This page has all of the homework for the first semester of College Physics. You can go directly to each section using the links above or by using the menu.
Here is a teaser problem from Click and Clack the Tappet Brothers (Car Talk on NPR).
ROAD TRIP TRIGONOMETRY (week of 8/23/14, Link)
RAY: Last Fall, Tommy and I decided that we were going to take a trip north to see the foliage. Tommy drove the first 40 miles. I drove the rest of the way. We looked at the foliage for three or four minutes, then decided to head home. We took the same route home. On the way back, Tommy drove the first leg of the trip and I drove the last 50 miles. I got home and my wife said, "Who did the driving?" I explained that Tommy drove the first 40 miles, then I drove the rest of the way. On the way back, Tommy drove the first leg of the trip, and I drove the last 50 miles. She said, "But who did most of the driving?" I told her, "You can figure it out. In fact, you can even figure out how much more of the driving was done by that person." And that's the question. Who drove the most -- and how many more miles did that person drive?
Here is another one. I have given you the short version here:
You start rowing a boat against the current and go one mile. Your hat falls into the water at the one mile mark. You continue to row for 10 minutes and then you turn around and row with the current. You catch up with the hat at the same point that you started rowing. How long was the hat floating along in the water?
They provide a nice solution on their website, but here is an
algebraic solution.
v = 0 m/sec v_{o} = 8.94 m/sec s = ? a = -6.86 m/sec² t = ?
v = v_{o} + at; t = (v - v_{o})/a = (0 + 8.94 m/sec)/(6.86 m/sec²) = 1.3 sec
v² = v_{o}² + 2as;
s = [v² - v_{o}²]/2a = [0 - (8.94 m/sec)²]/(2)(-6.86 m/sec²) = 5.8 m
v = 0 m/sec v_{o} = ? m/sec s = 30 m a = -3.43 m/sec² t = ?
v² = v_{o}² + 2as;
v_{o} = [v² - 2as]^{½} = [0 + (2)(3.43 m/sec²)(30 m)]^{½} = 14.35 m/sec
v = v_{o} + at; t = (v - v_{o})/a = (0 + 14.35 m/sec)/(3.43 m/sec²) = 4.18 sec
v = 0 m/sec v_{o} = ? m/sec s = ? m a = -4.86 m/sec² t = 2.1 sec
v = v_{o} + at; v_{o} = v - at = 0 + (4.86 m/sec²)(2.1 sec) = 10.2 m/sec
v² = v_{o}² + 2as;
s = [v² - v_{o}²]/2a = [0 - (10.2 m/sec)²]/(2)(-4.86 m/sec²) = 10.7 m
v = -20.0 m/sec v_{o} = ? m/sec s = ? m a = -9.8 m/sec² t = 2.0 sec
v_{o} = v - at = (-20.0 m/sec) + (9.8 m/sec²)(2.0 sec) = -0.4 m/sec
v² = v_{o}² + 2as;
s = [v² - v_{o}²]/2a = [(-20 m/sec)² - (-0.4 m/sec)²]/(2)(-9.8 m/sec²) = -20.4 m
From start to max:
v = 0 m/sec v_{o} = ? m/sec s = 10 m a = -9.8 m/sec² t = ? sec
v² = v_{o}² + 2as; v_{o} = [v² - 2as]^{½} = [0 - (2)(-9.8 m/sec²)(10 m)]^{½} = 14 m/sec
v = v_{o} + at; t = (v - v_{o})/a = (0 - 14 m/sec)/(-9.8 m/sec²) = 1.43 sec
Total time is the time to go to max plus the time to come back down:
2(1.43 sec) = 2.86 sec
v = ? m/sec v_{o} = 0 m/sec s = ? m a = -9.8 m/sec² t = 4.0 sec
s = v_{o}t + ½at² = 0 + ½(-9.8 m/sec²)(4 sec)² = -78.4 m
The well is 78.4 m deep.
Also:
v = v_{o} + at = (0 m/sec) - (9.8 m/sec²)(4.0 sec) = -39.2 m/sec
1st Time: v = ? m/sec v_{o} = 0 m/sec s = ? m a = -9.8 m/sec² t = 3.0 sec
s = v_{o}t + ½at² = 0 + ½(-9.8 m/sec²)(3 sec)² = -44.1 m
2nd Time: v = ? m/sec v_{o} = -3.1 m/sec s = ? m a = -9.8 m/sec² t = 2.7 sec
s = v_{o}t + ½at² = (-3.1 m/sec)(2.7 sec) + ½(-9.8 m/sec²)(2.7 sec)² = -44.1 m
Both calculations say the building is 44.1 m high. The model works!
v = -50 m/sec v_{o} = 0 m/sec s = ? m a = -9.8 m/sec² t = ? sec
t = (v - v_{o})/a = [(-50 m/sec) - 0]/(-9.8 m/sec²) = 5.1 sec
s = v_{o}t + ½at² = 0 + ½(-9.8 m/sec²)(5.1 sec)² = -127.5 m
It dropped 127.5 m.
Here are some mathematical relationships that will be helpful for some of these two dimensional problems. |
Calculate components of the initial velocity in x and y.
In y: v_{oy} = v_{o}sin25° = (3 m/sec)(sin25°) = 1.27 m/sec
In x: v_{ox} = v_{o}cos25° = (3 m/sec)(cos25°) = 2.72 m/sec
y | x | |
v | ? m/sec | 2.72 m/sec |
v_{o} | 1.27 m/sec | 2.72 m/sec |
s | -0.5 m | ? m |
a | -9.8 m/sec | 0 m/sec |
t | ? sec | ? sec |
Make a table from jump to water.
Solve for v in the y-direction, then solve for t in the y-direction. Use the time from the y-direction in the x-direction to determine the distance in the x-direction.
v_{y} = [v_{oy}² + 2a_{y}s_{y}]^{½} = [(1.27 m/sec)² + (2)(-9.8 m/sec²)(-0.5 m)]^{½}
= -3.38 m/sec
t = (v_{y} - v_{oy})/a_{y} = [(-3.38 m/sec) - 1.27 m/sec]/(-9.8 m/sec²) = 0.474 sec
s_{x} = v_{ox}t + ½at² = (2.72 m/sec)(0.474 sec) + 0 = 1.29 m
y | x | |
v | ? m/sec | ? m/sec |
v_{o} | 0 m/sec | ? m/sec |
s | -1000 m | 2000 m |
a | -9.8 m/sec | 0 m/sec |
t | ? sec | ? sec |
Make a table from drop to ground.
Solve for v in the y-direction, then solve for t in the y-direction. Use the time from the y-direction in the x-direction to determine the velocity in the x-direction.
v_{y} = [v_{oy}² + 2a_{y}s_{y}]^{½} = [0 + (2)(-9.8 m/sec²)(-1000 m)]^{½}
= -140 m/sec
t = (v_{y} - v_{oy})/a_{y} = [(-140 m/sec) - 0]/(-9.8 m/sec²) = 14.3 sec
v_{ox} = s_{x}/t = (2000 m)/(14.3 sec) = 140 m/sec
For the final velocity:
v = [(v_{x})² + (v_{y})²]^{½} = [(-140 m/sec)² + (140 m/sec)²]^{½} = 198 m/sec
θ = tan^{-1}[(v_{x})/v_{y})] = tan^{-1}(140/140) = 45°
v = 198 m/sec at an angle of 45° below horizontal.
Get the initial velocity from the initial components.
v_{0} = [(v_{ox})² + (v_{oy})²]^{½} = [(160 m/sec)² + (75 m/sec)²]^{½} = 176.7 m/sec
θ = tan^{-1}[(v_{oy})/v_{ox})] = tan^{-1}(75/160) = 25.1°
The initial velocity is 176.7 m/sec at 25.1° above horizontal.
y | x | |
v | 0 m/sec | 160 m/sec |
v_{o} | 75 m/sec | 160 m/sec |
s | ? m | ? m |
a | -9.8 m/sec | 0 m/sec |
t | ? sec | ? sec |
Make a table going from start to max.
Solve for s in the y-direction to get the maximum height.
s_{y} = [(v_{y})² - (v_{oy})²)]/[2a_{y}] = [(0 - (75 m/sec)²)]/[2(-9.8 m/sec²)] = 287 m
t = (v_{y} - v_{oy})/a_{y} = [(0 m/sec) - 75 m/sec]/(-9.8 m/sec²) = 7.65 sec (to max)
The total time is 2(7.65 sec) = 15.3 sec
For the distance in x (the range): s_{x} = v_{o}t = (160 m/sec)(15.3 sec) = 2449 m
Distance to Max = 287 m, Range = 2449 m
From part (a):
v_{ox} = v_{o}cosθ m/sec = (17.3 m/sec)cos40° = 13.3 m/sec
v_{oy} = v_{o}sinθ = (17.3 m/sec)sin40° = 11.12 m/sec
Make a table going from start to max.
y | x | |
v | 0 m/sec | v_{x} = v_{ox} m/sec |
v_{o} | 11.12 m/sec | 13.3 m/sec |
s | ? m | 15 m |
a | -9.8 m/sec | 0 m/sec |
t | t sec | t sec |
s_{y} = [(v_{y})² - (v_{oy})²)]/[2a_{y}] = [(0 - (11.2 m/sec)²)]/[2(-9.8 m/sec²)]
= 6.4 m
t = (v_{y} - v_{oy})/a_{y} = [(0 m/sec) - 11.2 m/sec]/(-9.8 m/sec²) = 1.14 sec (to max)
The total time is 2(1.14 sec) = 2.28 sec
y | x | |
v | vsinθ m/sec | vcosθ m/sec |
v_{o} | v_{o}sinθ m/sec | v_{o}cosθ m/sec |
s | 0 m | 30 m |
a | -9.8 m/sec | 0 m/sec |
t | t sec | t sec |
Make a table going from start to end.
This will require solving two equations and two unknowns. Solve for t in both the x-direction and the y-direction, then set those expressions equal to each other.
In x: 30 m = (v_{o}sinθ m/sec)(t) or t = (30 m)/(v_{o}sinθ m/sec)
In y: s_{y} = v_{oy}t + ½a_{y}t² = [v_{oy} + ½a_{y}t]t
0 = [v_{oy} + ½a_{y}t]t says that either t = 0 or [v_{oy} + ½a_{y}t] = 0
Time equal to zero is before it starts, so the second expression is needed to agree with the physical situation. So, from y:
t = -v_{oy}/½a = -(v_{o}sinθ)/½a
That means:
-(v_{o}sinθ)/½a = (30 m)/(v_{o}sinθ m/sec) or -(v_{o}²)(sinθ)(cosθ) = ½a(30)
v_{o} = -[½a(30)]/[(sinθ)(cosθ)]^{½} = [(15 m)(9.8 m/sec²)]/[(sin40°)(cos40°)]^{½}
= 17.3 m/sec
Make a table going from start to end.
y | x | |
v | vsinθ m/sec | v_{x} = v_{ox} m/sec |
v_{o} | v_{o}sinθ m/sec | v_{o}cosθ m/sec |
s | -15 m | 10 m |
a | -9.8 m/sec | 0 m/sec |
t | ? sec | ? sec |
Solve for the time in x and substitute the resulting expression into the distance expression for y.
In x: t = (10 m)/(v_{o}cos40°)
s_{y} = (v_{oy})t + ½at²
Substiute t from above and then solve for the initial velocity, v_{o}.
-15 m = (v_{o})(sin40°)[(10 m)/(v_{o}cos40°)] + ½(-9.8 m/sec²)[(10 m)/(v_{o}cos40°)]²
v_{o} = 5.975 m/sec; v_{oy} = 3.84 m/sec; v_{ox} = 4.58 m/sec;
t = (10 m)/(4.58 m/sec) = 2.18 sec
v_{y} = 3.84 m/sec - (9.8 m/sec²)(2.18 sec) = -17.52 m/sec
v = [(4.58 m/sec)² + (-17.52 m/sec)²]^{½} = 18.1 m/sec
θ = tan^{-1}(-17.52/4.58) = 75.3°
y | x | |
v | 0 m/sec | v_{x} = v_{ox} m/sec |
v_{o} | ? m/sec | ? m/sec |
s | ? m | ? m |
a | -9.8 m/sec | 0 m/sec |
t | 1.05 sec | 1.05 sec |
Symmetry makes this problem much easier.
Make a table going from start to max.
0 = v_{oy} + at; v_{oy} = -at = (9.8 m/sec²)(1.05 sec) = 10.29 m/sec
v_{o} = (v_{oy})/sinθ = (10.29 m/sec)/sin10° = 59.3 m/sec at 10°
v_{ox} = (v_{o})(cosθ) = (59.3 m/sec)(cos10°) = 58.4 m/sec
Because of the symmetry of this problem the final velocity in y will be the negative of the initial velocity in y. The velocity in x remains constant. θ = tan^{-1}(-10.29/58.4) = -10°
v = [(v_{x})² + (v_{y})²]^{½} = 59.3 m/sec at -10°
s_{x} = (58.4 m/sec)(2.1 sec) = 123 m
EXTRA PROBLEMS
Force up = Force down
T = mg = (3.6 kg)(9.8 m/sec²) = 35.28 N
Refer to the diagram below for the next few problems.
T_{1x} = T_{2x} and T_{1y} + T_{2y} = F
T_{1}cosθ_{1} = T_{2}cosθ_{2} and T_{1}sinθ_{1} + T_{2}sinθ_{2} = F
T_{2} = (T_{1}cosθ_{1})/cosθ_{2} and T_{1}sinθ_{1} + [(T_{1}cosθ_{1})/cosθ_{2}]sinθ_{2} = F
T_{1}[sinθ_{1} + [(cosθ_{1})(tanθ_{2})] = F ⇒ T_{1} = F/[sinθ_{1} + [(cosθ_{1})(tanθ_{2})]
T_{1} = (5.6 N)/[sin25° + [(cos25°)(tan25°)] = 6.63 N
T_{2} = (T_{1}cosθ_{1})/cosθ_{2} = (6.63 N)(cos25°)/(cos25°) = 6.63 N
T_{1x} = T_{2x} and T_{1y} + T_{2y} = F
F = T_{1y} + T_{2y} = T_{1}sinθ_{1} + T_{2}sinθ_{2}
F = (38 N)(sin15°) + (38 N)(sin15°) = 19.67 N
T_{1x} = T_{2x} and T_{1y} + T_{2y} = F
⇒ T_{1}cosθ_{1} = T_{2}cosθ_{2} and T_{1}sinθ_{1} + T_{2}sinθ_{2} = F
T_{1}cosθ_{1} = T_{2}cosθ_{2} and T_{1} = T_{2} ⇒ cosθ_{1} = cosθ_{2} ⇒ θ_{1} = θ_{2}
Tsinθ_{1} + Tsinθ_{1} = F ⇒ sinθ_{1} = F/(2T)
θ_{1} = sin^{-1}[F/2T] = sin^{-1}[(12 N)/{2(22 N)}] = 15.8° = θ_{2}
F = T_{1y} + T_{2y} = T_{1}sinθ_{1} + T_{2}sinθ_{2}
F = (8 N)(sin20°) + (5 N)(sin30°) = 5.24 N
T_{1x} = T_{2x} and F = T_{1y} + T_{2y}
⇒ T_{1}cosθ_{1} = T_{2}cosθ_{2} and F = T_{1}sinθ_{1} + T_{2}sinθ_{2}
⇒ T_{1} = T_{2}cosθ_{2}/cosθ_{1} and F = (T_{2}cosθ_{2}/cosθ_{1})sinθ_{1} + T_{2}sinθ_{2}
⇒ F = T_{2}cosθ_{2}tanθ_{1} + T_{2}sinθ_{2} ⇒ tanθ_{1} = (F - T_{2}sinθ_{2})/T_{2}cosθ_{2}
⇒ tanθ_{1} = [(40 N) - (30 N)sin56.31°]/[(30 N)cos56.31°] = 0.9037
θ_{1} = tan^{-1}0.9037 = 42.1°
T_{1} = [F - T_{2}sinθ_{2}]/sinθ_{1} = [(40 N) - (30 N)sin56.31°]/sin42.1° = 22.43 N
T_{1x} = T_{2x} and F = T_{1y} + T_{2y}
⇒ T_{1}cosθ_{1} = T_{2}cosθ_{2} and F = T_{1}sinθ_{1} + T_{2}sinθ_{2}
T_{1} = T_{2} = T and θ_{1} = θ_{2} = θ = 90° - 15° = 75°
⇒ F = Tsinθ + Tsinθ = 2Tsinθ
⇒ T = F/(2sinθ) = (600 N)/(2sin75°) = 311 N
Using the distances:
tanθ_{1} = (1 m)/(4 m) ⇒ θ_{1} = 14° and tanθ_{2} = (1 m)/(12 m) ⇒ θ_{2} = 4.76°
T_{1x} = T_{2x} and F = T_{1y} + T_{2y}
⇒ T_{1}cosθ_{1} = T_{2}cosθ_{2} and F = T_{1}sinθ_{1} + T_{2}sinθ_{2}
⇒ T_{1} = T_{2}cosθ_{2}/cosθ_{1} and F = (T_{2}cosθ_{2}/cosθ_{1})sinθ_{1} + T_{2}sinθ_{2}
⇒ F = T_{2}cosθ_{2}tanθ_{1} + T_{2}sinθ_{2} ⇒ T_{2} = F/[cosθ_{2}tanθ_{1} + sinθ_{2}]
⇒ T_{2} = (500 N)/[cos4.76°tan14° + sin4.76°] = 1509 N
⇒ T_{1} = T_{2}cosθ_{2}/cosθ_{1} = T_{1} = (1509 N)cos4.76°/cos14° = 1549 N
F = (mg)sinθ = (35 N)sin25° = 14.8 N
Whole System:
netForce = F_{g2} - F_{g1} = ma
= (0.4 kg)(9.8 m/sec²) - (0.15 kg)(9.8 m/sec²) = 2.45 N
= 2.54 N = ma
a = (2.45 N)/(0.55 kg) = 4.45 m/sec²
Isolate the 0.4 kg Mass:
netForce = F_{g2} - T_{2} = ma
T_{2} = F_{g2} - ma = (0.4 kg)(9.8 m/sec²) - (0.4 kg)(4.45 m/sec²)
T_{2} = 2.14 N
Whole System:
netForce = F_{g2} = (m_{1} + m_{2})a
= (2 kg)(9.8 m/sec²) = 19.6 N = (5 kg)a
a = (19.6 N)/(5 kg) = 3.92 m/sec²
Isolate the 2 kg Mass:
netForce = F_{g2} - T = m_{2}a
T = F_{g2} - m_{2}a = (2 kg)(9.8 m/sec²) - (2 kg)(3.92 m/sec²)
T = 11.76 N
F_{gx} = F_{g}sinθ ; F_{gy} = F_{g}cosθ ; F_{n} = F_{gy} ; f_{k} = μ_{k}F_{n}
m = (20 N)/(9.8 m/sec²) = 2.04 kg
x-axis (along the incline):
netForce = F_{gx} - f_{k} = ma
f_{k} = F_{gx} - ma = (20 N)sin30° - (2.04 kg)(2.34 m/sec²) = 5.22 N
y-axis (perpendicular to the plane):
f_{k} = μ_{k}F_{n}
μ_{k} = f_{k}/F_{n} = (5.22 N)/[(20 N)cos30°]
μ_{k} = 0.3
F_{12} = (12 kg)(9.8 m/sec²) = 117.6 N ; F_{5} = (5 kg)(9.8 m/sec²) = 49 N
F_{5x} = F_{5}sinθ = (49 N)sin25° = 20.71 N ; F_{5y} = F_{5}cosθ = (49 N)cos25° = 44.41 N
F_{n} = F_{5y} = 44.41 N ; f_{k} = μ_{k}F_{n} = (0.11)(44.41 N) = 4.88 N
Whole System:
netForce = F_{12} - f_{k} - F_{5x} = (m_{12} + m_{5})a
= 117.6 N - 4.88 N - 20.71 N = 92 N = (17 kg)a
a = (92 N)/(17 kg) = 5.41 m/sec²
Isolate the 12 kg Mass:
netForce = F_{12} - T = m_{12}a
T = F_{12} - m_{12}a = 117.6 N - (12 kg)(5.41 m/sec²) = 52.68 N
f_{s} = F_{c} ; f_{s} = μ_{s}F_{n} ; F_{c} = mv²/r ; F_{n} = mg
⇒ μ_{s}mg = mv²/r ⇒ μ_{s}g = v²/r
⇒ v = [rμ_{s}g]^{½} = [(0.5)(9.8 m/sec²)(20 m)]^{½} = 9.9 m/sec
See example #2 in the Centripetal Force section of Newton's Laws for a diagram.
F_{gx} = F_{cx}
F_{g}sinθ = F_{c}cosθ
⇒ tanθ = F_{c}/F_{g} = (m)(v²/r)/mg = v²/rg
= [(240 km/h)(1000 m/km)(1 h/3600 sec)]² /(9.8 m/sec²)(975 m)
= .466
⇒ θ = tan^{-1}(.466) = 24.9°
At the top of the hill the gravitational force pulling the car down must at least equal the centripetal force that would make the car go straight (at a tangent to the curve, away from the center). The maximum speed would be when those two forces are equal: F_{g} = F_{c}.
⇒ mg = mv²/r
⇒ v =[rg]^{½} = [(30 m)(9.8 m/sec²)]^{½} = 17.15 m/sec²
W = Fs = mgh = (0.75 kg)(9.8 m/sec²)(1.5 m) = 11.03 J
W = Fs = F_{x}s_{x} = [(1.5 kg)(9.8 m/sec²)(sin20°)](2 m) = 10.06 J
W = Fs ⇒ s = W/F = (18 J)/(25 N) = 0.72 m
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at the bottom of the incline.
⇒ ½m(7 m/sec)² + m(9.8 m/sec²)[(2 m)(sin20°)] = ½mv_{f}² + 0 + 0
The mass cancels out
⇒ v_{f}² = 2{½(7 m/sec)² + (9.8 m/sec²)[(2 m)(sin20°)]}
⇒ v_{f} = (62.41 m²/sec²)^{½} = 7.9 m/sec
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at 8 m. Friction-free ⇒ W = 0. From rest ⇒ v_{i} = 0.
⇒ 0 + m(9.8 m/sec²)(4 m) = ½mv_{f}² + 0 + 0
The mass cancels out
⇒ v_{f} = [2{9.8 m/sec²)(4 m)]^{½} = 8.85 m/sec
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + W
Set h = 0 at the bottom of the incline. F_{g} = mg
⇒ W = f_{k}(s) = µ_{k}F_{n}(s) = µ_{k}F_{gy}(s) = µ_{k}(mgcosθ)(s)
With this W the mass will still cancel out. h_{i} = (3.3 m)(sin50°) = 2.528 m.
⇒ 0 + (9.8 m/sec²)(2.528 m) = ½v_{f}² + 0 + (0.12)(9.8 m/sec²)(cos50°)(3.3 m)
⇒ 0 + 24.774 m²/sec² = ½v_{f}² + 0 + 2.495 m²/sec²
⇒ v_{f} = [2(24.774 m²/sec² - 2.495 m²/sec²)]^{½} = 6.68 m/sec
Part (b): Start: After the incline, End: where it stops on the horizontal section.
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + W
Set h = 0 at the bottom of the incline. h_{i} = h_{f} = 0. v_{i} = 5.42 m/sec. v_{f} = 0.
W = Fs = f_{k}s = F_{n}µ_{k}s = F_{g}µ_{k}s = (mg)(µ_{k})(s)
⇒ ½mv_{i}² + 0 = 0 + 0 + (mg)(µ_{k})(s)
Divide both sides by the mass and multiply both sides by 2.
⇒ v_{i}² + 0 = 0 + 0 + (2)(g)(µ_{k})(s)
⇒ s = (v_{i}²)/[(2)(g)(µ_{k})] = (5.42 m/sec)²/[(2)(9.8 m/sec²)(0.11)] = 13.6 m
Part (a):
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at the bottom of the incline. h_{i} = 1.75 m - 0.25 m = 1.5 m. W = 0.
F_{g} = mg F_{gx} = mgsinθ
With W = 0 the mass will cancel out.
⇒ 0 + (9.8 m/sec²)(1.5 m) = ½v_{f}² + 0 + 0
⇒ v_{f} = [2(9.8 m/sec²)(1.5 m)]^{½} = 5.42 m/sec
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at the bottom. Friction-free (ignore air friction) ⇒ W = 0.
v_{ix} = (20 m/sec)(cos30°) = 17.32 m/sec v_{iy} = (20 m/sec)(sin30°) = 10 m/sec
The mass again cancels out. In the vertical direction:
⇒ ½(10 m/sec)² + (9.8 m/sec²)(15 m) = ½v_{fy}² + 0 + 0
⇒ v_{fy} = {2[(50 m²/sec²) + (147 m²/sec²)]}^{½} = 19.8 m/sec
v_{f} = [v_{fx}² + v_{fy}²]^{½} = [(17.32 m/sec)² + (19.8 m/sec)²]^{½} = 26.3 m/sec
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at 20 m. h_{f} = 5 m. Assume friction-free ⇒ W = 0.
v_{i} = (75 km/hr)(1000 m/1 km)(1 hr/3600 sec) = 20.83 m/sec
The mass again cancels out. The velocity at the top of the second hill must be at least zero (a). The minimum speed is when the final velocity is zero (b).
⇒ ½(20.83 m/sec)² = ½v_{f}² + (9.8 m/sec²)(5 m) + 0
⇒ v_{f} = [2{(217 m²/sec²) - (49 m²/sec²)}]^{½} = 18.33 m/sec
⇒ The car will go over the top of the hill at 18.33 m/sec. (Part a)
Part b: v_{f} = 0 ⇒ v_{i} = [2gh_{f}]^{½} = 9.9 m/sec
KE_{i} + PE_{i} = KE_{f} + PE_{f} + W ⇒ ½mv_{i}² + mgh_{i} = ½mv_{f}² + mgh_{f} + Fs
Set h = 0 at 5 m down the ramp. h_{i} = (5 m)sin30° = 2.5 m. h_{f} = 0 m.
v_{i} = 0. v_{f} = 1 m/sec. m = F_{g}/g = (500 N)/(9.8 m/sec²) = 51 kg
W = Fs = (F_{gx} - f_{k})s = (F_{g}sinθ - F_{n}µ_{k})s = (F_{g}sinθ - F_{gy}µ_{k})s
= (F_{g}sinθ - [F_{g}cosθ]µ_{k})s = F_{g}[sinθ]s - F_{g}[cosθ]µ_{k}s
= (500 N)(sin30°)(5 m) - (500 N)(cos30°)(µ_{k})(5 m) = 1250 J - (2165 J)µ_{k}
⇒ 0 + mgh_{i} = ½mv_{f}² + 0 + W
⇒ 0 + (51 kg)(9.8 m/sec²)(2.5 m) = ½(51 kg)(1 m/sec)² + 0 + 1250 J - (2165 J)µ_{k}
⇒ µ_{k} = [½(51 kg)(1 m/sec)² + 1250 J - (51 kg)(9.8 m/sec²)(2.5 m)]/(2165 J) = 0.012
p_{i} = p_{f} ⇒ p_{1i} + p_{2i} = p_{1f} + p_{2f} ⇒ m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}
In this case: m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}
⇒ v_{f} = (m_{1}v_{1i} + m_{2}v_{2i})/(m_{1} + m_{2}) = [(0.4 kg)(0.12 m/sec) + 0]/(0.4 kg + 0.35 kg)
⇒ v_{f} = 0.064 m/sec
p_{i} = p_{f} ⇒ p_{1i} + p_{2i} = p_{1f} + p_{2f} ⇒ m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}
In this case: m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}
⇒ v_{f} = (m_{1}v_{1i} + m_{2}v_{2i})/(m_{1} + m_{2})
⇒ v_{f} = [(0.4 kg)(0.12 m/sec) + (0.35 kg)(-0.06 m/sec)]/(0.4 kg + 0.35 kg)
⇒ v_{f} = 0.036 m/sec
p_{i} = p_{f} ⇒ p_{1i} + p_{2i} = p_{1f} + p_{2f} ⇒ m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}
In this case: 0 = m_{1}v_{1f} + m_{2}v_{2f}
⇒ v_{2f} = -m_{1}v_{1f}/m_{2}
⇒ v_{2f} = -[(55 kg)(2 m/sec)]/(45 kg) = -2.44 m/sec
⇒ v_{2f} = 2.44 m/sec away from the boy
p_{i} = p_{f} ⇒ p_{1i} + p_{2i} = p_{1f} + p_{2f} ⇒ m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}
In this case: 0 = m_{1}v_{1f} + m_{2}v_{2f}
⇒ v_{2f} = -m_{1}v_{1f}/m_{2}
⇒ v_{2f} = -[(1.5 kg)(1 m/sec)]/(50 kg) = -0.03 m/sec
⇒ v_{2f} = 0.03 m/sec away from the book
Conservation of Energy ⇒ KE_{i} + PE_{i} = KE_{f} + PE_{f} + W
Set h = 0 at the height of the billiard balls. ⇒ All PE = 0. W = 0 (elastic).
⇒ ½m_{1}v_{1i}² + ½m_{2}v_{2i}² = ½m_{1}v_{1f}² + ½m_{2}v_{2f}² ⇒ m_{1}v_{1i}² + m_{2}v_{2i}² = m_{1}v_{1f}² + m_{2}v_{2f}²
⇒ (0.5 kg)(0.15 m/sec)² + (0.35 kg)(-0.2 m/sec)² = 0.02525 J = m_{1}v_{1f}² + m_{2}v_{2f}²
⇒ 0.02525 = (0.5)v_{1f}² + (0.35)v_{2f}² (From Cons. of Energy, two unknowns)
Conservation of Momentum ⇒ m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}
⇒ (0.5 kg)(0.15 m/sec) - (0.35 kg)(0.2 m/sec) = 0.005 kg-m/sec = m_{1}v_{1f} + m_{2}v_{2f}
⇒ 0.005 = (0.5)v_{1f} + (0.35)v_{2f} (From Cons. of Momentum, two unknowns)
Substitute v_{2f} = (0.005 - (0.5)v_{1f})/(0.35) into Cons. of Energy Eqn.
⇒ 0.02525 = (0.5)v_{1f}² + (0.35)[(0.005 - (0.5)v_{1f})/(0.35)]²
⇒ (1.2143)v_{1f}² - (0.01429)v_{1f} - 0.02518 = 0 ⇒ v_{1f} = -0.15 m/sec or 0.138 m/sec
⇒ v_{1f} = -0.15 m/sec and v_{2f} = 0.229 m/sec
See webpage for more details. Set h = 0 at bottom.
After bullet is in the block to maximum height.
⇒ ½(m_{1} + m_{2})v_{i}² + 0 = 0 + (m_{1} + m_{2})gh_{f} + 0
⇒ v_{i} = [(2)([m_{1} + m_{2}]gh_{f})/(m_{1} + m_{2})]^{½} = [(2)(9.8 m/sec²)(0.12 m)]^{½}
= 1.534 m/sec The velocity just after the bullet is embedded in the block.
Before the bullet hits to just after the bullet is embedded in the block.
Use conservation of momentum:
⇒ (0.01 kg)v_{1i} + 0 = (1.501 kg)(1.534 m/sec)
⇒ v_{1i} = (1.501 kg)(1.534 m/sec)/(0.01 kg) = 230.2 m/sec
Before bullet is in the block to maximum height.
To get W use conservation of energy from before collision to maximum height.
½(0.01 kg)(230.2 m/sec)² + 0 + 0 + 0 = 0 + (1.51 kg)(9.8 m/sec²)(0.12 m) + W
⇒ W = 529.9 J - 1.8 J = 528.1 J
From conservation of energy when W = 0 (elastic collision), initial and final height is at zero, m_{2} = 3m_{1}, v_{2i} = 0:
½m_{1}v_{1i}² + 0 + 0 + 0 = ½m_{1}v_{1f}² + ½(3m_{1})v_{2f}² + 0 + 0 + 0
v_{1i}² = v_{1f}² + 3v_{2f}² ⇒ (0.5)² = v_{1f}² + 3v_{2f}²
From conservation of momentum:
m_{1}v_{1i} + 0 = m_{1}v_{1f} + 3m_{1}v_{2f}
v_{1i} = v_{1f} + 3v_{2f} ⇒ 0.5 = v_{1f} + 3v_{2f} ⇒ v_{1f} = 0.5 - 3v_{2f}
Substitute v_{1f} into the energy equation, and then solve for v_{2f}:
⇒ (0.5)² = (0.5 - 3v_{2f})² + 3v_{2f}² = (0.5)² - (2)(0.5)(3v_{2f}) + (3v_{2f})² + 3v_{2f}²
⇒ 0 = -3v_{2f} + 12v_{2f}² = (3)(-1 + 4v_{2f})(v_{2f})
The two roots of the equation are v_{2f} = 0 (physically impossible), or v_{2f} = 0.25.
v_{2f} = 0.25 m/sec and v_{1f} = 0.25 - 3v_{2f} = 0.25 - 3(0.25) = -0.5 m/sec
Conservation of Energy ⇒ KE_{it} + KE_{ir} + PE_{i} = KE_{ft} + KE_{fr} + PE_{f} + W
⇒ ½mv_{i}² + ½Iω_{i}² + mgh_{i} = ½mv_{f}² + ½Iω_{f}² + mgh_{f}
Set h = 0 at the bottom of the incline. W = 0, v_{i} = 0, ω_{i} = 0, and h_{f} = 0.
⇒ 0 + 0 + mgh_{i} = ½mv_{f}² + ½Iω_{f}² + 0
⇒ I = [mgh_{i} - ½mv_{f}²]/[½ω_{f}²] ; ω_{f} = v_{f}/r = (12 m/sec)/(0.01 m) = 1200 rad/sec
⇒ I = [(0.02 kg)(9.8 m/sec²)(12 m) - ½(0.02 kg)(12 m/sec)²]/[½(1200 rad/sec)²]
⇒ I = 1.27 x 10^{-6} kg-m²
Conservation of Energy ⇒ KE_{it} + KE_{ir} + PE_{i} = KE_{ft} + KE_{fr} + PE_{f} + W
⇒ 0 + 0 + mgh_{i} = ½mv_{f}² + ½Iω_{f}² + 0 + 0
⇒ gh_{i} = ½v_{f}² + ½[⅔r²][v_{f}/r]² = ½v_{f}² + ⅓v_{f}² = ⅚v_{f}²
⇒ v_{f} = [(6/5)gh_{i}]^{½}
⇒ v_{f} = 3.07 m/sec
From conservation of energy: W = 0 (elastic collision), initial and final height is along the track where we assign h equal to zero and v_{2i} = 0:
½m_{1}v_{1i}² + 0 + 0 + 0 = ½m_{1}v_{1f}² + ½(m_{2})v_{2f}² + 0 + 0 + 0
the one-half and the mass will cancel out, so
v_{1i}² = v_{1f}² + v_{2f}²
From conservation of momentum:
m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} but m_{1} = m_{2}, so
v_{1i} = v_{1f} + v_{2f} ⇒ v_{1f} = v_{1i} - v_{2f} and substituting into energy equation:
⇒ v_{1i}² = (v_{1i} - v_{2f})² + v_{2f}² = v_{1i}² - 2v_{1i}v_{2f} + v_{2f}² + v_{2f}²
⇒ 0 = 2v_{2f}(v_{2f} - v_{1i})
The two roots of the equation are v_{2f} = 0 (physically impossible), or v_{2f} = v_{1i}.
⇒ v_{2f} = v_{1i}
From conservation of energy: W = 0 (elastic collision), initial and final height is along the track where we assign h equal to zero and v_{2i} = 0:
½m_{1}v_{1i}² + 0 + 0 + 0 = ½m_{1}v_{1f}² + ½(m_{2})v_{2f}² + 0 + 0 + 0
the one-half and the mass will cancel out, so
v_{1i}² = v_{1f}² + v_{2f}²
From conservation of momentum:
m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} but m_{1} = m_{2}, so
v_{1i} = v_{1f} + v_{2f} ⇒ v_{2f} = v_{1i} - v_{1f} and substituting into energy equation:
⇒ v_{1i}² = v_{1f}² + (v_{1i} - v_{1f})² = v_{1f}² + v_{1i}² - 2v_{1i}v_{1f} + v_{1f}²
⇒ 0 = 2v_{1f}(v_{1f} - v_{1i})
The two roots of the equation are v_{1f} = v_{1i} (physically impossible), or v_{1f} = 0.
⇒ v_{1f} = 0
L = 2 m
F_{b} = 20 N
F_{bx} = (20 N)sin50° = 15.32 N
F_{by} = (20 N)cos50° = 12.86 N
F_{mx} = (10 N)sin50° = 7.66 N
F_{my} = (10 N)cos50° = 6.43 N
F_{wy} = (15 N)sin50° = 11.49 N
Forces in x: F_{w} = F_{f} = 15 N
Forces in y: F_{v} = 20 N + 10 N = 30 N
Set l = 0 at the bottom. τ_{in} = τ_{out} ⇒ (F_{wy})l_{w} = (F_{by})l_{b} + (F_{my})l_{m}
⇒ (11.49 N)(2 m) = (12.86 N)(1 m) + (6.43 N)(x) ⇒ x = 1.57 m
L = 2.5 m; θ=35°
F_{b} = 40 N
F_{bx} = (40 N)sin55°
= 32.77 N
F_{by} = (40 N)cos55°
= 22.94 N
F_{m} = 600 N
F_{mx} = (600 N)sin55° = 491.49 N
F_{my} = (600 N)cos55° = 344.15 N
F_{wy} = (20 N)sin55° = 16.38 N
F_{wf} = 15 N (An upward frictional force at the wall, not shown in diagram.)
F_{wfy} = (15 N)cos35° = 12.28 N
Forces in x: F_{w} = F_{f} = 20 N
Forces in y: F_{v} + F_{wfy} = 40 N + 600 N = 640 N
Set l = 0 at the bottom. τ_{in} = τ_{out} ⇒ (F_{wy})l_{w} + (F_{wfy})l_{wf} = (F_{by})l_{b} + (F_{my})l_{m}
⇒ (16.38 N)(2.5 m) + (12.28 N)(2.5 m) = (22.94 N)(1.25 m) + (344.15 N)(x)
⇒ x = 0.125 m
F_{b} = 50 N;
F_{m} = ?
Forces in y: F_{v} = F_{b} + F_{m}
Set l = 0 at the F_{v}.
τ_{in} = τ_{out} ⇒ (F_{m})l_{m} = (F_{b})l_{b} ⇒ (F_{m})(0.6 m) = (50 N)(0.9 m)
⇒ F_{m} = 75 N
F_{b} = (10 kg)(9.8 m/sec²) = 980 N;
F_{m} = 500 N
Forces in y: F_{v} = F_{b} + F_{m}
Set l = 0 at the F_{v}.
τ_{in} = τ_{out} ⇒
(F_{m})l_{m} = (F_{b})l_{b}
⇒
(500 N)(0.8 m) = (980 N)(x)
⇒ x = 0.408 m
⇒ x_{cm} = 1.208 m from the end with the hanging mass.
5 kg ⇒ (5 kg)(9.8 m/sec²) = 49 N
8 kg ⇒ (8 kg)(9.8 m/sec²) = 78.4 N
Forces in y: T_{1} + T_{2} = 49 N + 5 N + 78.4 N = 132.4 N
Set l = 0 at the T_{2}.
τ_{in} = τ_{out} ⇒ (T_{1})(0.3 m) + (78.4 N)(0.15 m) = (5 N)(0.15 m) + (49 N)(0.45 m) ⇒ T_{1} = 36.8 N
⇒ T_{2} = 132.4 N - 36.8 N = 95.6 N
25 kg ⇒ (25 kg)(9.8 m/sec²) = 245 N
Set the x-axis along the supporting rod. T_{x} = Tsin50° and T_{y} = Tcos50°
Set l = 0 at the bottom of the supporting rod.
τ_{in} = τ_{out} ⇒
[(20 N)cos40°](L/2) + [(245 N)cos40°](L) = (T_{y})(L)
⇒
T_{y} = 195.3 N
⇒ T = T_{y}/cos50° = 303.9 N
25 kg ⇒ (25 kg)(9.8 m/sec²) = 245 N
Set the x-axis along the supporting rod. T_{x} = Tsin50° and T_{y} = Tcos50°
Set l = 0 at the bottom of the supporting rod.
τ_{in} = τ_{out} ⇒
[(20 N)cos40°](L/4) + [(245 N)cos40°](L) = (T_{y})(L)
⇒
T_{y} = 191.5 N
⇒ T = T_{y}/cos50° = 297.9 N