What Determines the Velocity in a Perfectly Inelastic Collision?

AI Thread Summary
In a perfectly inelastic collision involving two identical cars, each traveling at 12 m/s from perpendicular directions, the final velocity of the combined object is calculated using momentum conservation principles. The momentum vectors are combined, resulting in a final speed of approximately 8.4 m/s, derived from the equation √(6² + 6²). The confusion arises from the algebraic manipulation of the momentum formula, where the initial speeds are effectively halved due to the equal mass and direction components. Understanding the vector addition of momentum is crucial in solving such problems. Correctly applying these principles clarifies the calculations involved in determining the final velocity.
JinSu
Messages
8
Reaction score
0
This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.
 
Physics news on Phys.org
You have this

m \vec{v}_{1} + m \vec{v}_{2} = (m + m) \vec{v}_{both}

m \vec{v}_{1} + m \vec{v}_{2} = 2m \vec{v}_{both}

\frac{\vec{v}_{1}}{2} + \frac{\vec{v}_{2}}{2} = \vec{v}_{both}

then you have

\frac{-12 \vec{i}}{2} + \frac{12 \vec{j}}{2} = \vec{v}_{both}

Thus

\sqrt{6^2 + 6^2} = | \vec{v}_{both} |
 
Last edited:
JinSu said:
This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.
The momentum of the two cars is 12 m/s North and 12 m/s West. So the momentum is ((12m)^2 + (12m)^2)^1/2 = 2mv which can be written as ((6^2 x (2m)^2 + 6^2 x (2m)^2)^1/2 = 2mv

AM
 
JinSu said:
This is probably easy for most people, but I can't quiet seem to understand it.

Two automobiles have a perfectly inelastic collision. The automobiles, which are identical models except for the color and which contain only drivers of identical masses, meet at an intersection. Each was moving with speed 12m/s, one coming from the south, the other from the east. What is the velocity of the final composite object?

It was an old homework problem, just studying for my final exam and I am trying to understand it. The answer set from my prefessor says that the final speed is (6^2+6^2)^(1/2) = 8.4 m/s. I am not sure why its half the speed of the original... I know the formula is mv + Mv = (M+m)V. Any tips on why/how this problem is solved would be appreciated. Thanks.

You can use the properties of momentum to calculate this, so you can use vectors for the individual components. So just look up :biggrin:
 
Ah thanks, I see what I did wrong. Gosh, such a silly algebra mistake.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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...
Back
Top