Solving Motion of Block A on Wedge B

[hav0c]

Homework Statement

A block A of mass m is placed over a wedge B of same mass m. Assuming all surface to be smooth. The displacement of block A is 1 sec if the system is released from rest is :

Homework Equations

The Attempt at a Solution

i have made fbds and have the relations-
1. ma₁=(g-N₁)/m
2.N₁cos θ+mg cosθ=N₂
3.ma₂=sinθ(mg+N₁)
now I am stuck

 

[consciousness]

There is one more equation. The blocks stay in contact so their velocities perpendicular to the contact are same. What can you say about the relative acceleration perpendicular to the contact?

 

[hav0c]

There is one more equation. The blocks stay in contact so their velocities perpendicular to the contact are same. What can you say about the relative acceleration perpendicular to the contact?

What velocities are the same?
I don't get it
EDIT:Thanks i got it A1=a2sintheta

 

[Saitama]

Homework Statement

A block A of mass m is placed over a wedge B of same mass m. Assuming all surface to be smooth. The displacement of block A is 1 sec if the system is released from rest is :

Homework Equations

The Attempt at a Solution

i have made fbds and have the relations-
1. ma₁=(g-N₁)/m
2.N₁cos θ+mg cosθ=N₂
3.ma₂=sinθ(mg+N₁)
now I am stuck

Your first equation has a dimensional error.

I don't think the second equation would do any good. Your third equation is fine. Apply the constraint that blocks A and B stay together during the motion.

 

[Nickie]

I would like to commend you for taking the initiative to create free body diagrams and identifying the relevant equations for this problem. It shows that you have a good understanding of the basic principles involved.

To continue with your solution, I would suggest that you first try to simplify the equations you have identified. For example, equation 2 can be rewritten as N1(cosθ+sinθ)=mg. This can then be substituted into equations 1 and 3 to eliminate N1 and make the equations easier to solve.

Additionally, you can also consider conservation of energy in this system. Since the system is released from rest, the initial potential energy will be equal to the final kinetic energy. This can be expressed as mgh = 1/2mv^2, where h is the height of the wedge and v is the velocity of block A at the end of 1 second. This can be used to solve for v and then the displacement of block A can be calculated using the equation d=v*t.

I hope this helps guide you towards finding a solution to the problem. Remember, as a scientist, it is important to always approach problems with a critical and analytical mindset, and to continue to experiment and explore different possible solutions. Good luck!