Elastic Collision / Energy Preserved

[huntj106713]

Homework Statement

A block of mass m1= 5kg is released from point 1 which is 5 meters above the horizontal frictionless surface and slides down a frictionless track. It makes a head-on collision at point 2 with a block of mass m2= 10kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

I don't know where to start... The problem doesn't give the distance between m1 and m2, just how far m1 is above the ground.

Homework Equations

The Attempt at a Solution

 

[whybother]

Use energy conservation + conservation of momentum. You know that since the sliding block isn't losing energy to friction, that all it's energy must be in the form of kinetic and potential energy. At the point it's released, all of it's energy is in the form of gravitational potential energy. By the time it hits block 2, all of it's energy must be kinetic, so you'll be able to figure out the velocity of impact. Now, you combine what you know about energy-momentum conservation with the fact that block 1 is going to bounce a little back up the ramp, meaning you'll still have to look at gravitational potential energy and that should do it.

 

[nlightner]

As a scientist, it is important to start by identifying the known and unknown variables in this problem. From the given information, we know the masses of both blocks (m1=5kg and m2=10kg) and their initial positions (point 1 and point 2). We also know that the surface and track are both frictionless, meaning there is no external force acting on the blocks during their motion.

The unknown variable in this problem is the maximum height to which m1 rises after the collision. To solve for this, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another.

In an elastic collision, both kinetic energy and momentum are conserved. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. We can set up an equation to represent this:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1u1^2 + (1/2)m2u2^2

Where v1 and v2 are the velocities of m1 and m2 before the collision, and u1 and u2 are the velocities after the collision.

Since m2 is initially at rest, v2 = 0. We can also use the fact that the blocks are on a frictionless surface, meaning there is no external work done on the system, and therefore the total mechanical energy is conserved. This can be represented by the equation:

(1/2)m1v1^2 + m1gh = (1/2)m1u1^2 + (1/2)m2u2^2

Where h is the maximum height that m1 rises after the collision.

We can solve for u1 in the first equation and substitute it into the second equation:

u1 = [(m1-m2)/(m1+m2)]v1

Substituting this into the second equation and solving for h, we get:

h = [(m1-m2)/(m1+m2)]^2(5m)

Plugging in the values for m1 and m2, we get:

h = (1/9)(5m) = 5/9 m = 0.56 m

Therefore, the maximum height to which m1 rises after the collision is 0.56 meters.