Block sliding down an incline and hitting a smaller block

  • Thread starter 12boone
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Homework Statement


In a physics lab, a cube slides down a frictionless incline as shown in the figure (Intro 1 figure) , and elastically strikes another cube at the bottom that is only 1/6 its mass. The picture is located in the attachement. If the incline is 30 cm high and the table is 90 cm off the floor, where does each cube land? [Hint: Both leave the incline moving horizontally.]


Homework Equations


kinematics and MVi+MbVi=MVf+MbVf


The Attempt at a Solution



I have done several things. I tried to find the final velocity using kinematics and then plug it in to find the other final velocity. I tried using the 1/6 to cancel all masses so that i get vi=vf+1/6vf. I am getting wrong answers everytime though.
 

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Answers and Replies

  • #2
Doc Al
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Show exactly what you did. Your first goal should be to find the speeds at which they leave the table.

Hint: You'll need both energy and momentum conservation.
 
  • #3
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i used mgh=1/2mv^2 to get the speed of the big block which is 2.42 m/s I plugged it in like this 9.8(.30m)=1/2v^2 then used algebra to solve. then i did the same thing for the final velocity of the lil block so i did 9.8(.90)=1/2v^2 and i found it was 4.2 m/s. then i found the va' by doing 2.42=Va'+1/6(4.2) and that is 1.72.
 
  • #4
Doc Al
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45,089
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i used mgh=1/2mv^2 to get the speed of the big block which is 2.42 m/s I plugged it in like this 9.8(.30m)=1/2v^2 then used algebra to solve.
This sounds good.
then i did the same thing for the final velocity of the lil block so i did 9.8(.90)=1/2v^2 and i found it was 4.2 m/s.
I don't understand what you did here. It looks like you found the vertical component of the block speed when it hits the floor. You'll need this later, but first you need to find the horizontal speed with which both blocks leave the table.

Once you have the speed of the big block just before the collision, apply conservation of momentum and energy to find the block speeds just after the collision.
 
  • #5
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if i use that i get 2.42=Va'+1/6Va'b i have two variables
 
  • #6
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i mean 2.42=Va'+1/6Vb'
 
  • #7
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I have no idea how to do this part. i keep getting two variables no matter what i try to manipulate. what is the eq for con of energy. 1/2 mv^2 +mgh=1/2mv^2+mgh?
 
  • #8
Doc Al
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45,089
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i mean 2.42=Va'+1/6Vb'
OK. That's conservation of momentum.
i keep getting two variables no matter what i try to manipulate. what is the eq for con of energy. 1/2 mv^2 +mgh=1/2mv^2+mgh?
The potential energy doesn't change during the collision, so just compare the initial KE (of the big block) to the final KE (of both blocks). That's how you'll get your conservation of energy equation, which is the second equation that you need.
 

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