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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

**V**

_{2}f=(2m1/(m1+m2))V_{1}i+(m2-m1)/(m1+m2)V_{2}iSince V2i of the smaller mass is zero, I'm only left with the first part, which is.

**V**

_{2}f=(2m1/(m1+m2)V_{1}isubstituting 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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