = 2.025 kgm/s = 2.0 kgm/s
To find the force,
FDt = mDv
We know that in 5.00x10^{3} second, a .045 kg golf ball undergoes a change in velocity of 45 m/s. (Change is always final minus initial, and I am making the direction of the original pitch positive)
FDt = mDv
F(5.00x10^{3} s) = (.045 kg)(45 m/s)
F = 405 N
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18. A 115kg fullback is running at 4.0 m/s to the east and is stopped in 0.75s by a head on tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted in the tackler, and (d) the average force exerted on the tackler.
Let's make to the East the positive direction. Since the Fullback is running toward the East (+4.0 m/s), and is stopped, the change in velocity is 4.0 m/s.
The original momentum of the fullback:
p = mv = (115 kg)(4.0 m/s) = 460 kgm/s (East)
Since the Fullback is running toward the East (+4.0 m/s), and is stopped, the change in velocity is 4.0 m/s.
Since Impulse = Dp = FDt, then the impulse exerted on the Fullback is
Dp = mDv = (115 kg)(4.0 m/s) = 460 kgm/s (West)  so to stop something going to the East, you exert a force to the West. Radical.
The tackler would undergo a change in momentum due to the reaction force, which would be exactly equal, but opposite, according to Newton's third law. So the tackler would have the opposite impulse exerted on them as the Fullback:
Dp = 460 kgm/s (East)
And to find the force exerted on the tackler:
Impulse = Dp = FDt
460 kgm/s = F(.75 s)
F = 613.3 N = 610 N
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