Question

2Kg के 1 पिस्टल से 20g द्रव्यमान की एक गोली 150 m/s के क्षेतिज वेग से छोड़ी जाती है पिस्टन के पीछे हटने के वेग का परिचालन करो​

Answer 1

Answer:

Initially gun ko aur bullet ko velocity Zero hoga

To, law of conservation of linear momentum se

banayenge

M1V1+M2V2=MV

Mg×Vg+Mb×Vb=( Mg+Mb)×0

(Here M=Mg+Mb)

Mg×Vg= -Mb×Vb

Vg= -Mb×Vb/Mg

Vg= - 20×10^(-3)×150/2

Vg= -1. 5m/s

Answer 2

The recoil velocity of the pistol is 1.5 m/s.

Explanation:

From a pistol of 2Kg, a bullet of mass 20g is released with a horizontal velocity of 150 m/s. Operate the retreat velocity of the piston.

It is given that,

Mass of the pistol, $m_1=2\ kg$

Mass of the bullet, $m_2=20\ g=0.02\ kg$

Velocity of the bullet, $v_2=150\ m/s$

We need to find the recoil speed of the pistol. Initially, the pistol and the bullet both are at rest. The recoil speed of the pistol can be calculated using conservation of linear momentum as :

$m_1v_1+m_2v_2=0$

$2v_1+0.02\times 150=0$

$v_1=-1.5\ m/s$

So, the recoil velocity of the pistol is 1.5 m/s. Hence, this is the required solution.

Learn more

Conservation of momentum

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