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Abullet of mass 20 g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg .what is the recoil velocity of the pistol

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  • Réponse publiée par: reyquicoy4321

    Hello!

    A bullet of mass 20 g is horizontally fired with a velocity of 150 m/s from a pistol of mass 2 kg .what is the recoil velocity of the pistol

    We have the following data:  

    ΔP (momentum before impact) = ?    

    mA (mass) = 20 g → 0.02 kg  

    vA (velocity) = 150 m/s  

    mB (mass) = 2 kg  

    vB (velocity) = ? (in m/s)

    We apply the conservation law to the amount of motion:

    *** Note¹: Law of conservation of momentum, since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.

    *** Note²: They will have velocities in opposite directions and it can be observed that if we multiply the mass of each one by their respective velocity, the result of this operation will be the same for both.

    m_A*v_A = m_B*v_B

    0.02*150 = 2*v_B

    3 = 2\:v_B

    2\:v_B = 3

    v_B = \dfrac{3}{2}

    \boxed{\boxed{v_B = 1.5\:m/s}}\Longleftarrow(velocity\:of\:the\:pistol)\:\:\:\:\:\:\bf\green{\checkmark}

    The recoil velocity of the pistol is 1.5 m/s

    _______________________

    \bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

  • Réponse publiée par: nelspas422

    { \huge \bf { \mid{ \overline{ \underline{Correct \: Question}}} \mid}}

    A bullet of mass 20 grams is horizontally fired with a velocity 150m per second from a pistol of mass 2 kg what is the recoil velocity of the pistol ?

    \huge{\bold{\underline{\underline{....answer....}}}}

    \huge{\bold{\underline{Given:-}}}

    Mass of Bullet (m) = 20 grams. Velocity of the Bullet after Firing (u) = 150 m/s. Mass of The Pistol (M) = 2 kg.Let the recoil velocity of the Pistol be "v".

    \huge{\bold{\underline{Explanation:-}}}

    \rule{300}{1.5}

    From Law of Conservation of momentum,

    Initial Momentum = Final Momentum.

    \large{\boxed{\tt  P_i = P_f}}

    But Before Firing, the whole system (Gun & Bullet) is at Rest.

    Therefore, Initial momentum will be zero.

    Then the equation becomes,

    \large{\leadsto {\underline{\underline{\tt P_f = 0}}}}

    Now, Formulating the Values of Final Momentum.

    \large{\tt \leadsto mu + Mv = 0}

    M = Mass of Pistol.m = Mass of Bullet. u = Velocity of bullet. v = Velocity of bullet.

    Substituting the values

    \large{\tt \leadsto 20 \times 10^{-3}  \times 150 + 2 \times v = 0}

    [Here Mass of bullet (m) is taken as 20 grams = 20 × 10⁻³ Kg.]

    \large{\tt \leadsto 20 \times 150 \times 10^{-3} + 2v = 0}

    \large{\tt \leadsto 3000 \times 10^{-3} + 2v = 0}

    \large{\tt \leadsto 3\times 10^3 \times \dfrac{1}{10^3}+ 2v = 0}

    \large{\tt \leadsto 3 \times \cancel{10^3} \times \dfrac{1}{\cancel{10^3}}+ 2v = 0}

    \large{\tt \leadsto 3 \times 1 + 2v = 0}

    \large{\tt \leadsto 3 + 2v = 0}

    \large{\tt \leadsto 2v = - 3}

    \large{\tt \leadsto v = \dfrac{- 3}{2}}

    \large{\tt \leadsto v = \cancel{\dfrac{- 3}{2}}}

    \huge{\boxed{\boxed{\tt v = - 1.5 \: m/s}}}

    So, The Recoil velocity of the Pistol is 1.5 m/s.

    Note:-

    Negative sign of Recoil velocity indicates that the recoil velocity of the Pistol is opposite to the Motion of the Bullet.

    \rule{300}{1.5}

  • Réponse publiée par: smith21

    1.5m/s

    Explanation:

    Mass of bullet=M₁=20g=0.02kg

    Mass of gun=M₂=2kg

    velocity of bullet initially at rest=u₁=0m/s

    velocity of gun initially=u₂=0m/s

    velocity of bullet after firing=v₁=150m/s

    velocity of gun after firing=v₂

    according to law of conservation of momentum,

    M₁u₁+M₂u₂=M₁v₁+M₂v₂

    =0.02×0+2×0=0.02×150+2×v₂

    =0+0=3+2v₂

    = -3=2v₂

    v₂=\frac{-3}{2}

    v₂=-1.5m/s

    the recoil velocity of the bullet is 1.5m/s

    Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet

  • Réponse publiée par: joyce5512

    m1 = 20g (= 0.02 kg)

    Mass of pistol, m2 = 2 kg

    Initial velocity of the bullet (u1) and pistol (u2) = 0

    Final velocity of the bullet, v1 = +150m s-1

    Let v be the recoil velocity of the pistol.

    Total momentum of the pistol and bullet after it is fired is

    = (0.02 kg x 150 m s-1) + (2 kg x v m s-1)

    = (3 + 2v) kg m s-1

    Total momentum after the fire = Total momentum before the fire

    3 + 2v = 0

    →v = -1.5 m/s  

  • Réponse publiée par: homersoncanceranguiu
    Law of conservation of momentum
    20*150=2000*x is equals 1.50 m/s
    if any doubt than ask me in comment section
  • Réponse publiée par: tayis

    1.5m/s

    Explanation:

    Mass of bullet=M₁=20g=0.02kg

    Mass of gun=M₂=2kg

    velocity of bullet initially at rest=u₁=0m/s

    velocity of gun initially=u₂=0m/s

    velocity of bullet after firing=v₁=150m/s

    velocity of gun after firing=v₂

    according to law of conservation of momentum,

    M₁u₁+M₂u₂=M₁v₁+M₂v₂

    =0.02×0+2×0=0.02×150+2×v₂

    =0+0=3+2v₂

    = -3=2v₂

    v₂=

    v₂=-1.5m/s

    the recoil velocity of the bullet is 1.5m/s

    Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet

  • Réponse publiée par: Laurenjayshree

    answer:

    James Clerk Maxwell FRS FRSE (13 June 1831 – 5 November 1879) was a Scottish scientist in the field of mathematical physics.[2] His most notable achievement was to formulate the classical theory of electromagnetic radiation, bringing together for the first time electricity, magnetism, and light as different manifestations of the same phenomenon. Maxwell's equations for electromagnetism have been called the "second great unification in physics" [3] after the first one realised by Isaac Newton.

    With the publication of "A Dynamical Theory of the Electromagnetic Field" in 1865, Maxwell demonstrated that electric and magnetic fields travel through space as waves moving at the speed of light.[4] He proposed that light is an undulation in the same medium that is the cause of electric and magnetic phenomena.[5] The unification of light and electrical phenomena led his prediction of the existence of radio waves. Maxwell is also regarded as a founder of the modern field of electrical engineering.[6]

    He helped develop the Maxwell–Boltzmann distribution, a statistical means of describing aspects of the kinetic theory of gases. He is also known for presenting the first durable colour photograph in 1861 and for his foundational work on analysing the rigidity of rod-and-joint frameworks (trusses) like those in many bridges.

    His discoveries helped usher in the era of modern physics, laying the foundation for such fields as special relativity and quantum mechanics. Many physicists regard Maxwell as the 19th-century scientist having the greatest influence on 20th-century physics. His contributions to the science are considered by many to be of the same magnitude as those of Isaac Newton and Albert Einstein.[7] In the millennium poll—a survey of the 100 most prominent physicists—Maxwell was voted the third greatest physicist of all time, behind only Newton and Einstein.[8] On the centenary of Maxwell's birthday, Einstein described Maxwell's work as the "most profound and the most fruitful that physics has experienced since the time of Newton".[9] Einstein, when he visited the University of Cambridge in 1922, was told by his host that he had done great things because he stood on Newton's shoulders; Einstein replied: "No I don't. I stand on the shoulders of

    Explanation:

  • Réponse publiée par: candace08

    size

    Explanation:triangel or shape cm

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Abullet of mass 20 g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg .wha...