Solve Energy & Momentum Problem: Urgent Homework

[nns91]

Homework Statement

A 2kg block and an 8kg block are both attached to an ideal spring for which k=200 N/m and both are initially at rest on a horizontal frictionless surface. In an initial experiment a 0.1kg ball of clay is thrown at the 2kg block. The clay is moving horizontally with speed v when it hits and sticks to the block. The 8kg block is held still by a removable stop. As a result, the spring compresses a maximum distance of 0.4m.

In a second experiment, an identical ball of clay is thrown at another identical 2kg block, but this time the stop is removed so that the 8kg block is free to move.

e. State the principle(s) that can be used to calculate the velocity of the 8kg block at the instant that the spring regains its original length. Write the appropriate equation(s) and show the numerical substitutions, but do not solve for velocity.

Homework Equations

Pi=Pf
Ei=Ef

The Attempt at a Solution

So I use conservation of momentum and get

m_clay*v_clay + m_blocks * 0 = ( m_clay+m_blocks)* vf

However, the rubric said I need to include also conservation of energy so how will it fit in here ??

 

[LowlyPion]

Until the first weight returns to normal relaxed spring detent, the spring contains PE. Only as it crosses the equilibrium point is it not to be accounted for.

What happens to the speed of the larger block after the clay laden block crosses that point going backward?

 

[nlightner]

I would approach this problem by using the principles of conservation of momentum and conservation of energy. These principles state that in a closed system, the total momentum and energy remain constant.

In the first experiment, the clay ball transfers its momentum to the 2kg block, causing it to move with a final velocity of vf. The 8kg block remains at rest due to the stop. Therefore, the initial momentum of the system is equal to the final momentum, which can be expressed as:

Pi = Pclay + P2kg
mclay * v + 0 = (mclay + 2kg) * vf

We can also consider the conservation of energy in this system. Initially, the system has only kinetic energy due to the motion of the clay ball. After the collision, the system has both kinetic energy and elastic potential energy stored in the spring. This can be expressed as:

Ei = Ef
(1/2) * mclay * v^2 = (1/2) * (mclay + 2kg) * vf^2 + (1/2) * k * x^2

Where x is the maximum compression of the spring, which is given as 0.4m.

In the second experiment, the 8kg block is free to move after the collision. Therefore, we can use the same principles to calculate the velocity of the 8kg block at the instant the spring regains its original length. The equations would be the same, but with a different mass (mclay + 8kg) and a different final velocity (v8kg).

Pi = Pclay + P2kg
mclay * v + 0 = (mclay + 8kg) * v8kg

Ei = Ef
(1/2) * mclay * v^2 = (1/2) * (mclay + 8kg) * v8kg^2 + (1/2) * k * x^2

These equations can be solved simultaneously to find the final velocity of the 8kg block. By using both principles, we can obtain a more accurate and comprehensive understanding of the system and its behavior.