Let's imagine an ideal scenario where you're lifting your own weight in its entirety. Let's say a woman weighing 100 lbs. Suppose she's doing an idealized handstand and pushup from that position. So she's lifting 100 lbs. Let's say ideally all of the forces are on her arms only. Do these forces...
Here we are talking about non-relativistic quantum physics. So we all know kinetic energy T = E - V = \frac{1}{2}mv^2 in classical physics. Here V is the potential energy of the particle and E is the total energy. Now what I am seeing is that this exact same relation is being used in quantum...
Maybe a silly question but on the above question using the conservation of momentum:
momentum before firing (0) = momentum after firing (55*35)+(M*2.5)
If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)?
Thanks
Textbook solution:
##v## is the instantaneous velocity,
$$P(t)=(M+b t) v$$
Then $$impulse = \Delta P = (M+b t) v = \int^{t}_{0} F dt'$$
Thus $$v=\frac{F t}{(M + bt)}$$
What I did instead was:
Let ##M## be the instantaneous mass, and ##M_0## be the initial mass, then $$M=M_{0} + b t$$...
For the sake of this question, I am primarily concerned with the position wave function. So, from my understanding, the wave function seems to 'collapse' to a few states apon measurement. We know this because, if the same particle is measured again shortly after this, it will generally remain in...
I have trouble solving this problem any help would be appreciated.
Problem statement
##J=\frac{mr^2}{2}##
a) Determine the motion of yoyos for ##n=1,2,3##
The case for ##n=1## is simple, however, I am having trouble with ##n=2## and ##n=3##.
for ##n=2## I started by drawing all the forces...
Why, in lagrangian mechanics, do we calculate: ##\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}## to get the (generalised) momentum change in time
instead of ##\frac{d T}{dq}##?
(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external...
I honeslty don't quite know how to start. It seems like the Hooke's coefficent k is independent of the answer to this problem.
I would also appreciate any clue of expressing the condition when "balls will collide again". The fact that all balls can keep moving make this rather difficult.
It...
A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity of the particle toward the north-west relative to the observer?
I ask because while the particles total...
So I've managed to confuse myself on this problem :)
Since the problem says we can assume ##m_p << m_b##, I'm assuming that the velocity of the bowling ball will be unchanged, such that ##\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i##
I started out using the energy-momentum principle, ##(\vec...
So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated,
Oscar.
I am wondering why heavier bullets have a higher momentum than lighter bullets when using similar powder charges?
At the muzzle, a typical 150 grain bullet fired from a 30-'06 will travel at around 3000 ft/s.
A 200 grain bullet from the same rifle will travel around 2600 ft/s.
(The velocities...
1. When an object attached to a fixed point with a string, is given a velocity and the string goes taut.
So it says in this book (Applied Mathematics 1 by L. Bostock and S. Chandler) that when the string goes taut, the component of the velocity of the particle becomes zero in the direction...
When A hits B,
COLM
mV = -mVa + 2mVb
V = 2Vb - Va
COKE
0.5mv^2 = 0.5mVa^2 + 0.5(2m)Vb^2
V^2 = Va^2 + 2Vb^2
When B hits C
COLM
2mVb=4mVc
Vc = 0.5Vb
COE
0.5(2m)Vb^2 = 0.5kx^2 +0.5(4m)Vc^2
sub Vc = 0.5b
mVb^2 = KX^2
After that im stuck, cause i cant find V in terms of Vb only
So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity
at B ,
$$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$
or,
$$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$
Now,
Suppose I take the limit as mass of ramp goes to...
Let's say you have two masses on either side of a spring. Mass 1 is connected to the end of a spring. The spring itself has no mass. Mass 2 is free in space. So you have:
[M1]-[spring] [M2]
So it's more descriptive, I'll name the variables like you might in programming. Let's define...
I first got the velocity of the combined mass with conservation of momentum and as it was in the mean position the velocity can be written as v = wA ( w= angular frequency , A = amplitude ) as we have to take it back to natural lenght i put A as the initial extension but i am getting a wrong ans...
I have done question 1. But I'm struggling with the other one. So since the only thing I know about the rocket is the mass and the velocity, I guess I have to use momentum to solve this problem. From the first question, I found out that the x-velocity of the projectile is ##v_x=5...
Here is my calculation:
F = ma
50N = 1050kg * a
a = 0.0476m/s²
S = ut + ½at ²
1000m = 0t + ½(0.0476)t²
t = 204.980s
y = 204.980s (time to travel 1000m)
since impulse = momentum,
F * t = mv
F * x = m * distance covered/y
50N * x = 1050kg * 1000m/204.980s
50N * x = 5122.450N⋅s
x = 102.440s...
So far I found the answer for a and b, but when I attempted to do the other ones I was completely lost.
A.) P= MV
M = 25g = .025kg
V = 18
.025 * 18 = .45kg*m/s
B.) KE= 1/2 mv^2
1/2 (.025)(18)^2
4.05 J
Change in KE = Change in thermal energy
0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1
vblock = 0.885
By Conservation of Momentum,
(0.05)(854) = (0.05)*vbu + (6)(0.885)
I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the...
why can't we know where electron goes after it was hit by light? Light has a travel direction, can't we assume that electron bounces to the same direction that the light was headed??
In A.P. French's Special Relativity, the author said the following,
As I understand, photons are massless, so I don't think the last equation above applies to photons, but then, when deriving it, he used an equation proper to photons (##E=pc##).
So in which context is ##m=p/c## valid?
Since Pi = Pf,
0 = MbVbg + McVcg
I just need to express Vbg in terms of Vbc and Vcg (that is, I need to express the velocity of the ball relative to the ground in terms that I know/want to solve for):
by reference frames:
Vbc = Vbg + Vcg
so Vbg = Vbc -Vcg
Now I can sub in and solve
0 =...
I really want to know which answer is correct. I don’t really know if I should include velocities to the left as negative velocities in the equation. Is it -1 or 4.33? Please help! Thanks!!!
Hints given:
-Start with free body diagram. Use the relationship between impulse and momentum to find the final velocity of the car after he has pushed for time t.
-Use a kinematic equation to relate the final velocity and time to the distance traveled.
-What is his initial velocity?
My...