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A bullet of mass \(20\, g\) is horizontally fired with a horizontal velocity \(150\,ms^{-1}\) from a pistol of mass \(2\, kg\). What is the recoil velocity of the pistol?
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A bullet of mass \(20\, g\) is horizontally fired with a horizontal velocity \(150\,ms^{-1}\) from a pistol of mass \(2\, kg\). What is the recoil velocity of the pistol?

Answer

As per question-
Mass of bullet, \(m_b=20\,g=0.02\,kg\)
Mass of pistol, \(m_p=2\,kg\)
Initial velocity of bullet, \(u_b=0\)
Final velocity of bullet, \(v_b=150\,ms^{-1}\)
Initial velocity of pistol, \(u_b=0\)
Now let us find out momentum before firing
Answer for A bullet of mass 20, g is horizontally fired with a horizontal velocity 150,ms^{-1} from a pistol of mass 2, kg. What is the recoil velocity of the pistol?
As, momentum \(p=mv\)
So, \(p_1=m_bu_b+m_pu_p\)
\(=0.02\times 0+2\times 0\)
\(p_1=0\ kg\ m\ s^{-1}\)
Now, let us find out momentum after firing
Answer for A bullet of mass 20, g is horizontally fired with a horizontal velocity 150,ms^{-1} from a pistol of mass 2, kg. What is the recoil velocity of the pistol?
\(p=m_pv_p+m_bv_b\)
\(=2\times v_p+0.02\times 150\)
\(p_2=3+2v_p\)
From law of conservation of momentum,
Momentum after firing \(=\)Momentum before firing
\(3+2v_p=0\)
So, we get \(v_p=-1.5\,ms^{-1}\)
Here, negative sign denotes that the direction in which the pistol would recoil is opposite to that of bullet.
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