Please Help with Physics!

  • #1
Can anyone help with one or more of these problems? I need them to study with for an exam by Thursday evening. Thank you in advance!

1. A .30-Kg object with a velocity if 0.60 m/s in the +x direction makes a head on elastic collision with a 0..40 Kg object initially at rest. What is the final velocity of the 0.30-Kg object after collision?

2. A railroad car, of mass 220Kg, rolls with no friction on a horizontal track with a speed of 15 m/s. A 80 Kg stunt man drops straight down a distance of 6.0 m, and lands in the car. How fast will the car be moving after this happens?

3.A small bomb, of mass 12 Kg, is moving toward the North with a velocity of 5.0 m/s. It explodes into three segments: a 6.0 Kg fragment moving west with a speed of 9.0 m/s; a 4.0 Kg fragment moving East with a speed of 10 m/s; and a third fragment with a mass of 1.0 Kg. What is the velocity of the third fragment? (neglect friction)

4. A pendulum of length 60cm is pulled 35cm away from the vertical axis and released with a speed of 2.0 m/s. What will be its speed at the bottom of the swing
 

Answers and Replies

  • #2
berkeman
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Welcome to PF, skittles. You need to show your work so far on each of these problems in order to get help from us. What have you done so far on each of these problems? What equations are you going to use to work with these questions about acceleration, velocity and position?
 
  • #3
Well, i do know that there is an elastic head-on collision which means change in momentum and conservation of KE.
 
  • #4
Hootenanny
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skittlesj526 said:
Well, i do know that there is an elastic head-on collision which means change in momentum and conservation of KE.

Can you form an expression based on those comments. HINT: In an elastic collision both, kinetic energy and momentum are conserved.
 
  • #5
Well, i figured the first one out. I found the velocity of mass B prime and used it to find the velocity of mass A prime. I'm working on number 4 right now. I'm thinkin that total energy at point A is equal to total energy at point B but there is no mass or angle and I don't know how to determine the height for the PE equation unless I should be using a different equation, but which?
 
  • #6
Hootenanny
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You need to use trig to determine the change in height. As for the unknown mass, if you use the equations for potential and kinetic energy, you will find that the masses will cancel out.

Reagrds,
-Hoot
 
  • #7
yeah i see that the masses will cancel out....but there is one problem....i've never had trig and don't know where to begin with a trig formula...is there a name?
 
  • #8
Hootenanny
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You've never done trigonometry? Sine, cosine, tangent etc?
 
  • #9
yeah, SOH CAH TOA, right? but i don't even have an angle
 
  • #10
Hootenanny
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You can calculate an angle, in this case you have a triangle. At the maximum displacement the hypotenues is the string length, the base is the x displacement.

So you have;

[tex]\theta = \left( \frac{0.35}{0.6} \right) \cos^{-1}[/tex]

You can then use either sine or tan to find the opposite side, which will then allow you to calculate the change in height. You may find it useful to draw a diagram.

Regards,
-Hoot
 
  • #11
so now that i have the angle, i can set TEa=TEb. Does it start with PE and KE? so that....
TEa=PE+KE and TEb=KE ?
 
  • #12
Hootenanny
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Yes, it starts with both potential and kinetic energy. I'm not sure what your TE are though. You should get a change in height of about 0.487m.

Regards,
-Hoot
 
  • #13
yeah, i got 48.73 cm which converts about the same but now i have:
mgh+1/2 mv2=1/2 mv2

how will the masses cancel?
 
  • #14
Hootenanny
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[tex]mgh + \frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2}[/tex]

Simply divide throughout by [itex]m[/itex]

Regards,
-Hoot
 
  • #15
i got an ending velocity of 15.42 m/s
 
  • #16
Hootenanny
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skittlesj526 said:
i got an ending velocity of 15.42 m/s

I don't think thats correct. Can you show your working?

Regards,
-Hoot
 
  • #17
just seemd kinda large, doesn't it?
 
  • #19
Hootenanny
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Still wrong I'm afraid, can you show your working please.
 
  • #20
should i convert that to 0.1542m/s?
9m
(m)(-9.8)(.4873)+1/2(m)(v)2=1/2(m)(v)2
 
  • #21
Hootenanny
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Here's my working

[tex]mgh + \frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2}[/tex]

[tex] 2gh + v_{i}^{2} = v_{f}^{2}[/tex]

Can you take it from here? Note: you need to be in meters.

Regards,
-Hoot
 
  • #22
i now got l2.35l m/s (abs value)
 
  • #23
Hootenanny
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Well, I got 3.68 m/s (absolute values are not required as we are discussing speed). I think your using -9.81? You should just use 9.81 as we are considering a change in potential energy not absolute potential energy.

Regards,
-Hoot
 
  • #24
how did you get 3.68m/s?
 

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