Understanding Velocity Ratios: Derivation and Confusion Explained

  • Thread starter Thread starter Manasan3010
  • Start date Start date
  • Tags Tags
    Confused Ratio
AI Thread Summary
The discussion clarifies the derivation of velocity ratios, concluding that if v1 equals 2v2, then the correct ratio is 2:1. This indicates that the speed of the 1 kg mass is twice that of the 4 kg mass. There is confusion regarding whether to interpret the ratio as 1:2 based on coefficients or 2:1 based on velocity. To avoid ambiguity, it is emphasized that stating the mass speeds directly clarifies the relationship. Overall, the key takeaway is the importance of clear communication in expressing velocity ratios.
Manasan3010
Messages
38
Reaction score
3
Homework Statement
Two particles of mass 1kg and 4kg have same kinetic energy at a time. What is the ratio of their velocities at this time?
Relevant Equations
$$\frac{1}{2}mv^2$$
I was able to derive,
##1/2*1*v_1^2 = 1/2*4*v_2^2 \\
v_1=2v_2
##
But I am confused whether the ratio is 1:2(Looking at the coefficient) or 2:1
 
Physics news on Phys.org
V1 is twice V2, so the ratio is 2:1
if you want to be unambiguous, say: the 1 kg mass speed is twice as high as that of the 4kg mass
 
  • Like
Likes Manasan3010
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanged mass'

Thread 'A cylinder connected to a hanged mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Collision between two uniformly moving particles
Replies
7
Views
2K
Using conservation of energy and momentum in projectile motion
Replies
18
Views
1K
A rod that falls while rotating from the end of a table
Replies
11
Views
2K
What minimum speed do we need to give a charged ball?
Replies
12
Views
1K
Method of mirroring - metal plates
Replies
11
Views
2K
Find velocities when a mass leaves a system of a rod with two masses
Replies
10
Views
2K
Shortest distance along the shore and into the lake
Replies
6
Views
894
Closed cycle of an ideal gas
Replies
3
Views
1K
The speed of the point of intersection of two uniformly moving lines
Replies
24
Views
1K
The efficiency of a heat engine
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top