Physics Exercises Prepolitécnico

María Gabriela Ruiz Hinojosa
Ingeniería Química




Exercises
Leyes de Newton
Given the forces acting on the object and the magnitude of its acceleration, find the object's mass (in "kg").

Fx=m ax 3 N=m ax ax=3 NmFy=m ay 4 N-m g=m ay ay=4 Nm-g
a2=ax2+ay2 2,52=3m2+4m-9,82 6,25=9m2+16m2-78,4m+96,04
89,79 m2=25-78,4 m 89,79 m2+78,4 m-25=0
m=-78,4±78,42-4 89,79 -252 89,79 m=0,248279405 kg
Given the forces acting on the object and the magnitude of its acceleration, find the object's mass (in "kg"). No forces, other than the ones shown, are present.

Fx=m ax 3 N=m ax ax=3 NmFy=m ay 4 N=m ay ay=4 Nm
a2=ax2+ay2 2,52=3m2+4m2 6,25=9m2+16m2
6,25 m2=25 m=2 kg
A block of mass m is hanging by two strings of negligible mass as shown in the figure. String 1 is at an angle of 60o with the horizontal and string 2 at an angle of 45o with respect to the horizontal. Find the magnitude of the force of tension in strings 1 and 2. Express your answers in terms of m and g. (For the values of the trigonometric functions input sqrt(3)/2 for cos(30) or sin(60), 1/2 for cos(60) or sin(30) and sqrt(2)/2 for cos(45)or sin(45) as needed).

Fy=0 T1 sin60+T2 sin45-m g=0 32 T1+22 T2=m gFx=0 T1 cos60=T2 cos45 12 T1=22 T2 T1=2 T2
32 2 T2+22 T2=m g 32 2 T2+22 T2=m g
T2=2 m g2 3+1 T2=2 m g3+1
T1=2 2 m g3+1 T1=2 m g3+1
A block of mass m is at rest on an inclined plane of angle θ with respect to the horizontal. There is friction between the block and the inclined surface (μk and μs).

Imagine that you start increasing the angle of the incline, then at some point, the block will start to slide down. The angle at which this will happen is:
Px=Fr m g sinθ=μS m g cosθ μS=tanθ
Now, assume that the value of θ is less than the value in part A) (so that the block will remain stationary if no other forces act on it). What is the maximum magnitude of a horizontal Force F that can be applied without causing the block to move up the incline?

Fy=0 N=Py+Fy N=P cosθ+F sinθ
Fx=0 FR+Px=Fx μS N+P sinθ=F cosθ
μS P cosθ+F sinθ+P sinθ=F cosθ
μS P cosθ+P sinθ=F cosθ-μS F sinθ F=P sinθ+μS cosθcosθ-μS sinθ
Finally, assume that the value of θ is Greater than the value in part A) (so that the block will slide down if no other forces act on it). What is the minimum magnitude of a horizontal Force F that is needed so that the block will not slide down the incline?

Fy=0 N=Py+Fy N=P cosθ+F sinθ
Fx=0 Px=Fx+FR P sinθ=F cosθ+μS N
P sinθ=F cosθ+μS P cosθ+F sinθ
P sinθ-μS P cosθ=F cosθ+μS F sinθ F=P sinθ-μS cosθcosθ+μS sinθ
A block of mass m is at rest on an incline plane of angle α with respect to the vertical. There is friction between the block and the incline surface (μk and μs).

What is the value of the normal force exerted by the incline on the block? Express your answer in terms of some or all of the variables m, g, μs, μk, and α.
Fy=0 N=Py N=m g sinα
What is the value of the force of friction exerted by the surface on the block? Express your answer in terms of some or all of the variables m, g, μs, μk, and α.
Fx=0 Fr=Px Fr=m g cosα
If you were able to change the angle of the incline plane, at what angle will the block start to slide?
Px=Fr m g cosα=μS m g sinα μS=cotα
A block is moving. For an angle smaller than the one calculated in part c), a steeper slope, the block will slide down. What is the block's acceleration? Express your answer in some or all of the variables m, g, μs, μk, and α.
Fx=m a Px-Fr=m a m g cosα-μk m g sinα=m a
a=g cosα-μk sinα
In the system shown, the strings and the pulleys are ideal. There is no friction anywhere in the system. As the system is released, block m1 moves downward. Use Newton's laws to find the magnitude of the acceleration of block m1.

F=m a m1 g-T=m1 a1T2=m2 a2
m1 g-2 m2 a2=m1 a1
a2=2 a1 m1 g-2 m2 2 a1=m1 a1 m1 g=m1 a1+4 m2 a1
a1=m1m1+4 m2 g
Two blocks of masses MA and MB are connected by a light cord passing around a fixed, frictionless pulley. The coefficient of kinetic friction between all surfaces is μk. Find the magnitude of the horizontal force necessary to drag block B to the left at constant speed. 

FxB=F-FRBG-FRBA-T=0 F=FRBG+FRBA+T
F=μk NBG+μk NBA+T
FyB=NBG-NBA-mB g=0 NBG=NBA+mB g
F=μk NBA+mB g+μk NBA+T
FyA=NAB-mA g=0 NAB=mA g
F=μk mA g+mB g+μk mA g+T
FxA=FRAB-T=0 T=FRAB=μk NAB=μk mA g
F=μk mA g+mB g+μk mA g+μk mA g
F=3 μk mA g+μk mB g
Cinemática
Two cyclists are in a race. One cyclist knows that he is slower, so he cheats: he removes the faster cyclist's bike chain. The cheater starts from rest immediately with acceleration 2.2 m/s2. The faster cyclist has to take 6 seconds to replace her bike chain. She then follows (also from rest) with acceleration 2.9 m/s2. Assume that both cyclists accelerate smoothly and that they do not reach their maximum speeds during this race. What is the maximum length that the race can be (in meters) in order for the slower cyclist to win?
x=v0 t+12 a t2 x=12 2,2 ms2 t2 x=12 2,9 ms2 t-62
12 2,2 ms2 t2=12 2,9 ms2 t-62
t2-12 t+36=0,76 t2 0,24 t2-12 t+36=0
t=3,2 s x=11,3 mt=46,8 s x=2408,7 m
Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reaches the maximum height of 60 m. The acceleration due to gravity is 10 m/s2.
Part 1. Find the time t the projectile spends in the air. Enter the answer in "s".
ymax=v02 sin2θ2 g 60 m=v02 sin60º22 10 ms2 v0=40 ms
tV=2 v0 sinθg=2 40 ms sin60º10 ms2 tV=6,93 s
Part 4. Find the horizontal range Δx of the flight. Enter the answer in "m".
xmax=v02 sin2θg=40 ms2 sin120º10 ms2 xmax=138,56 m
Starting from rest, a child throws a ball of mass m with an initial speed v, at an angle B with the horizontal direction. The child then chases after the ball, accelerating at a constant acceleration a. If the child wants to catch the ball at the same height as it was thrown, what must be the child's acceleration a? 

Ball:
tV=2 v0 sinθg=2 v sinBg
x=vx tV=v cosB 2 v sinBg=2 v2 sinB cosBg
Child:
d=v0 t+12 a t2 2 v2 sinB cosBg=12 a 2 v sinBg2
a=g cosBsinB a=g cotB
A cannon is shooting emergency packets to people stranded on the roof of a flooded building of height H = 108 meters relative to the cannon, the corner of which is located a distance D = 55 meters from the cannon. It is desired that the incoming packets are flying tangent to the roof as shown so that they land gently with as little impact as possible and slide along to a stop. Find the initial speed v0 and at what angle θ (in radians) the cannon should be aimed to achieve the above scenario.
x=vx t D=v0 cosθ t
vyf=vy0-a t 0=v0 sinθ-g t t=v0 sinθg
D=v0 cosθ v0 sinθg D=v02 sin2θ2 g
vyf2=vy02-2 g d 0=v0 sinθ2-2 g H H=v02 sin2θ2 g
ymax=H=v02 sin2θ2 gxmax=2 D=v02 sin2θg
H2 D=v02 sin2θ2 gv02 sin2θg HD=sin2θ2 sinθ cosθ tanθ=2 HD
tanθ=2 108 m55 m θ=75,71º=1,32 rad
108 m=v02 sin75,712 9,8 ms2 v0=46,74 ms
Two masses, m1 and m2, are hung over a pulley as shown. Assume that m1 is heavier, that the pulley is massless and frictionless, and that the rope does not slip. The blocks are held motionless and then released. Determine the magnitude of the acceleration and the velocity of m1 after it has fallen a distance of d meters.
m1 g-T=m1 a
T-m2 g=m2 a
m1 g-m2 g=m1 a+m2 a
a=m1-m2m1+m2 g
vf2=v02+2 a d=2 m1-m2m1+m2 g d vf=2 m1-m2m1+m2 g d
Centro de masa
The diagram above shows a uniform density hollow cylindrical shell with an open top. It has radius R and height h. Find the height for the center of mass of this cylinder, taking the origin of the coordinate system at the center of the bottom.
ym1=h2m1=V1 ρ=2 π R h e ρym2=0m2=V2 ρ=π R2 e ρ
ym=ym1 m1+ym2 m2m1+m2
ym=h2 2 π R h e ρ+02 π R h e ρ+π R2 e ρ ym=h22 h+R
Two small particles of mass m1 and mass m2 attract each other with a force that varies with the inverse cube of their separation. At time t0, m1 has velocity v directed towards m2, which is at rest a distance d away. At time t1, the particles collide. How far does m1travel in the time interval (t0 and t1).

xcm0=m1 x10+m2 x20m1+m2=m1 0+m2 dm1+m2 xcm0=m2 dm1+m2
Dado que no existen fuerzas externas al sistema conformado por m1 y m2, la velocidad del centro de masa se mantiene constante. Ésta es:
vcm=m1 v1+m2 v2m1+m2=m1 v+m2 0m1+m2 vcm=m1 vm1+m2
xcmf=xcm0+vcm t=m2 dm1+m2+m1 vm1+m2 t1-t0
xcmf=m2 d+m1 v t1-t0m1+m2
Dado que al tiempo t1 las masas m1 y m2 colisionan, la posición del centro de masa corresponde a la posición de las masas m1 y m2.
Conservación de momentum lineal y de energía
A cue ball (white) hits another billiards ball (red). The red ball is initially at rest. The white ball's motion is at a 90° angle to the motion of the red ball after the collision. After the collision, the ratio of speed of the white ball to that of the red ball is 1: 3. What is the ratio of the initial speed to the final speed of the white ball? All balls have the same mass.

m v0Bx=m vfBx+m vfRx v0Bx cosα=vfB
m v0By=m vfBy+m vfRy v0By=vfR=3 vfB
v0B=v0Bx2+v0By2=vfB2+3 vfB2
v0B=vfB2+3 vfB2 v0B=2 vfB
Two cars are approaching a perpendicular intersection without a stop sign. Car 1 has a mass 750 kg and is heading north and car 2 has mass 700 kg and is heading west. The two cars collide at the intersection, and stick together as a result of the collision. The police report stated that after the collision, the two cars were moving in a direction 40º west of north. What is the ratio of the initial speed of car 1 to car 2?
m1 v10x+m2 v20x=m1+m2 vfx 700 kg v20=750 kg+700 kg vfx
v20=7570+1 vf cos140°
m1 v10y+m2 v20y=m1+m2 vfy 750 kg v10=750 kg+700 kg vfy
v10=7075+1 vf sin140°
v10v20=7075+1 vf sin140°7570+1 vf cos140°=5660 tan140° v10v20=-0,78
A spring-like trampoline dips down 0.09 m when a particular person stands on it. If this person jumps up to a height of 0.48 m above the top of the uncompressed trampoline, how far will the trampoline compress after the person lands?
Fe=m g k x0=m g k=m g0,09
Ee=Ep 12 k xf2=m g h 12 m g0,09 xf2=m g 0,48
xf2=0,0864 xf=0,29 m
The Lennard-Jones equation shown below gives the potential energy of two atoms when they are separated by a distance r.
What is the distance between atoms at the condition of stable equilibrium?
Ur=4 U0 r0r12-r0r6=4 U0 r012r12-r06r6
dUdr=4 U0 -12 r012r13+6 r06r7=0 -12 r012r13-6 r06r7=0
2 r012r13=r06r7 2 r06r6=1 r6=2 r06 r=62 r0
What is the minimum work needed to completely dissociate the molecule by separating the atoms from each other if they are initially at equilibrium?
UE=4 U0 r062 r012-r062 r06=4 U0 14-12
UE=-U0 W=U0
A small block of mass 1.2 kg is launched by a compressed spring with force constant k=800 N/m. The initial compression of the spring is 0.12 m. The block slides along a horizontal frictionless surface and then up an inclined plane that makes an angle θ=40° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is μk=0.3. Use g=9.8m/s2.

Ee=Ep+TR 12 k x2=m g h+FR hsinθ
FR=μk N=μk m g cosθ
12 k x2=m g h+μk m g cosθ hsinθ
12 k x2=m g h 1+μktanθ
12 800 Nm 0,12 m2=1,2 kg 9,8 ms2 h 1+0,3tan40°
h=0,36 m
Movimiento rotacional
The system shown in the diagram contains two blocks, of masses 1.4 kg and 4.2 kg, connected by a light string over a pulley of radius 0.2 m and rotational inertia 2.8 kg m2. The block of mass 4.2 kg is free to slide on a horizontal frictionless surface and the pulley is free to rotate on a frictionless fixed horizontal axle passing through its center. Find the magnitude of the angular acceleration α of the pulley once the system is released from rest. The string connecting the blocks does not slip on the pulley. Enter your answer in rad/s2 and use g = 9.8 m/s2.
T=m1 a=m2 g-a
3 m2 a=m2 g-a a=g4 a=2,45 ms2
T=4,2 kg 2,45 ms2 T=10,29 N
τTotal=α I π2 T r=α I π2 10,29 N 0,2 m=α 2,8 kg m2
α=1,15 rads2
You drop a block from rest. The block lands on the spring (spring constant = 160 Nm) and compresses it .5 m before the block and spring momentarily come to rest. (The spring then pushes the block upward.) What is the mass of the block?
m g h0=12 k x2+m g hf
m g h0-hf=12 k x2
m 10 ms2 3 m-1 m=12 160 N m 0,5 m2
m=1 kg
Two pool balls, each of mass 0.2 kg, collide as shown in the figure above. Before the collision, the black ball's velocity makes an angle of 30º with the horizontal line. After the collision, the white ball's velocity makes an angle of 60º with the vertical line. What is the black ball's speed after the collision?

m vB0x+m vW0x=m vBfx+m vWfx
10 ms cos30°=5 ms sin60°+vWfx vWfx=4,33 ms
m vB0y+m vW0y=m vBfy+m vWfy
10 ms sin30°-8 ms=5 ms cos60°+vWfy vWfy=-5,5 ms
vWf=vWfx2+vWfy2 vWf=7 ms