Physics Experiment Notes: Finding Bullet Speed Using Spring

Experimental Setup

• Bullet fired from a gun hits a wooden block connected to a spring.
• Mass of bullet: $m_b = 10 ext{ grams} = 0.01 ext{ kg}$.
• Mass of block: $m = 5 ext{ kg}$.
• Compression of spring: $x = 10 ext{ cm} = 0.1 ext{ m}$.
• Spring constant: $k = 200 ext{ N/m}$.

Conservation of Energy

• Inelastic collision: bullet and block move together after impact.
• Total kinetic energy converts to elastic potential energy.
• Kinetic energy of block and bullet: $KE = \frac{1}{2} (m + m_b) v^2$.
• Elastic potential energy of spring: $PE = \frac{1}{2} k x^2$.
• Set $KE = PE$.

Velocity Calculation

• After cancelling $\frac{1}{2}$: $(m + m_b) v^2 = k x^2$.
• Rearranging gives: $v = \sqrt{\frac{k x^2}{m + m_b}}$.
• Substitute values:
  • $x = 0.1 ext{ m}$,
  • $k = 200 ext{ N/m}$,
  • $m = 5 ext{ kg}$, and $m_b = 0.01 ext{ kg}$.
• $v = \sqrt{\frac{200 \times (0.1)^2}{5 + 0.01}} = 2 ext{ m/s}$.

Conservation of Momentum

• Momentum conservation for the collision:
• Before collision: only bullet has momentum: $p<em>{initial} = m</em>b v_b$,
• After collision: block and bullet together: $p<em>{final} = (m + m</em>b)v$.
• Use conservation: $m<em>b v</em>b = (m + m_b)v$.
• Solve for bullet's initial speed: $
  vb = \frac{(m + mb)}{m_b} v$.
• Substitute known values and solve.

Final Result

• Final computed speed of the bullet: $v_b \approx 1000 ext{ m/s}$.

Summary

• Experiment effectively demonstrates conservation of energy and momentum principles to find bullet speed.