Question:
A bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?
A bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?
Updated On: Jul 2, 2022
- $3.0\,cm$
- $2.0\,cm$
- $1.5\,cm$
- $1.0\,cm$
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The Correct Option is D
Solution and Explanation
According to work-energy theorem,
$W = \Delta K$
$-F\times3=\,\frac{1}{2} m\left(\frac{v_{0}}{2}\right)^{2}-\frac{1}{2}\, mv^{2}_{0}$
where, $F$ is resistive force and $v_{0}$ is initial speed.
Let, the further distance travelled by the bullet before coming to rest is s.
$\therefore -F \left(3+ s\right) = K_{f}-k_{i} =\frac{1}{2}\,mv^{2}_{0}$
or $\frac{1}{4} \left(3+s\right)=1$
or $\frac{3}{4}+\frac{s}{4}=1$
$\therefore s = 1\, cm$
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Concepts Used:
Work-Energy Theorem
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.