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


by lillybeans
Tags: block, collides, incline, sliding
lillybeans
lillybeans is offline
#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!
Phys.Org News Partner Science news on Phys.org
Simplicity is key to co-operative robots
Chemical vapor deposition used to grow atomic layer materials on top of each other
Earliest ancestor of land herbivores discovered
lillybeans
lillybeans is offline
#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
Hyperfluxe is offline
#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.


Register to reply

Related Discussions
Block 1 slides down and collides with block two pushing it onto a friction surface. Introductory Physics Homework 1
Block 1 collides with block two which is attached to a spring. How far do they go? Introductory Physics Homework 1
Block sliding down incline, hitting another block, find distance. Introductory Physics Homework 0
Block sliding down an incline and hitting a smaller block Introductory Physics Homework 7
block sliding up incline Classical Physics 5