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A 8.0 g bullet is fired into a 9.00 kg wood cube that is initially at rest. The bullet lodges in the cube, and the combined system moves forward with a velocity of 40 cm/s. What was the initial veloci

Physics

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Answer

450.4 m/s\text{450.4 m/s}

Step-by-step solution

3 steps

Here is the solution:

Given:
Mass of bullet, mb=8.0 g=0.008 kgm_b = 8.0 \, \text{g} = 0.008 \, \text{kg}
Mass of wood cube, mw=9.00 kgm_w = 9.00 \, \text{kg}
Initial velocity of wood cube, uw=0 m/su_w = 0 \, \text{m/s} (at rest)
Final velocity of the combined bullet and wood cube, V=40 cm/s=0.40 m/sV = 40 \, \text{cm/s} = 0.40 \, \text{m/s}
Initial velocity of bullet, ub=?u_b = ?

  1. Apply the principle of conservation of momentum.

    Since the bullet lodges in the wood cube, this is an inelastic collision, and the total momentum before the collision equals the total momentum after the collision.

    mbub+mwuw=(mb+mw)Vm_b u_b + m_w u_w = (m_b + m_w)V
  2. Substitute the known values into the equation.

    (0.008 kg)ub+(9.00 kg)(0 m/s)=(0.008 kg+9.00 kg)(0.40 m/s)(0.008 \, \text{kg})u_b + (9.00 \, \text{kg})(0 \, \text{m/s}) = (0.008 \, \text{kg} + 9.00 \, \text{kg})(0.40 \, \text{m/s}) (0.008)ub+0=(9.008)(0.40)(0.008)u_b + 0 = (9.008)(0.40) (0.008)ub=3.6032(0.008)u_b = 3.6032
  3. Solve for ubu_b.

    ub=3.60320.008u_b = \frac{3.6032}{0.008} ub=450.4 m/su_b = 450.4 \, \text{m/s}

    The initial velocity of the bullet is 450.4 m/s\text{450.4 m/s}.

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