Question

A bullet fired into a target loses half its velocity after penetrating 25 cm. How much further will it penetrate before coming to rest?​

Answer 1

Explanation:

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Answer 2

Concept

Work energy theorem: Let a force F acts on a body and the body moves through some distance, and the kinetic energy of the body changes from K₁ to K₂. The kinetic energy change of the particle is equal to the work done by the force. That is

                               W₁₂ = K₂ - K₁ ,                              ...(1)

where W₁₂ is the work done in moving the particle from position 1 to position 2,

K₁ is the kinetic energy of the particle at position 1,

K₂ is the kinetic energy of the particle at position 2.

Given

• The bullet penetrates through a distance of 25cm.
• The velocity of the bullet is reduced to half its initial value.

Find

The distance through which the bullet will further penetrate the target.

Solution

Assumption about the force involved

To proceed further we have to assume the nature of the force that deaccelerates the bullet i.e. the profile of the force exerted by the target on the bullet.

Assume the force exerted by the target on the bullet is constant and has a magnitude F.

Work done on the bullet in penetrating 25cm

The work done by the force F on the bullet as it penetrates 25cm through the target is

          W₁₂ = F × 0.25m = 0.25F.

Using the equation (1)

          W₁₂ = K₂ - K₁ = $\frac{1}{2}m\left(\frac{v}{2}\right)^2 - \frac{1}{2}mv^2$ = $-\frac{3}{8}mv^2$.

The force on the bullet

Using the above two equations

          0.25 F = $-\frac{3}{8}mv^2$

⇒                F = $-\frac{3}{2}mv^2$.                      ...(2)

Work done in further penetration of the target

Let the bullet further penetrates the target through a distance x.

The work done by the force W₂₃ = Fx.

Using the equation we write,

     W₂₃ = K₃ - K₂ = $0 -\frac{1}{2}m\left( \frac{v}{2} \right)^2$ = $-\frac{1}{8}mv^2$.

Using the above two equations

Fx = $-\frac{1}{8}mv^2$

Using the value of F from equation (2)

x = $\left( -\frac{1}{8}mv^2 \right)\cdot \left(- \frac{2}{3mv^2} \right)$ = 1/12 m = 100/12 cm

⇒  x  = 8.33 cm.

The bullet will further penetrate the target through a distance of 8.33cm.

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