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

In summary: Good job on finding the correct answer!In summary, the question asks for the velocity of a block of mass m=4 kg after it collides elastically with another block of mass M=17 kg that is initially sliding down a frictionless inclined plane at a height of 6 m. Using the conservation of momentum formula, V2f=(2m1/(m1+m2))V1i, and the conservation of energy formula, mgh=1/2mv^2, the final velocity of m is calculated to be 17.57 m/s. However, due to slight differences in digits, the computer program may not approve the answer.
  • #1
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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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  • #2
lillybeans said:

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
CAPA sucks eh...haha. I attempted the problem the same way too and got an incorrect answer, my digits were slightly off.
 

1. What is the physics behind a block sliding down an incline and colliding with another block?

The motion of a block sliding down an incline and colliding with another block is governed by Newton's laws of motion. The first law states that an object will remain at rest or continue to move at a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The third law states that for every action, there is an equal and opposite reaction.

2. How does the mass of the blocks affect their motion when colliding?

The mass of the blocks affects their motion when colliding in two ways. First, the heavier block will have a greater inertia and will require a larger force to be accelerated. Second, the heavier block will exert a greater force on the lighter block, causing it to accelerate more. This results in a transfer of momentum between the two blocks.

3. What factors influence the speed of the blocks after the collision?

The speed of the blocks after the collision is influenced by several factors. These include the masses of the blocks, the angle of the incline, the coefficient of friction between the blocks and the incline, and the elasticity of the collision (whether it is an elastic or inelastic collision).

4. How does the angle of the incline affect the motion of the blocks?

The angle of the incline affects the motion of the blocks in two ways. First, a steeper incline will result in a greater gravitational force acting on the blocks, causing them to accelerate faster. Second, the angle of the incline will affect the normal force acting on the blocks, which will in turn affect their frictional force and ultimately their acceleration.

5. What is the difference between an elastic and an inelastic collision?

An elastic collision is one in which the total kinetic energy of the system is conserved, meaning the objects bounce off each other without any loss of energy. In contrast, an inelastic collision is one in which the objects stick together after colliding and some kinetic energy is lost in the form of heat or deformation. In the case of blocks colliding on an incline, the collision is most likely inelastic due to the friction and deformation of the blocks.

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