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Worked Examples · Example 5.4

Q.In a ballistics demonstration a police officer fires a bullet of mass 50.0 g50.0\ \text{g} with speed 200 m s−1200\ \text{m s}^{-1} on soft plywood of thickness 2.00 cm2.00\ \text{cm}. The bullet emerges with only 10%10\% of its initial kinetic energy. What is the emergent speed of the bullet?

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The problem uses the Work-Energy Theorem to relate the initial and final kinetic energies of a bullet. Given that the bullet retains 10%10\% of its initial kinetic energy, we calculate its emergent speed to be 2010 m s−1\boxed{20\sqrt{10}\ \text{m s}^{-1}}.

The core concept for solving this problem is the Work-Energy Theorem. This theorem is a fundamental principle in physics that connects the work done on an object to its change in kinetic energy. It states that the net work done on an object is equal to the change in its kinetic energy.

Wnet=ΔKE=KEf−KEiW_{\text{net}} = \Delta KE = KE_f - KE_i

Here, KEiKE_i is the initial kinetic energy and KEfKE_f is the final kinetic energy.

In this scenario, as the bullet penetrates the soft plywood, the plywood exerts a resistive force on the bullet. This force acts in the direction opposite to the bullet's motion, doing negative work on the bullet. This negative work causes the bullet to decelerate, meaning its kinetic energy decreases.

The problem simplifies things by directly telling us the relationship between the final and initial kinetic energies: the bullet emerges with 10%10\% of its initial kinetic energy. This means we do not need to calculate the work done by the resistive force explicitly (which would require knowing the force and the distance over which it acts). Instead, we can directly calculate the initial kinetic energy, then find the final kinetic energy, and finally use that to determine the emergent speed.

Notice that the thickness of the plywood (2.00 cm2.00\ \text{cm}) is given but is not required to find the emergent speed. This information would be necessary if we were asked to calculate the average resistive force exerted by the plywood.

Here is the step-by-step solution:

  1. Identify given quantities and convert units.

    First, we list the information provided in the problem and ensure all units are consistent with the SI system (kilograms, meters, seconds).

    • Mass of the bullet, m=50.0 gm = 50.0\ \text{g}. To convert grams to kilograms: m=50.0 g×1 kg1000 g=0.050 kgm = 50.0\ \text{g} \times \frac{1\ \text{kg}}{1000\ \text{g}} = 0.050\ \text{kg}.
    • Initial speed of the bullet, vi=200 m s−1v_i = 200\ \text{m s}^{-1}.
    • The bullet emerges with 10%10\% of its initial kinetic energy. This means the final kinetic energy, KEfKE_f, is 0.100.10 times the initial kinetic energy, KEiKE_i.
  2. Calculate the initial kinetic energy (KEiKE_i).

    The kinetic energy of an object is defined by the formula:

    The kinetic energy (KEKE) of an object with mass mm and speed vv is:

    KE=12mv2KE = \frac{1}{2}mv^2

    Using the given mass and initial speed:

    KEi=12(0.050 kg)(200 m s−1)2KE_i = \frac{1}{2} (0.050\ \text{kg}) (200\ \text{m s}^{-1})^2

    KEi=12(0.050)(40000) JKE_i = \frac{1}{2} (0.050) (40000)\ \text{J}

    KEi=0.025×40000 JKE_i = 0.025 \times 40000\ \text{J}

    KEi=1000 JKE_i = 1000\ \text{J}

  3. Calculate the final kinetic energy (KEfKE_f).

    The problem states that the bullet emerges with 10%10\% of its initial kinetic energy. …

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