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Robin fired a bullet of mass 100 gm from a gun of mass 5 kg. The bullet leaves the gun with a speed of 400 m/s. After penetrating 4 cm of a plank of 10 cm, the bullet loses one third of its initial velocity.
a. Calculate the backward velocity of the gun?
b. Can the bullet penetrate the plank of the wood completely? Explain mathematically.

please help

Sagot :

onyebuchinnaji

(a) The recoil or backward velocity of the gun is 8 m/s.

(b) The bullet cannot penetrate the plank of the wood completely.

The given parameters include;

  • mass of the bullet, m₁ = 100 g = 0.1 kg
  • mass of the gun, m₂ = 5 kg
  • initial velocity of the bullet, u₁ = 400 m/s
  • thickness of the plank, x = 10 cm

(a) The backward or recoil velocity of the gun is calculated by applying the principle of conservation of linear momentum.

  • let the backward or recoil  velocity of the gun = u₂

m₁u₁  + m₂u₂ = 0

m₂u₂ = -m₁u₁

[tex]u_2 = -\frac{m_1u_1}{m_2} \\\\u_2 = - \frac{0.1 \times 400}{5} \\\\u_2 = -8 \ m/s[/tex]

Thus, the recoil or backward velocity of the gun is 8 m/s.

(b) determine if the bullet can penetrate the plank of the wood completely

  • the bullet traveled 4 cm and lost ¹/₃ of u₁
  • the remaining distance to completely penetrate the plank = 6 cm
  • the final velocity of the bullet at 4 cm, v = 400 - ¹/₃ x 400 m/s  = 266.67 m/s

the acceleration of the bullet is calculated as;

v² = u₁² + 2as

2as = v² - u²

[tex]a = \frac{v^2 -u_1^2}{2s} \\\\a = \frac{(266.67)^2 -(400)^2}{2\times 0.04} = -1.111 \times 10^6 \ m/s^2[/tex]

Finally, determine the distance traveled by the bullet when it comes to a complete stop, that is the final velocity = 0

[tex]v_f^2 = v^2 + 2ad\\\\2ad = v_f^2 - v^2\\\\d = \frac{v_f^2 - v^2}{2a} \\\\d = \frac{(0) - (266.67)^2}{2(-1.111\times 10^6)} \\\\d = 0.032 \ m[/tex]

d = 3.2 cm

The total distance traveled by the bullet inside the plank = 4 cm + 3.2 cm = 7.2 cm

Therefore, the bullet cannot penetrate the plank completely.

To learn more about linear momentum visit: https://brainly.com/question/15869303

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