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

by lillybeans
Tags: block, collides, incline, sliding
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lillybeans
#1
Nov4-11, 03:30 PM
P: 68
1. The problem statement, all variables and given/known data

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?



2. Relevant equations

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

3. 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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lillybeans
#2
Nov4-11, 03:53 PM
P: 68
Quote Quote by lillybeans View Post
1. The problem statement, all variables and given/known data

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?



2. Relevant equations

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

3. 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.
Hyperfluxe
#3
Nov8-11, 06:07 PM
P: 35
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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