www.lsmsa.edu teacher Robert Dalling's Physics Lectures (see also www.ushumans.net)
Momentum
Newton’s Second Law can be written
F = ma = mΔv / Δt = Δ( mv ) / Δt,
where the vector momentum P = mv. Momentum equals mass times velocity. When two objects collide, the force on one is equal and opposite to the force on the other.
F1 = - F2
or
ΔP1 / Δt = - ΔP2 / Δt.
We can cancel Δt to get
ΔP1 / Δt = - ΔP2 / Δt,
or
ΔP1 = - ΔP2.
A change Δ is always given by final minus initial values. Using a prime to indicate final values and no-prime to indicate initial values, this is written
P1‘ - P1 = -( P2’- P2 ) = P2 - P2’.
We can gather initial values on the left side and final values on the right side.
P1 + P2 = P1’ + P2’,
or,
Σ P = Σ P’.
This is the statement of the conservation of momentum This means that the sum of momentum before an interaction is equal to the sum of momentum after the interaction, no matter how many objects interact and no matter how complicated is the interaction. The vector quantity mass times velocity turns out to be very useful because its total never changes during a collision or interaction of particles. Momentum is mass times velocity, p = mv. It is a vector having direction and magnitude. Its (un-named) units are kg m/s. As usual, we can also write the vector equation in terms of its individual components
Σ Px = Σ Px’
and
Σ Py = Σ Py’.
In terms of velocities, for two masses we have
m1v1 + m2v2 = m1v1’ + m2v2’.
Since the velocities are vectors, some can be positive while others might be have negative vlaues.
Example:
A 2000 kg car moving at 20 m/s has momentum = 40,000 kg m/s. This is twice the momentum of a 1000 kg car moving at 20 m/s but the same momentum as a 1000 kg car moving at 40 m/s.
During a collision of two objects–for example, metal spheres or bowling balls–the surfaces in contact are bending inward, storing energy elastically, and then rebounding. Initially the energy is entirely kinetic. A portion becomes elastic energy for a moment until and then that portion mostly returns to kinetic energy. If 100% of the initial kinetic energy is regained as final kinetic energy then the collision is said to be “100% elastic,” as here. When two objects collide and then stick together, as two balls of soft clay typically do, then the collision is said to be “completely inelastic.” In this case, the two objects move off with a mutual, final speed. During a collision, damage is caused by the conversion of kinetic energy into bent metal and such. The “punch” or force of a collision is due to momentum since F = Δp / Δt. Kinetic energy causes damage and momentum provides the punch.
Example:
A billiard ball with mass m1 = 0.1 kg moving right at a speed of 0.5 m/s collides with an identical ball m2 that is motionless. After the collision, the second ball moves away but the first ball is motionless. What is the speed of m2 after the collision?
m1v1 + 0 = 0 + m2v2
or v2 = v1.
This exchange of momentum can be seen in Newton’s Cradel. If one mass is pulled aside and released, its subsequent collision with the group of identical masses will cause a single mass to move away at the other end of the group. Pulling two masses aside results in two masses emerging from the other side.
Example.
Here is an animation of a pendulum bouncing off the wall of its freely moving support.
Example.
Mass m1 = 2 kg moving right runs into m2 = 1 kg moving left. If both objects stop after the collision, what was the ratio of their incoming speeds? Let’s choose the rightward direction to be positive.
Σ Pbefore = ΣPafter
m1v1 - m2v2 = 0
v1 / v2 = m2 / m1 = ½
Example.
A 2-kg ball of clay moving right at a speed of 0.2 m/s collides with a 1-kg clay ball moving left with a speed of 0.1 m/s. The two stick together and move off with a final speed v'. This is a "totally inelastic collision."
Σ Pbefore = ΣPafter
m1v1 - m2v2 = (m1+m2)v'
v' = ( m1v1 - m2v2 ) / (m1+m2) = ( 2 kg )( 0.2 m/s ) - ( 1 kg )( 0.1 m/s ) / ( 2kg +1kg ) = 0.1 m/s.
Example.
A gun and a bullet within the gun are both at rest. The momentum of the gun and of the bullet are both zero before the explosion that causes the gun to move in one direction and the bullet to travel in the opposite direction.
Pgun = Pbullet
mgun vgun = mbullet vbullet
( 8 kg )( vgun ) = ( 0.020 kg )( 1000 m/s )
vgun = 0.16 m/s
When a warship fires a projectile to its right, the entire boat moves to its left.
Example.
A firecracker is placed on the ground and the explodes, sending two fragments in opposite directions. Before the event, the sum of the momentum is zero.
P = 0 = P’ so m1v1 = m2v2.
If one fragment has twice the mass of the other then it will have half the other's speed.
A stationary astronaut in space, or a person on a frictionless ice-pond, can throw an object one way to be pushed in the opposite direction. When another person catches that thrown object, they'll have an equal change in momentum. Here is a video clip of two skaters tossing a pillow back and forth. See that the momentum of a person changes each time the pillow is either thrown or caught. This is the way that subatomic particles exert forces on one another.
An unstable atom becomes more stable by radiating either light, electrons, or helium nuclei (These are called gamma, alpha, and beta radiation.) The atom recoils as it emits radiation. Subatomic particles collide in a particle accelerator. Knowing the mass and speed of one particle and the speed of the second enables scientists to determine the mass of the second.
CQ.
Name some collisions that are seen every day: a bug and a windshield, a bat with a baseball, a golf club and a golf ball, see here. In an example of the conservation of momentum, we feel a force when water is ejected from a hose that we are holding.
Example.
A 1-kg chunk of putty moving at 1 m/s collides with and then sticks to a 5-kg bowling ball that had been sitting still before the collision. After the putty collision, the putty and ball move with a combined momentum of 1 kg m/s. After that collision, the bowling ball and putty are moving with a speed of 1/6 m/s
Example.
A rifle of mass M = 2 kg is suspended by strings and fires a bullet of mass m = 0.01 kg with a speed of v = 200 m/s. What is the recoil speed V of the gun? We have MV = mv or
V = mv / M = ( 0.01 kg )( 200 m/s ) / 2 m/s.
Example.
A collisional force F is doubled when bouncing straight back rather than sticking. Let’s compare the collisional force when an object either sticks to or bounces off a wall. When the incoming object collides and sticks we have Δp = pf - pi = -pi but when the object bounces back with the same but reversed speed then Δp = pf - pi = -pi - pi = -2pi = -2mvi. We see that the impact FΔt = Δp is then double. For example, twice the force occurs when a waterwheel reverses the direction of the colliding water compared to just stopping the water.
Example.
A large fish with mass m1 = 5 kg is swimming with speed v1 = 1 m/s towards a small fish of mass m2 = 1 kg that is swallowed. What must be the speed v2 of the smaller fish if both are stopped by the collision?
Pbefore = Pafter
+m1v1 - m2v2 = 0
v2 = m1v1 / m2 = (5 kg)(1 m/s ) / 1 kg = 5 m/s.
Example.
A bullet of mass m1 = 0.01 kg moving at speed v1 = 300 m/s is fired into a block of wood of mass m2 = 5 kg and remains within the block as it slides a distance x along the ground that has a coefficient of friction μ = 0.7. What is the distance x?
The conservation of momentum gives
m1v1 = ( m1 + m2 )v’
or
v’ = m1v1 / ( m1 + m2 ) = ( 0.01 kg )( 300 m/s ) / ( 0.01 kg + 5 kg ) = 6 m/s.
The initial kinetic energy of the joined items is lost to work done against friction.
½ ( m1 + m2 )v’2 = μ( m1 + m2 )gx.
We get x = v’2 / (2μg) = ( 6 m/s )2 / ( 2 )( 0.7 )( 9.8 m/s 2 ) = 0.026 meters.
Example.
Here is an example of the conservation of momentum.
Demonstration.
Video of a person lying on “bed of nails” with a concrete block resting on the person's chest. A sledge hammer smashes the concrete. The sledge hammer's kinetic energy does damage to the block which flies apart. The sledge hammer's momentum change = force gives a jolt to the person. That jolt is spread over the combined area of all the nails. Bunch 500 nails together and they form a block with 1/4 m2 area. We could easily lie on that block of metal. It doesn't matter that the area is spread out into lots of nail points.
Demonstration.
The physics of a Karate chop.
Example.
An ice-skater with mass m1 = 70 kg is heading in the +x direction at speed v1 = 1.67 m/s and grasps and embraces another skater with mass m2 = 50 kg who was heading in the +y at direction at speed v2 = 2.2 m/s. The two skaters will then move off together at what angle θ and with what speed V?
Conserving momentum in the x-direction gives
ΣPx = ΣPx’
or
m1v1 + 0 = ( m1 + m2 ) Vx
We have
Vx = m1v1 / ( m1 + m2 ) = 0.974 m/s
Conserving momentum in the y-direction gives
ΣPy = Σpy’,
or
0 + m2v2 = ( m1 + m2 ) Vy.
Vy = m2v2 / ( m1 + m2 ) = 0.92 m/s.
We find the magnitude and angle of the resultant, final velocity from
V = √( Vx2 + Vy2 ) = 1.34 m/s.
Since tan θ = Vy / Vx, we have
θ = 43°.
What portion of the initial kinetic energy was lost in this interaction?
( Ki - Kf ) / Ki = 0.5.
Example.
Here is a 2D, elastic collision.
Here is another simulation.
Example.
The coefficient of restitution e is a measure of the elasticity of a collision. We have e = 1 when the collision is 100% elastic and we have e = 0 when the collision is 100% inelastic. In a collision of two moving objects, the coefficient is defined through the equation
v2f - v1f = -e ( v2i - v1i ).
If on each successive bounce, a ball rebounds with a fraction f of its previous speed, what is the total time the ball will spend in the air after the first bounce occurs? The speed before the nth bounce is vn. We have vn / g. We have vn+1 = e vn. The time needed to fall from the peak of motion before the nth bounce is vn / g. During this bounce, the total time in the air is double that, tn = 2 vn / g. We have tn+1 = 2 vn+1 / g = 2 e vn /g = e tn The total time in the air is the sum of each of these bounce times,
T = to + e to + e2 to + e3 to + ... = to Σ en = to / ( 1 - e ).
Impulse
A force F applied for a period of time t is said to be an impulse I, where
I = F Δt = ΔP.
The impulse I changes the momentum of the mass through the product F Δt. The same change in momentum occurs when applying either a strong force for a short time or a weak force for a long time. To move something we can either push lightly for a long time or strongly for a short time.
Example.
A 5-kg box sliding across ice at 2 m/s takes 0.5 second to cross a stretch of rough ground that applies a frictional force of 2 N. What is its speed after reaching the end of the rough ground? The force is in the negative direction, opposite the direction of the momentum, P. Since
FΔt = ΔP = Pf - Pi = mvf - mvi,
we have
vf = ( FΔt + mvi ) / m = [ (-2 N )( 0.5 s ) + ( 5 kg )( 2 m/s ) ] / 5 kg = 9/5 = 1.8 m/s.
Example.
A 1000-kg car moving at 10 m/s brakes to a stop in 5 seconds. The average breaking force is
F = Δp / Δt = mΔv / Δt = ( 1,000 kg )( 10 m/s ) / 5s = 2000 N.
Example.
What is the impulse and what is the force that a seatbelt applies to stop a 20-kg child in 0.005 seconds?
I = FΔt = Δp = m( vf - vi ) = ( 20 kg )( 0 - 10 m/s) = -200 kg m/s.
F = Δp / Δt = ( -200 kg m/s ) / 0.005 s = 4000 N.
How does this force compare to the weight of the child?
F / mg = 4000 N / ( 20 kg )( 9.8 m/s2 ) = 20.4
Could you keep that child held in your hands during the collision? No. The force needed to hold the 45-pound child would be 900 pounds. Can you imagine holding a 20-lb bowling ball as several of your friends jerked you backwards?
Example.
Here and here and here are some other examples.
Demonstration.
Here are some demonstrations.
Here are some impact times that are encountered in sports: golf ball or baseball =1 millisecond, tennis hit = 5 ms, football kick = 8 ms, handball = 13 ms, soccer ball hit with head = 23.
Example.
A pitched baseball is hit with a bat. What is the impact? The ball comes toward the batter at 90 mph = 40 m/s and is sent away from the batter at 100 mph = 44.4 m/s. The weight mg of the ball is 5.25 ounces or mass m = 0.14 kg. Before the collision, the momentum of the ball is towards batter with magnitude
P = mv = ( 0.14 kg )( 40 m/s ) = 5.6 kg m/s.
After the collision, it is away from the batter with magnitude
P’ = mv = ( 0.14 kg )( 44.4 m/s ) = 6.2 kg m/s.
So that Δp = p - p’ = ( -6.2 - 5.6 ) kg m/s = 11.82 kg m/s
and
F = Δp / Δt = ( 11.82 kg m/s ) / ( 0.001 s ) = 11,820 N = 2660 lb.
Without really noticing it, we naturally try to extend the time over which a collision lasts.
1) We lift our arms upward while making a large step down because our center of mass raises a bit and this causes us to take a greater amount of time to reach the ground
2) When we jump down we bend our knees as we land to lengthen the impact duration time and decrease the force of the impact. We want to extend the impact time when landing on concrete. Compared to landing on concrete, the impact time is greater when landing on dirt. The impact time is most extended when jumping into water–but less so with increasing entry speeds.
3) When we catch a ball our hands move inward to lengthen the impact duration time and decrease the force of the impact.
4) When we catch a heavy object dropped to us from above we might try to swing it around.
5) A glass dropped on a carpet might not break but always breaks when dropped onto a wooden floor because the carpet increases the collision time Δt.
6) Impacts are lessened when gymnasts land on cushioned rather than on wooden floors. Impacts are decreased for boxers hitting with gloves not bare fists, padded automobile dashboards, cushioned running shoes, automobile airbags, and safety helmets. The impact from catching a baseball is reduced when catchers allow the elbows to bend inward. There would be a much greater impact if catchers instead placed their hands against a wall. Here are calculations of forces and impulses occurring in a train collision.
CQ.
Name some situations in which a force is applied for an amount of time.
Demonstration.
Analyze this collision between two cats.
Demonstrations.
The Physics Force group from the University of Minnesota demonstrate equal and opposite forces in people who are riding in bumper cars, and they stop a 50-mph egg without breaking it.
Try this 2D collision in which the table can be tilted and the masses can be magnetized.
Here is a simulation of billiard collisions.
Student.
For the physics of billiards, see here.
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