When a Stone of mass 20 kg falls on the floor from a height of 1.2 meters how much momentum will it transfer to the floor
Answers
Explanation:
There are many many ways to do this sort of problem. I’ll assume you don’t know calculus, and go through things right from the basics.
Momentum is given by p=mvp=mv. You know mass=m=20kgmass=m=20kg, but you don’t know velocity=vvelocity=v yet. So you need to calculate the velocity of the mass at the time it hits the floor and stops, transferring its momentum into the floor.
It is important to note here that we are using the conservation of momentum. If the mass stops when it hits the floor, then it will have no momentum (as p=mv=20×0=0p=mv=20×0=0). As the conservation of momentum says the total amount must stay constant, the amount of momentum transferred to the floor must equal the momentum of the mass as it impacts with the floor.
So how do we calculate v?
v=v0+atv=v0+at
The velocity of an object is equal to its starting velocity (at time=t=0time=t=0), plus the acceleration of the object multiplied by the time over which the experiment is occurring.
This is because the acceleration is the “change in velocity over time”; so acceleration×timeacceleration×time is equal to the change in velocity (in that time).
This equation works for uniform, or constant, acceleration. This is (approximately) the case for gravitational acceleration at the surface of the Earth, so we’re good to go.
In this case, I’m assuming v0v0 is 0 - in other words, I’m assuming the object started from rest, from the way the question is phrased.
We know the acceleration due to Earth’s gravitational pull; it’s 9.81 metres per second squared. However, we don’t know the time over which this is taking place. Now we need to figure out the time it takes for the object to fall 1.2 metres, starting from rest, while being accelerated at 9.81ms−2−2.
To keep this answer a reasonable length, I’m just going to link to a proof of the below equation elsewhere on Quora: derivation.
The equation to use is s=v0t+12at2s=v0t+12at2, where “s” is the displacement, in this case 1.2 metres. As v0=0v0=0, we can rearrange the above as t=2sa−−√t=2sa.
You now have all the equations you need to solve the problem. Using s = 1.2 metres and a = 9.81ms−2−2, you can find the time taken for the mass to fall from the above equation. You can then multiply this value by the acceleration again to get the velocity, v, at the time of impact, via v=v0+atv=v0+at. You can then multiply this by m = 20kg to get the momentum of the mass as it hits the floor, via p=mvp=mv. Then, by using the concept of conservation of momentum, you can say this is equivalent to the momentum transferred into the floor
Answer:
There are many many ways to do this sort of problem. I’ll assume you don’t know calculus, and go through things right from the basics.
Momentum is given by p=mv . You know mass=m=20kg , but you don’t know velocity=v yet. So you need to calculate the velocity of the mass at the time it hits the floor and stops, transferring its momentum into the floor.
It is important to note here that we are using the conservation of momentum. If the mass stops when it hits the floor, then it will have no momentum (as p=mv=20×0=0 ). As the conservation of momentum says the total amount must stay constant, the amount of momentum transferred to the floor must equal the momentum of the mass as it impacts with the floor.
So how do we calculate v?
v=v0+at
The velocity of an object is equal to its starting velocity (at time=t=0 ), plus the acceleration of the object multiplied by the time over which the experiment is occurring.
This is because the acceleration is the “change in velocity over time”; so acceleration×time is equal to the change in velocity (in that time).
This equation works for uniform, or constant, acceleration. This is (approximately) the case for gravitational acceleration at the surface of the Earth, so we’re good to go.
In this case, I’m assuming v0 is 0 - in other words, I’m assuming the object started from rest, from the way the question is phrased.
We know the acceleration due to Earth’s gravitational pull; it’s 9.81 metres per second squared. However, we don’t know the time over which this is taking place. Now we need to figure out the time it takes for the object to fall 1.2 metres, starting from rest, while being accelerated at 9.81ms −2 .
To keep this answer a reasonable length, I’m just going to link to a proof of the below equation elsewhere on Quora: derivation.
The equation to use is