Block sliding down an incline then collides with another block question.

  • Thread starter lillybeans
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Homework Statement



In the diagram below, a block of mass M=17 kg slides down a frictionless inclined plane and collides elastically with another block of mass m=4 kg. If the mass at the top of the wedge is initially at a height of 6 m above the horizontal plane, what is the velocity of m after the collision?

[URL]http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/LernerCh10/graphics/lerner10_13.gif[/URL]

Homework Equations



conservation of energy formula, conservation of momentum formula. shown below in my attempt at the solution.

The Attempt at a Solution



I used the momentum formula to find V2f

V2f=(2m1/(m1+m2))V1i+(m2-m1)/(m1+m2)V2i

Since V2i of the smaller mass is zero, i'm only left with the first part, which is.

V2f=(2m1/(m1+m2)V1i

substituting the masses in I get V2f=34/21 V1i

To solve for V1i, I used mgh=1/2mv^2. (m=17kg, g=9.8, h=6m) and I got V1i=10.84m/s.

Then substituting the value i got back into the previous equation, V2f=34/21 * 10.84 = 17.56 m/s. This answer is obviously shown by the computer to be NOT correct. I don't understand where I went wrong, because I even used the conservation of kinetic energy to check that before and after the kinetic energy is the same! Please help!
 
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Answers and Replies

  • #2
68
1

Homework Statement



In the diagram below, a block of mass M=17 kg slides down a frictionless inclined plane and collides elastically with another block of mass m=4 kg. If the mass at the top of the wedge is initially at a height of 6 m above the horizontal plane, what is the velocity of m after the collision?

[URL]http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/LernerCh10/graphics/lerner10_13.gif[/URL]

Homework Equations



conservation of energy formula, conservation of momentum formula. shown below in my attempt at the solution.

The Attempt at a Solution



I used the momentum formula to find V2f

V2f=(2m1/(m1+m2))V1i+(m2-m1)/(m1+m2)V2i

Since V2i of the smaller mass is zero, i'm only left with the first part, which is.

V2f=(2m1/(m1+m2)V1i

substituting the masses in I get V2f=34/21 V1i

To solve for V1i, I used mgh=1/2mv^2. (m=17kg, g=9.8, h=6m) and I got V1i=10.84m/s.

Then substituting the value i got back into the previous equation, V2f=34/21 * 10.84 = 17.56 m/s. This answer is obviously shown by the computer to be NOT correct. I don't understand where I went wrong, because I even used the conservation of kinetic energy to check that before and after the kinetic energy is the same! Please help!

Nevermind I got it. the right answer is 17.57 m/s, which is extremely close to what i got but the computer program did not approve my first answer for reasons god knows why.
 
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  • #3
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CAPA sucks eh...haha. I attempted the problem the same way too and got an incorrect answer, my digits were slightly off.
 

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