Chapter 8
A 25 kg child is initially riding in a 50 kg cart moving at a speed of 2.0 m/s. The child manages to jump off the back of the cart in such a way that he has no final velocity. What is the resulting velocity of the cart? b) Compare the child's change in momentum to the cart's change in momentum. c) What additional information would be needed in order to determine the force the child used to jump off the cart?
(m+m) v = mv + mv (25 + 50 kg)(2.0 m/s) = 0 + (50 kg)(v) v = 3.0 m/s b) They are equal. (conservation of momentum) You can run the math to prove it if you prefer. c) You have enough information to determine the change in momentum for either the child or the cart. The change in momentum is equal to the impulse. To solve for the force, you need the time.
A student holds a 2.0 kg air rifle, unbraced with a very loose grip. The muzzle velocity of a 2.0 g bullet fired from the rifle is about 140 m/s. What is the recoil speed of the gun? b) What could the student do to reduce the recoil speed of the air rifle? Explain.
(m+m)v = mv + mv 0 = (2.0 kg) v + (2x10^-3 kg)(140 m/s) v = -0.14 m/s b) The student could brace the rifle against her shoulder. The overall change in momentum remains the same, but the mass of the student is added to the mass of the gun resulting in a lower change in velocity.
A 60 kg person and an 75 kg person each jump off opposites sides of a 40 kg raft with a speed of 3.0 m/s. If the raft was initially at rest, what is the final speed of the raft? b) Was this collision elastic? Explain.
(m+m+m) v = mv + mv + mv 0 = (60 kg)(-3 m/s) + (75 kg)(3 m/s) + (40 kg)(v) v = 1.13 m/s in whichever direction the 60 kg person jumped (in the negative direction in this case) b) Kinetic energy is conserved in elastic collisions. Since there was no kinetic energy in the system at the start and there IS kinetic energy at the end (all three objects are in motion), this collision was NOT elastic.
J = F(delta t) Jext = 0 Pi = Pf
Exterior impulse Initial momentum = Final momentum.
ΣF
F + - mg or F + -W
A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg.
Fnet = (m)(delta v) / (delta t) v1 = 10m/s v2 = 0m/s delta v = v2 - v1 delta v = 0m/s - 10m/s delta v = -10m/s magnitude = 10m/s delta t = 0.26s Fnet = (70kg)(10m/s) / (0.26s) Fnet = 2.7*10^3N
A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)?
Fnet = (m)(delta v)/(delta t) V1 = 0m/s V2 = 600m/s Change in speed = V2-V1 (600m/s) - (0m/s) = 600m/s magnitude = 600m/s time = (2ms) x (1s/1000ms) = 2*10^-3s Fnet = (0.03kg)(600m/s) / (2.00*10^-3s) F net = 9*10^3N
Three point masses start from rest at a position very nearly (0,0). When released from rest, they apply an initial repelling force on each other. Particle A has a mass of 0.1 kg and moves at a speed of 2 m/s to the west. Particle B has a mass of 0.2 kg and moves at a speed of 3 m/s at an angle 30° below the horizontal. Particle C has a mass of 0.3 kg. What is the velocity of particle C (speed and direction)?
Horizontal: 0 = mv + mvcosangle + mvx 0 = (0.1 kg)(-2 ms) + (0.2 kg)(3 m/s)(cos 30) + (0.3 kg)(vx) vx = -1.07 m/s Vertical: 0 = mvsinangle + mvy 0 = (0.2 kg)(-3 m/s)(sin 30) + (0.3 kg)(vy) vy = 1.0 m/s Combine the components: (-1.07 m/s)2 + (1.0 m/s)2 = v2 v = 1.46 m/s tan θ = (1.0 m/s) / (1.07 m/s) θ = 43°, except the angle should be in the second quadrant θ = (180 - 43) = 137°
______ is maximized from applying a large force for a long time on an egg, breaking the egg and sending bits of pieces flying.
Impulse
Δplarge = Δpsmall (mΔv)large = (mΔv)small
In order for the change in momentum to be the same for objects (i.e., for momentum to be conserved), smaller masses must always experience larger changes in velocity.
Impulse equation
J = Ft J- impulse F- force on object t- amount of time the force is experienced J = Δp
A 20 kg rocket experiences an engine thrust of 300 N for 5.0 seconds. What is the impulse provided by the engine? What is the net impulse on the rocket?
Jengine = Fenginet J = (300 N)(5.0 s) J = 1500 N-s ΣJ = ΣFt ΣJ = (Fengine - W)t ΣJ = (300 N - (20 kg)(9.8 N/kg))(5.0 s) ΣJ = 520 N-s It is the net impulse on the rocket that will determine the rocket's motion.
Delta t = 0.02s Vf = 5m/s Vi = -10m/s Mass = 0.2kg N = 3kg*m/s What was the impulse on the ball from the person? What was the average force on the person's face from the ball?
Jnx = Delta Px = Fn(delta t) Jnx = Pf - Pi Jnx = Mvf - Mvi Jnx = (0.2kg)(+5m/s) - (0.2kg)(-10m/s) Jnx = 1kg*m/s - (-2kg*m/s) Jnx = 3kg*m/s Impulse *on the ball* has a positive force to the right. Jnx = Fnx * Delta t 3kg*m/s = Fnx (0.02s) Fnx = 3kg*m/s / 0.02s = -150N Force *on the face* from the ball has a negative force to the left.
Two cars moving crash into each other. m1 = 1000kg m2 = 2000kg M1 + M2 = 3000kg momentum and velocity?
Mc * Vc + Mt * Vt 1000kg * 9 + 2000 * 0 9000 + 0 = 9000kg*m/s 3000Vn = 9000 Vn = 3m/s East
impulse units
N-s
What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 10.0 m/s ? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?
P = mv P = (1.20*10^4kg)(10m/s) P = 1.20*10^5kg*m/s V = P/M V = (1.20*10^5kg*m/s)/(8.00kg) V = 1.50*10^4m/s
(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s . (b) Compare the elephant's momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s . (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant?
Pe = (Me)(Ve) Pe = (2000kg)(7.50m/s) Pe = 1.50x10^4kg*m/s Pd = (Md)(Vd) Pd = (0.0400kg)(600m/s) Pd = 24.0kg*m/s Pe/Pd = 1500kg*m/s / 24kg*m/s = 625 times larger than momentum of a dart. Pn = (Mn)(Vn) Pn = (90kg)(7.40m/s) Pn = 666kg*m/s
In the original Superman movie, Lois Lane falls off of the top of a skyscraper (a very tall building). She falls for about half the height of the skyscraper before Superman catches her and carries her back to the top of the building. In the movie, Lois goes from freefalling to moving upward in less than one second. In terms of momentum and impulse, explain why this is not good for Lois.
Since the change in velocity includes a direction change, this means she experiences a large change in velocity. Since this large change in velocity occurs over a very short amount of time, it means the force that causes this change will be extremely large, which is not good for Lois. To estimate the value of the force: v2 = 2aΔy + vo2 You'll assume a conservative 800 ft (244 m) tall building, which means she fell for about 122 m. For simplicity's sake you'll ignore air resistance. v2 = 2(-9.8 m/s/s)(-122 m) v = -49 m/s (terminal velocity occurs around 53 m/s for a human body although this depends on how the person is falling) You'll assume a very conservative upwards speed of 5 mph after she's caught (2.2 m/s). The overall change in velocity is then about 50 m/s. Next, you'll give Lois a weight of about 120 lbs, which corresponds to a mass of 54 kg. ΣFt = mΔv (Fsuperman - (54 kg)(9.8 N/kg))(1.0 s) = (54 kg)(50 m/s) F = 3229 N or about 730 pounds of force This is on the conservative side.
Use principles of impulse and momentum to explain why it's generally better to hit an airbag rather than a dashboard during a vehicle collision.
The change in momentum the passenger experiences will be the same regardless of whether they hit the airbag or the dashboard. However, by hitting the airbag, the change in momentum takes longer to occur, which means that the airbag will apply a smaller force to the passenger.
A large truck collides with a small car that is initially at rest. Which vehicle experiences the largest collision force? Which vehicle experiences the largest change in momentum? Which vehicle experiences the largest change in velocity?
The force of the truck on the car is equal to the force of the car on the truck. (Newton's third law) The momentum lost by the truck is equal to the momentum gained by the car. (conservation of momentum) The small car will experience the largest change in velocity. This is due to the conservation of momentum between two objects with unequal masses
A 0.02 kg ball falling at a speed of 10 m/s bounces off the ground. The final speed of the ball is 5 m/s upwards. If the ball was in contact with the ground for 0.003 seconds, what was the normal force on the ball?
The initial velocity is down which means it is negative. ΣFt = mΔv (N + - mg) t = m (vf - vo) (N - (0.02 kg)(9.8 N/kg))(0.003 s) = (0.02 kg)(10 m/s - -5 m/s) N = 100 N
An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?
The larger mass object would have more kinetic energy. 1) its heavier 2) it covers a larger area 3) the more mass an object has, the larger the kinetic energy because of its weight.
A student drops a 10 g super bouncy ball from an initial height 2.0 m above a sensor plate. The plate measures that the ball was in contact with the plate for 0.004 seconds. Assuming that the speed of the ball just after the collision was the same as just before the collision, what was the force of the sensor plate on the ball during the collision?
Ug = K mgh = ½ mv2 (9.8 N/kg)(2.0 m) = ½ (v2) v = -6.3 m/s ΣFt = mΔv (N - W)t = m(vfinal - vinitial) (N - (10x10^-3 kg)(9.8 N/kg))(0.004 s) = (10x10^-3 kg)(6.3 m/s - -63 m/s) N = 31.6 N
Class B model rocket engines can apply 5.0 N-s of impulse to a rocket. If a class B engine that has a burn time of 1.2 s is attached to a 0.15 kg rocket, what maximum speed will the rocket reach when it is launched? (Ignore air resistance.)
Use the rocket's impulse to solve for the average force of the engine on the rocket: J = Ft (5.0 N-s) = F(1.2 s) F = 4.2 N ΣFt = mΔv ((4.2 N) - (0.15 kg)(9.8 N/kg))(1.2 s) = (0.15 kg)(Δv) Δv = 21.8 m/s
An astronaut on a spacewalk pushes off of the outside of the spaceship. a) Compare the force of the astronaut on the ship to the force of the ship on the astronaut. b) Compare the astronaut's change in momentum to the spaceship's change in momentum. c) Compare the astronaut's change in velocity to the spaceship's change in velocity.
a) They are equal. (Newton's third law) b) They are equal. (conservation of momentum) c) The astronaut has a larger change in velocity. The two objects have the same change in momentum, which is equal to mass x change in velocity. Since the astronaut has less mass, she must have a larger change in velocity.
a) A 0.145 kg baseball goes from a speed of 40 m/s to the right to a speed of 30 m/s to the left. What was the baseball's change in momentum? b) What is the impulse on the baseball? c) If the change in momentum occurred in 0.003 seconds, what was the net force acting on the ball?
a) Δp = mΔv Δp = (0.145 kg)(-30 m/s - 40 m/s) Δp = -10.2 kg-m/s b) J = Δp J = -10.2 N-s c) ΣFt = mΔv ΣF(0.003 s) = (-10.2 kg-m/s) ΣF = 3400 N
a) A 0.5 kg model rocket is given an engine that can apply a thrust force of 8.0 N for 3.0 seconds. What is the maximum speed of the rocket when it is launched upwards? b) What was the rocket's change in momentum? c) What was the net impulse on the rocket?
a) ΣFt = mΔv (Fengine - W) = m(vfinal - vinitial) (8.0 N - (0.5 kg)(9.8 N/kg))(3.0 s) = (0.5 kg)(v - 0) v = 18.6 m/s b) Δp = mΔv Δp = (0.5 kg)(18.6 m/s - 0) Δp = 9.3 kg-m/s c) ΣJ = Δp ΣJ = 9.3 N-s
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×103 m/s , given the collision lasts 6.00×10 - 8 s
change in momentum = delta p = (f)(delta t) change in momentum of paint chip = delta p = (m)(delta v) (delta p) = (0.100mg)(10^-6kg/1mg)(4.00*10^3m/s) (delta p) = (1*10^-8kg)(4.00*10^3m/s) (delta p) = 4.00*10^5N*s F = (delta p)/(delta t) F = (4.00*10^-5 N*s)/(6.00*10^-8 s) F = 666.67N --> force on the window
What is an elastic collision?
elastic collision is the type of collision in which kinetic energy is conserved. KEi = 1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1^2 + 1/2m2v2^2
All collisions can be classified as either _______ collisions or _______ collisions.
elastic; inelastic.
momentum
how difficult it is to stop a moving object.
both cars must experience the same _______ (same horizontal net force and same time)
impulse ΣJ = ΣFt
what is an inelastic collision?
inelastic collision is the type of collision in which kinetic energy is not conserved. KE1 = 1/2m1v1^2 + 1/2m2v2^2 KE2 = 1/2m1v1^2 + 1/2m2v2^2 KE1 does not equal KE2, so the collision is inelastic.
if an object has a large _____ (measured by mass), it will be difficult to change the ______ of the object.
inertia; motion.
units of momentum
kg-m/s
momentum lost by the _____ car must equal the change in momentum of the smaller car
large
Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of −0.120 m/s . (The minus indicates direction of motion.) What is their final velocity?
m1v1 + m2v2 = (m1+m2)(v) (150000kg)(0.300m/s) + (110,000kg)(-.120m/s) = (150000kg + 110000kg)(v) v = (45,000kg*m/s - 13,200kg*m/s)/(260,000kg) v = (31,800kg*m/s) / (260,000kg) vf = 0.12m/s
Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00×103 kg , and the second a mass of 7.50×103 kg . If the two satellites collide elastically rather than dock, what is their final relative velocity?
m1v1 + m2v2 = (m1+m2)v' v1' = ((m1-m2)/(m1+m2))(v1) v1' = ((4.00*10^3kg - 7.50*10^3kg)/(4.00*10^3kg + 7.50*10^3kg))(0.250m/s) v1' = (-3.50*10^3kg/11.50*10^3kg)(0.250m/s) v1' = -7.61*10^-2m/s v2' = (2m1/(m1+m2))(v1) v2' = ((2(4.00*10^3kg)/(4.00*10^3kg + 7.50*10^3kg))(0.250m/s) v2' = (8.00*10^3kg/11.50*10^3kg)(0.250m/s) v2' = 1.74*10^-1m/s v'net = v2' - v1' v'net = (1.74*10^-1m/s) - (-7.61*10^-2m/s) v'net = 0.25m/s
Impulse is the same as change in ________. means that both cars must also experience the same changes to their _______ during the collision.
momentum ΣFt = mΔv
the total _______ before a collision will equal the total _______ after the collision.
momentum Σpinitial = Σpfinal
A 0.30 kg toy car with an initial velocity of 4 m/s collides with a 0.5 kg car with an initial velocity of 4 m/s in the opposite direction. If the cars stick together during the collision, what is the final velocity of the two cars? Assume negligible friction. b) If the collision took 0.005 seconds, what was the force applied by the cars on one another?
mv + mv = (m+m)v (0.3 kg)(4 m/s) + (0.5 kg)(-4 m/s) = (0.3 + 0.5 kg) v v = -1 m/s b) ΣFt = mΔv The only horizontal force is the force of each car on the other. Pick one of the cars to focus on. F (0.005 s) = (0.5 kg)(-1 m/s - -4 m/s) F = -500 N If you had picked the other mass, you would have calculated the same size force in the opposite direction.
A 0.1 kg arrow with an initial velocity of 30 m/s hits a 4.0 kg melon initially at rest on a frictionless surface. The arrow emerges out the other side of the melon with a speed of 20 m/s. What is the speed of the melon? b) Why would we normally not expect to see the melon move with the is speed after being hit by the arrow?
mv + mv = mv + mv (0.1 kg)(30 m/s) + 0 = (0.1 kg)(20 m/s) + (4.0 kg) v v = 0.25 m/s b) The melon would not normally be resting on a frictionless surface. Remember that momentum is only conserved if there are no external forces acting on the system.
Newton's second law notes that in order for an object to change velocity (whether that involves a change in speed, a change in direction, or both) a _______ must be applied. ΣF = ma
net force
impulse-momentum theorem
net impulse = object's change in momentum
As long as no ______ ________ act on the system, the total ________ will be conserved regardless of whether the collision is elastic or inelastic.
outside forces; momentum.
momentum equation
p = mv p- momentum m- mass (inertia) v- velocity
Sphere A has a mass of 2.0 kg and moves at a speed of 5 m/s. Sphere B has a mass of 1.0 kg and moves at a speed of 7 m/s. Which sphere has more momentum (i.e., which will be more difficult to stop)?
p = mv pA = (2.0 kg)(5 m/s) = 10 kg-m/s pB = (1.0 kg)(7 m/s) = 7 kg-m/s Sphere A has more momentum.
Which of the following has more momentum: a 60 kg person moving at 4 m/s or an 80 kg person moving at 3 m/s?
p = mv p = mv p = (60 kg)(4 m/s) p = (80 kg)(3 m/s) p = 240 kg-m/s p = 240 kg-m/s The two people have the same momentum.
Can objects in a system have momentum while the momentum of the system is zero? Explain your answer.
p1 = m1v1 p2 = m2v2 p1 + p2 = 0 p1 = -p2 m1v1 = -m2v2 if momentum of two objects has equal value and direction of velocity is opposite, system can have momentum of zero.
A battleship that is 6.00×107 kg and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship's recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs.
pf = 0 mu + mv = 0 v = -mu/M v = -(1100kg)(575m/s)/(6.00*10^7kg) v = -1.1*10^-2m/s b) K = 1/2mu^2 + 1/2mv^2 K = (1/2)(1100kg)(575m/s)^2 + (1/2)(6.00*10^7kg)(-1.0542*10-2m/s)^2 K = 1.82*10^8 J
definition of acceleration
rate of change of velocity a = Δv / Δt
Impulse definition
relative measurement of the effort that is put into changing the velocity of some object. (This is not the same as the force—the push or pull.)
How can a small force impart the same momentum to an object as a large force?
small force can impart the same momentum as a large force if that small force is applied for a small interval of time compared to the time which the large force is applied.
Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not.
the energy is not always conserved whenever momentum is conserved.
An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum?
the object with the larger mass will have the larger momentum.
inelastic collision
the total kinetic energy of the system is NOT conserved. Some kinetic energy is lost during the collision. Usually the lost kinetic energy is the result of internal friction, heat, deformation of the objects, etc.
elastic collision
the total kinetic energy of the system is conserved. This means that the total kinetic energy of the system remains constant; the kinetic energy that the system has before a collision is equal to the kinetic energy the system has after the collision.
Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. tile floor for a day care center.
to reduce the time of impact (contact), effect of padding is necessary. padding means using a softer material to stop the motion of an object. thus, carpeted or sand floor would cause less effect of impulsive force and thereby reduces the occurence of injuries.
Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed?
v1 = 10m/s v2 = 0m/s Delta v = 0m/s - 10m/s = -10m/s Delta v = magnitude = 10m/s Fnet = (m)(delta v)/(delta t) Fnet = (70kg)(10m/s) / (0.26s) Fnet = 3*10^3N m1v1 + m2v2 = (m1 + m2)(v) (70kg)(10m/s) + (70kg)(-10m/s) = (70kg + 70kg)(v) 700kg*m/s - 700kg*m/s = (140kg)(v) 0kg*m/s = (140kg)(v) v = (0kg*m/s)/(70kg) v = 0m/s final velocities are the same so there is not change in force.
If an object is moving with high _____, it will be difficult to stop. You have to apply a small ________ for a long time or a large _______ for a small time.
velocity; net force.
A 1200 kg car goes from a speed of 30 m/s to 20 m/s. What was the car's change in momentum? b) If the change in momentum occurred due to a net force of 1000 N acting on the car, how long did it take for the change in momentum to occur?
Δp = mΔv Δp = (1200 kg)(20 m/s - 30 m/s) Δp = 12000 kg-m/s ΣFt = mΔv (1000 N)(t) = (12000 kg-m/s) t = 12 seconds
change in object's momentum
Δp = pfinal - pinitial
A 5.0 kg box moving with an initial velocity of 3.0 m/s slows to a speed of 1.0 m/s due to friction. What was the box's change in momentum?
Δp = pfinal - pinitial Δp = (5.0 kg)(1.0 m/s) - (5.0 kg)(3.0 m/s) Δp = -10 kg-m/s Note that since the mass is constant, you could have also written the equation as: Δp = mΔv
A 0.145 kg baseball approaches a batter with an initial speed of 30 m/s. The batter hits the ball, and the ball leaves the bat moving in the opposite direction with a speed of 20 m/s. If the ball and the bat were in contact for 0.01 s, what was the force of the bat on the ball?
Δv = vfinal - vinitial Δv = (-20 m/s) - (30 m/s) = - 50 m/s ΣFt = mΔv ΣF(0.01 s) = (0.145 kg)(-50 m/s) ΣF = -725 N Since the only horizontal force acting on the ball was the bat, the net force in this case is the final answer.
A 30 kg rocket is initially at rest on the ground. The rocket experiences an upward 400 N of force from the engine during launch. If the engine fires for 6.0 seconds, what is the final speed of the rocket? (Assume that the mass of the fuel is negligible.)
ΣFt = mΔv (Fengine + - mg) t = m (vfinal - vinitial) [(400 N) - (30 kg)(9.8 N/kg)] (6.0 s) = (30 kg) (vfinal - 0) v = 21.2 m/s
Newton's second law
ΣFt=mΔv net impulse = object's change in momentum
An astronaut on a space-walk has become untethered. The astronaut and her equipment come to a total mass of 150 kg. She is initially floating with zero velocity relative to her ship. In order to get to her ship, she throws some of her equipment in a direction away from the ship. The equipment has a mass of 5.0 kg and a velocity of 6 m/s. What is the resulting speed of the astronaut?
Σpinitial = Σpfinal (m+m)v = mv + mv (150 kg)(0) = (145 kg)(v) + (5.0 kg)(6 m/s) v = -0.21 m/s
Two identical 0.25 kg toy cars travel toward each other on a frictionless track. Car A has an initial velocity of +4 m/s. Car B has an initial velocity of -6 m/s. The cars collide and stick together. What is the final velocity of the combined cars?
Σpinitial = Σpfinal mv + mv = (m+m)v (0.25 kg)(4 m/s) + (0.25 kg)(-6 m/s) = (0.25 + 0.25 kg)(v) v = -1.0 m/s
A 0.5 kg toy car traveling to the right on a frictionless track with an initial velocity of 4 m/s hits a 0.25 kg toy car initially at rest on the track. The two do not stick together. The final speed of the larger car is 2.0 m/s to the right. What is the final speed of the second car?
Σpinitial = Σpfinal mv + mv = mv + mv (0.5 kg)(4 m/s) + 0 = (0.5 kg)(2.0 m/s) + (0.25 kg)(v) v = 4 m/s
A 0.75 kg toy car travels to the right on a frictionless track with an initial speed of 6 m/s. The car collides with a 0.5 kg car traveling to the left on the same track with a speed of 6 m/s. The collision is elastic. What is the final velocity of both cars?
Σpinitial = Σpfinal mv + mv = mv + mv (0.75 kg)(6 m/s) + (0.5 kg)(-6 m/s) = (0.75 kg)(v1) + (0.5 kg)(v2) ½ mv2 + ½ mv2 = ½ mv2 + ½ mv2 ½ (0.75 kg)(6 m/s)2 + ½ (0.5 kg)(6 m/s)2 = ½ (0.75 kg)(v1)2 + ½ (0.5 kg)(v2)2 v1 = -3.6 m/s v2 = 8.4 m/s
