Understanding Conservation of Momentum Through Examples
Momentum, defined as mass times velocity, is conserved in isolated systems. This video demonstrates how to apply the law of conservation of momentum to solve practical physics problems involving collisions, explosions, and momentum transfer. For a solid foundation on forces at play, reviewing Newton's Laws of Motion Explained with Real-Life Examples can be very helpful.
Example 1: Astronaut Throwing a Ball in Space
- Scenario: A 70 kg astronaut throws a 5 kg ball east at 20 m/s.
- Goal: Find the astronaut's velocity immediately after throwing.
- Key points:
- Initial momentum is zero since both are at rest.
- After throwing, total momentum remains zero (momentum before = momentum after).
- Using the equation: 0 = (mass_astronaut * velocity_astronaut) + (mass_ball * velocity_ball)
- Calculation:
- Using known values, astronaut's velocity = -1.43 m/s (westward).
- The negative sign indicates direction opposite to the ball's motion.
- Insight: A heavy mass moves slower than a lighter one when momentum is conserved.
Example 2: Exploding Mass Breaking into Two Fragments
- Scenario: A 50 kg mass at rest explodes into 30 kg and 20 kg fragments.
- The 30 kg fragment moves west at 40 m/s.
- Goal: Determine the velocity of the 20 kg fragment (eastward).
- Key points:
- Total initial momentum is zero (mass at rest).
- Conservation of momentum gives: -(mass_fragment_A * velocity_A) = mass_fragment_B * velocity_B
- Calculation:
- Velocity of 20 kg fragment = 60 m/s east.
- Insight: The fragment with lesser mass moves faster to conserve overall momentum.
Example 3: Railroad Cart with a Dropped Rock
- Scenario: A 200 kg railroad cart moves east at 15 m/s; a 50 kg rock is dropped vertically into it.
- Goal: Find the final combined speed of the cart and rock.
- Key points:
- Rock initially has zero velocity in x-direction; no initial momentum in x.
- Momentum is conserved in the horizontal direction.
- The rock accelerates to the cart's new speed, causing the cart to slow.
- Calculation:
- Initial momentum = 200 kg * 15 m/s = 3000 kg·m/s
- Final mass = 200 + 50 = 250 kg
- Final velocity = 3000 / 250 = 12 m/s
- Insight: Momentum transfer from cart to rock slows the cart and accelerates the rock.
Key Takeaways
- Momentum is conserved in all isolated systems.
- Heavy objects move slower than light objects when exchanging momentum.
- Direction is crucial: velocities in opposite directions have opposite signs.
- Forces during interactions transfer momentum between objects.
- Using Newton's third law helps understand action-reaction forces involved.
For further context on energy considerations in motion, you might find Conservation of Energy: Calculating Speeds and Heights in Physics Problems insightful.
By mastering these examples, you can confidently solve various conservation of momentum problems relevant to physics and engineering.
in this video we're going to talk about how to solve conservation of momentum problems
so let's start with this one a 70 kilogram astronaut throws a 5 kilogram ball
with a velocity of 20 meters per second east what is the velocity of the astronaut
so let's draw a picture let's say this is the astronaut and he throws a ball
in this direction now based on newton's third law that means the astronaut is going to
fill a force that's going to propel him in the opposite direction so we know the direction of his velocity is going to be
west how can we calculate that velocity well based on the law of conservation of
momentum the momentum before he throws the ball
is going to equal to the total momentum after he throws it so p initial is equal to p final
so before the event before he throws the ball we have
the momentum of the person or the astronaut plus the momentum of the ball
and then after he throws it we're gonna have a new momentum of the astronaut and the ball
so m a v a plus m b v b
is equal to m a v a final plus m b
v b final now before he throws the ball the astronaut and the ball are at rest
they're not moving so the initial momentum on the left side is zero
now after he throws it the astronaut and the ball will both have momentum
so therefore we have this equation zero is equal to m a v a plus
m b v b so i'm going to take this term move it to this side
so negative m a v a prime which stands for final velocity
is equal to m b v b prime so let's plug in what we know
the mass of the astronaut is 70 kilograms and we're looking for the velocity of
the astronaut after he throws the ball the mass of the ball is 5 kilograms and the velocity of the ball is positive
20. it's positive because the ball is moving towards the right in
the positive x direction so this is going to be 5 times 20 which
is 100 so we got negative 70 v a prime is equal to 100 and now we get to divide
both sides by negative 70. so the final velocity the astronaut is negative 1.43
meters per second now the reason why the answer is negative is because
he's moving in the negative x direction now notice that the velocity of the ball is significantly greater than the
velocity of the astronaut because the astronaut has more mass
he's going to move backward with a lower speed the ball has less mass than the
astronaut so it's going to move to the right with a greater velocity so as the mass increases
the velocity decreases assuming if the momentum is constant if it's conserved
if you decrease the mass the velocity increases so whenever momentum is constant
objects with a very large mass tend to move slower and objects with a low mass
objects that tend to be light are the ones that move faster now here's another similar question
a 50 kilogram mass initially at rest explodes into two fragments so let's say
this is the 50 kilogram object and then after some time period it explodes into two parts
so let's say this is the first fragment and
this is the second fragment now the 30 kilogram fragment that's going to move west
at 40 meters per second and our goal is to find the velocity of the 20 kilogram fragment
now we know it has to move east so the velocity has to be positive so once again we have the total initial
momentum is equal to the total final momentum before the event this object was at rest
so the total initial momentum is zero and the final momentum is going to be the momentum
of the first fragment plus the momentum of the segment fragment after the collision
so let's call this fragment a and fragment b so just like before we're going to have
this equation negative m a v a prime is equal to mb
vb prime so fragment a has a mass of 30 kilograms and the velocity of negative 40.
now you need to put the sign of the velocity because this is moving to the west so it has a negative value
the mass of the second fragment fragment b is 20 and we need to calculate vb so negative
30 times negative 40 is positive 1200 and that's equal to 20
times vb prime so 1200 divided by 20 is
60. and we have a positive answer to indicate that it's going east in the positive x direction
so that's the final velocity of fragment b now fragment b
has less mass than fragment a and therefore fragment b has a greater speed
fragment a has more mass and so the speed is less it's 40 meters per second as opposed to 60 meters per
second now let's move on to our next problem a 200 kilogram empty railroad cart
moves east at 15 meters per second a 50 kilogram rock is dropped straight down into the moving cart
what is the final speed of the railroad cart so let's say this is the cart
it has a mass of 200 kilograms and it's currently moving at a speed of 15 meters per second
now what's going to be the new speed if we take a rock
and drop it into it now the momentum of this object in the x direction is fixed
and this rock is moving in a negative y direction so it's not going to affect
the momentum of this cart in the x direction so the momentum is going to be conserved
now if the momentum is constant and if the mass increases by in this case adding a rock
we should expect that the speed should decrease so our answer should be less than 15.
now the momentum that we had before p initial has to equal the momentum after p final
so let's calculate the initial momentum it's going to be mass times velocity so that's 200
times 15. so that's 3 000. now that should equal the final momentum
so that's m v
now the new mass is no longer 200 it's the 200 kilograms of the rebel cart plus
50 kilograms of the rock so the total mass of this object now this system
is going to be 250 kilograms and now we gotta calculate the new speed so it's 3 000 divided by 250
and so the rebel cart with the rock is going to move at a combined speed of 12 meters per second
so now let's talk about what's happening here so what happens
as soon as the rock falls into the cart so let's say
this is the rock now the speed of the rock in the x direction
initially was zero now it's 12. so the rock accelerated in the x
direction so once you drop the rock the rebel card is moving and so
this portion of the rebel cart will exert a horizontal force
on the rock causing the rock to accelerate from 0 to 12
in the x direction now based on newton's second law for every action force there is an equal and
opposite reaction force so when a rebel car exerts a force on the rock
the rock exerts a force on a rebel card and so that force slows down a verbal
card from 15 meters per second to 12 meters per second
so that makes sense the rebel cart accelerates the rock and the rock decelerates the rebel cart
but what's really happening here is that these forces are acting in such a way
to transfer momentum from the railroad cart to the rock
if we calculate the new momentum of the rebel card the original momentum is 3000 but the new momentum is going to be the
mass of 200 times the final speed of 12. so the rebel cart has
a momentum of 2400 and the rock which has a mass of 50 and a speed of 12 is 50 times 12
which is 600 so the total momentum is still the same so basically
the railroad cart lost 600 units of momentum and that 600 units was transferred to
the rock so these forces they acted in such a way to take some of the momentum from the
railroad cart and transfer it to the rock until the speeds of the rock and the cart
were the same until they reached equilibrium so to speak
so hopefully this gives you a better understanding of how forces
can transfer momentum from one object to another you
The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In physics problems, this means the sum of the momentum before an event (like a collision or explosion) equals the sum after the event. To apply this, identify all masses, velocities before and after, and set total initial momentum equal to total final momentum to solve for unknowns.
Since the astronaut and object start at rest, the total initial momentum is zero. When the astronaut throws a ball in one direction, to conserve momentum, the astronaut moves in the opposite direction. Use the equation: 0 = (mass_astronaut × velocity_astronaut) + (mass_ball × velocity_ball). Rearranging gives the astronaut’s velocity = -(mass_ball × velocity_ball) / mass_astronaut. This calculation shows that heavier masses move slower in the opposite direction to lighter, faster-moving objects.
If the original mass is at rest, its initial momentum is zero. Upon explosion, fragments move in opposite directions to conserve momentum. You can use the formula: (mass_fragment1 × velocity_fragment1) + (mass_fragment2 × velocity_fragment2) = 0. Knowing one fragment’s mass and velocity allows you to solve for the other fragment’s velocity by rearranging the equation, ensuring opposite directions are accounted for by sign conventions.
The rock initially has no horizontal momentum, while the cart is moving. When the rock drops into the cart, momentum is conserved horizontally, causing the combined mass to move at a new velocity. Calculate by summing initial momentum of the cart alone, then divide by the total combined mass (cart + rock) to find the final velocity. The cart slows down, transferring momentum to accelerate the rock horizontally.
Momentum is the product of mass and velocity. To conserve momentum in an interaction, if a lighter object moves quickly, the heavier object must move slower in the opposite direction so that the total momentum remains constant. This inverse relationship means mass and velocity adjust reciprocally to maintain conservation laws, explaining why heavier masses have lower velocities in such scenarios.
In momentum calculations, direction is crucial and typically represented by positive and negative signs. Velocities in opposite directions have opposite signs (e.g., east is positive, west is negative). Properly assigning these signs ensures accurate calculation of momentum sums, which are vector quantities. Ignoring direction can lead to incorrect answers, so maintaining consistent sign conventions is essential.
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