Calculation Force?

Question

This feels like something I should know, but am just too stupid to remember. Can someone here help me out?

I am not looking for Newton's second law of f=ma. Instead what I want is if given one objects velocity and mass, what is the force that it applies to another object that it collides with (eg a rock with a known mass and final velocity impacts the ground, what is the force it hit's the Earth with?).

I'm still confused, so I'll provide an example you can solve for me.

A .5 kg rock free falls and lands on my head. It collides with my head with a velocity of 5m/s. What is the force that the rock applies to my head when it collides with it?

If I simply do p=vm like you suggested, I end up with 2.5kg *m/s, but that's not F in Newtons. I would need units with kg *m/s^2 in order for that to be the force.

So what am I doing wrong, or still need to do?

Answer 1

Here you are talking about Collision of two bodies. Now when this occurs in a very small interval of time and during which the two bodies exert large force on each other is called an impact.

The phenomenon of impact consists of two phases:

The Period of Deformation: Just after the impact, the two bodies deform. The time interval from the first contact to the maximum deformation is called the period of deformation. At the end of this period both bodies move with the same velocity v.

The period of Restitution: At the end of period of deformation the two bodies either regain their original shapes fully or partially, or remain permanently deformed. Also at the end of restitution period the two bodies separate and travel with different velocities except in the case of plastic impact.

Now let us consider body A with

mass= Ma
velocity= Va

collides with body B.

Body B exerts an impulsive force during deformation period is Fd, and with this, body A changes its velocity to V.

Applying principle of impulse and momentum,

MaVa - [Fd dt = MaV

Where the integral extends over the period of deformation.

Now let body B exerts an impulsive force Fr and changing the velocity of body A again from V to Va'.
For this,

MaV-[ Fr dt = MaVa'
Where the integral extends over the period of restitution.

Fd and Fr both have different values.

So this is regardind finding force as you ask.

But usually in such cases we determine either coefficient of restitution or the velocity with which the body rebounded after the impact.

coefficient of restitution = [Fr dt/ [Fd dt

e=(V-Va')/(Va-V) for body A

e=(V-Vb')/(Vb-V) for body B

When both the bodies are moving.

So combining both the equations from definition of e, and law of conservation of momentum,
we get,
e= - (Vb'-Va')/(Vb-Va)

This is the parameter which indicates energy loss during an impact and can be determined experimentally. It is 1 for perfectly elastic impact and 0 for perfectly plastic impact.

In the case you mentioned:
Rock is A,
Floor is B,

Vb= velocity before impact=0
Vb'=velocity after impact=0
Impact type is elastic,
e=1, we get from the formula,
Va'= -Va
The rock would rebound with the same velocity with which it strikes the immovable body.

For finding force we must know about period of deformation and period of restitution or coeefficient of restitution and both the velocities.

I would suggest you to refer Mechanics for Engineers by Beer and Johnston, and also by Timoshenko and Young, or Irving Shames on same area of Applied Mechanics.

Too long an answer? But this topic can not be explained by single formula. :)

Write the specific problem with data, I would be pleased to solve it for you.

Good luck.

DETAILS ADDED:-

Now in answer to your details added:Since this is plastic collision,it can be solved by Newton's law only.

here, F=m.g = 0.5*9.8 = 4.9 Kg.m/sec^2,
as when body is falling freely, its force depends upon mass only. If a pillow falls on your head, it will strike with same velocity as the rock strikes, as it depends on gravitational acceleration only. So the FORCE depends on mass only and not with the striking velocity.

Answer 2

If a rock moving at a fixed velocity (where KE = 1/2 mv2) strikes an immovable (stone?) wall, all of its energy is released instantaneously and deceleration takes an instant. If the stone strikes a softer (or smaller) object (a basketball?), the stone partially penetrates, deforms or moves the target and slows down (decelerates) over a greater period of time releasing its energy during the interval. Therefore the fixed amount of kinetic energy is released more slowly with less impact. When race cars hit the wall they crumple absorbing energy and allowing the race car driver to decelerate less suddenly. Because F = ma, force will be greater as acceleration (or deceleration) increases. Acceleration is a change in velocity with time. Compare running into a tree vs running into a hedge, etc.

Answer 3

Of course this is pretty easy, but I won't be looking down on you. Because you are not looking forward to Newton's second Law of Motion. Instead, you should be looking to the principle of momentum. If you know the mass and velocity, you should have an equation like this:

p=mv

p: Momentum
m: mass
v: velocity

Once you got that, that's your answer!!!

Answer 4

To answer this question, you still need Newton's second law of motion: F = ma

But a = (Vf - Vi)/t where Vf is final velocity and Vi is initial velocity

Substituting that into the second law of motion, we get:

F = m (Vf - Vi)/t
F = (mVf - mVi)/t
F=(pf - pi)/t where pf = mvf and pi = mvi

pf = final momentum, pi = linear momentum

Answer 5

Mass m = 6 kg Displacement = top h = 5 m Weight = mg = 6 x 9.8 = fifty 8.8 N Weight and displacement are interior the comparable course (downward). artwork by ability of weight = weight x displacement = fifty 8.8 x 5 J = 294 J Ans: 294 J _____________________.

Answer 6

impulse and momentum