Question

A ballistic spring system can be used to measure the speed of bullets. A bullet of mass 0.005 kg is fired into a block of mass 2 kg. The block, with the embedded bullet, then slides across a frictionless table and collides with a horizontal spring whose spring constant is 50 N m . The opposite end of the spring is anchored to a wall. The spring’s maximum compression is measured to be 0.1 m. Calculate the initial speed of the bullet. Along the way, identify which model(s) you are using and why. Show your work clearly.

Answer 1

Here we will use the concept of conservation of energy and the conservation of momentum :

Let the initial speed of bullet was v m/s .

Given, mass of bullet (m) = 0.005 kg

mass of block (M) = 2 kg

Spring constant of the spring (k) = 50 N/m

and the maximum compression in the spring (x) = 0.1 m

Applying conservation of momentum between point A and B, we get

  

or, $v_{1}= \frac{mv}{M + m}$

Where, v₁ is velocity of block with embedded bullet after collision.

Now, applying conservation of energy between block with embedded bullet and the spring system.

Initial energy of the system = kinetic energy of the block with embedded bullet = K

And    $\textbf{K}= \frac{1}{2} (m + M) \, v_{1}^{2} = \frac{1}{2} \frac{(mv)^{2}}{M + m} = \frac{1}{2}\times \frac{(0.005)^{2}}{2 + 0.005} \times v^{2}$

or, $\textbf{K}= 6.23\times 10^{-6} \times v^{2}$

The final energy of the system = Potential energy stored in the spring = U

And $\textbf{U} = \frac{1}{2}kx^{2} = \frac{1}{2}\times 50 \times (0.1)^{2} = 0.25 \, \, J$

From conservation of energy

   $\textbf{K}= \textbf{U}$

or,    $6.23\times 10^{-6} \times v^{2} = 0.25$

or,

or,   

Hence, the initial speed of the bullet was 200.25 or nearly 200 m/s.

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