Physics - elastic collision

[physixnot4me]

(3) two blocks are free to slide along the frictionless wooden track. a block of mass m1=5.00kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2=10.0kg, initially at rest. the two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Do you use conservation of kinetic energy and momentum to figure out the final velocity of the block M1 after the collision. Then let all that kinetic energy be converted to gravitational potential energy to determine how high it rises?

what equations are appplied to a question like this?

 

[Chi Meson]

Do you use conservation of kinetic energy and momentum to figure out the final velocity of the block M1 after the collision. Then let all that kinetic energy be converted to gravitational potential energy to determine how high it rises?
what equations are appplied to a question like this?

Yes. Since you have two unknowns (v1 and v2, both "after" the collision) you need two equations. The equations you use are the "total momentum before = total momentum after" and the "total KE before = total KE after" equations:

 

[quicknote]

In order to calculate the maximum height to which m1 rises after the elastic collision, we can use the equations of conservation of kinetic energy and momentum.

First, we can calculate the final velocity of the block m1 after the collision using the equation:

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 and m2 are the masses of the two blocks, v1i and v2i are the initial velocities of the blocks, and v1f and v2f are the final velocities.

Since the two blocks are initially at rest, v1i = 0 and v2i = 0. Also, since the collision is elastic, kinetic energy is conserved, so we can use the equation:

1/2 m1v1i^2 + 1/2 m2v2i^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2

Solving for v1f, we get:

v1f = (m1 - m2) / (m1 + m2) * v1i

Next, we can use the equation for conservation of momentum to calculate the final velocity of m2:

m1v1i + m2v2i = m1v1f + m2v2f

Solving for v2f, we get:

v2f = 2m1 / (m1 + m2) * v1i

Now, we can use the equation for gravitational potential energy to calculate the maximum height h to which m1 rises:

m1gh = 1/2 m1v1f^2

Solving for h, we get:

h = v1f^2 / (2g)

Substituting the values for v1f and g, we get:

h = [(m1 - m2) / (m1 + m2) * v1i]^2 / (2g)

Therefore, we can use these equations to calculate the maximum height to which m1 rises after the elastic collision. It is important to note that these equations assume a frictionless wooden track and no external forces acting on the system, so they may not be accurate in real-world scenarios. However, they provide a good estimate for theoretical calculations.