Answer:
(a) The velocity is 12 m/s.
(b) The average force is 18 000N.
Explanation:
(a) The principle of conservation of momentum,
total momentum before = total momentum after
[tex]m_{1}u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]
But, the initial velocities equal zero,
[tex]u_{1}[/tex] = [tex]u_{2}[/tex] = 0
So that,
0 = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]
Substituting the values of variables given in the question,
0 = 1 × [tex]v_{1}[/tex] + 0.02 × 600
0 = [tex]v_{1}[/tex] + 12
[tex]v_{1}[/tex] = - 12 m/s
Thus, the velocity is 12 m/s.
(b) F = ma
From the third equation of motion,
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
a = - [tex]\frac{u^{2}}{2s}[/tex]
= [tex]-\frac{600^{2} }{2(0.2)}[/tex]
= -900, 000 m/s^2
a = 900, 000 m/s^2
The acceleration is 900 Km/s^2.
F = 0.02 × 900000
= 18 000 N
Thus, the average force is 18 000N.
The recoil velocity of the plunger is 12 m/s.
The average force exerted upon the bullet by the gun is 18,000 N.
The given parameters;
The recoil velocity of the plunger is calculated by applying the principle of conservation of linear momentum;
m₁u₁ = m₂u₂
[tex]u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.02 \times 600}{1} \\\\u_1 = 12 \ m/s[/tex]
The acceleration of the bullet before stopping in 20 cm is calculated as follows;
[tex]v^2 = u^2 +2as\\\\0 = u^2 + 2as\\\\-2as = u^2 \\\\-a = \frac{u^2 }{2s} \\\\-a = \frac{600^2 }{2 \times 0.2 } \\\\-a = 900,000 \ m/s^2\\\\a = -900,000 \ m/s^2[/tex]
The average force exerted by the bullet on the gun is calculated as;
F = ma
F = (0.02)(-900,000)
F = - 18,000 N
Thus, the average force exerted upon the bullet by the gun is 18,000 N.
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