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The man fire a 50-g arrow that moves at an unknown speed. It hits and embeds in a 350-g block that slides on an air track. At the end, the block runs into and compresses a 4000-N/m spring 0.10 mm.

Part A
How fast was the arrow traveling?

Part B
Indicate the assumptions that you made and discuss how they affect the result.

User Voreny
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1 Answer

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Answer:

a) vAix = 80 m/s

b) The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

Step-by-step explanation:

Arrow embeds into block

Take in the first instance the system to be the arrow + block (isolated). Establish reference coordinate system with the +x axis running horizontally in the direction of the arrow’s motion. The initial state (i) is the arrow travelling with velocity vAix and the final state (f) is the arrow embedded in the block. Now, apply the component form of the Generalized Impulse Momentum Equation to this system:

pAi + pBi + JonA + JonB = pAf + pBf

pAix + pBix + Jx = pAfx + pBfx

mA*vAix + mB*vBix + 0 = (mA + mB)*vfx

0.05*vAix + 0 = (0.05 + 0.35)*vfx

vAix = 8*vfx (1)

Arrow embeds into block

Now consider the next phase of motion and take as the system the arrow + block + spring. The initial state (i) is the arrow and block travelling with velocity equivalent to the final velocity from equation 1 (final state velocity in first phase becomes initial velocity in next phase); vix’ = vfx and the final state (f) is the arrow + block brought to rest and the spring compressed an amount, Δx = 0.1 m. Now, apply the Generalized Work Energy Principle to the system

Ei + W = Ef

Ki + Usi + W = Kf + Usf

0.5*(mA + mB)*vix’² = 0.5*k*Δx²

(0.05 + 0.35)*vfx² = 4000*(0.1)²

vfx = √(40/0.4) = 10 m/s

Substituting above back into equation 1:

vAix = 8* 10 m/s = 80 m/s

Arrow embeds into block

The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

User Greg Michalec
by
6.8k points