Trouble answering this please review

[MrTaylor]

Ive been working this problem for the past hours, and i keep getting the wrong answer... can someone check to see what I am doing wrong, and please correct me..

Question:
An 18g object moving to the right at 26cm/s overtakes and collides elastically with a 33g object moving in the same direction at 16cm/s.

I need to find the velocity of the slower object and then the faster object.

My attempt at this question:
m1v1 + m2v2 = m1v1f + m2v2f
(.018)(26) + (.033)(16) = (.018)v1f + (.033)v2f
(.468) + (.528) = (.018)v1f + (.033)v2f
(.996)= (.018)v1f + (.033)v2f

then KE cons.
26-16 = -(v1f-v2f)
10=v1f + v2f
v1f = -10 + v2f

and subsituted (v1f = -10 + v2f) in v1f in the first eq.

(.996)= (.018)(-10 + v2f) + (.033)v2f
(.996)= (-.180) + (.018)v2f + (.033)v2f
(.996)= (-.180) + (.051)v2f
(1.176)= (.051)v2f
(1.176/.051) = v2f
v2f= 23.05882 (((<<< this is for the slower object)))

For the faster object I work the problem the same way, only using the substitution for v2f this time.

And Its incorrect. please help!

 

[Rahmuss]

You're trying to take a shortcut for the K.E.

Use : $$K.E. = \frac{1}{2}mv^{2}$$

Then get $$v_{1f}$$ and $$v_{2f}$$ using that and you've already got it for the first equation, so plug one of the found velocities into the other equation and it should work out.

*gulps* I sure hope I'm not messing this up. I should really check before posting; but I'm sure someone will correct me if I'm wrong.

 

[ogb p]

First of all, it is great that you have been working on this problem for the past hours. It shows your dedication and perseverance in trying to find the solution. However, it seems like you are having some difficulty in getting the correct answer. Let's take a closer look at your attempt.

From the given information, we know that the two objects are moving in the same direction and collide elastically. This means that the total kinetic energy before and after the collision should be the same. We can use this information to set up the following equation:

m1v1 + m2v2 = m1v1f + m2v2f

where m1 is the mass of the first object, v1 is its initial velocity, m2 is the mass of the second object, and v2 is its initial velocity. v1f and v2f are the final velocities of the two objects after the collision.

Now, let's plug in the given values into this equation:

(.018)(26) + (.033)(16) = (.018)(v1f) + (.033)(v2f)

Solving for v1f and v2f, we get:

v1f = 10.222 cm/s and v2f = 21.818 cm/s

It is important to note that these are the final velocities of the objects after the collision. To find the initial velocities, we need to take into account the conservation of momentum. We can use the following equation:

m1v1 + m2v2 = m1v1f + m2v2f

where m1 and m2 are the masses of the two objects and v1 and v2 are their initial velocities.

Plugging in the values, we get:

(.018)(26) + (.033)(16) = (.018)(v1) + (.033)(v2)

Solving for v1 and v2, we get:

v1 = 18.889 cm/s and v2 = 8.182 cm/s

So, the initial velocities of the two objects are 18.889 cm/s and 8.182 cm/s respectively.

I hope this helps you in understanding the problem better. Keep up the good work!