Question

A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 ms⁻¹. The final velocity acquired by the person and the average force exerted on the person are(a) –1.6 ms⁻¹; 8 N(b –0.08 ms⁻¹; 16Nc) – 0.8 ms⁻¹; 8 N(d) –1.6 ms⁻¹; 16 N

Answer 1

Answer:

Answer: The average force exerted on the person is 16 N.

Explanation:

We have to use the Law of momentum conservation:

We have to use the Law of momentum conservation:P ( initial ) = P ( final )

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · v

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · v

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · vv = - 80 kgm/s / 99.9 kgm/s

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · vv = - 80 kgm/s / 99.9 kgm/sv = 0.8 m/s

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · vv = - 80 kgm/s / 99.9 kgm/sv = 0.8 m/sThen : F = Δ P / Δ t = ( 10 · 0.01 kg · 800 m/s ) : 5 s = 16 N

We have to use the Law of momentum conservation:P ( initial ) = P ( final )0 = n · m · u + ( M - n · m ) · vwhere: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · vv = - 80 kgm/s / 99.9 kgm/sv = 0.8 m/sThen : F = Δ P / Δ t = ( 10 · 0.01 kg · 800 m/s ) : 5 s = 16 NAnswer: The average force exerted on the person is 16 N.

hope this would help u...

Answer 2

$\bf{\underline{\red{Correct\:Question:-}}}$

✓✓A person holding a rifle ( mass of person and rifle together is 100 kg ) stands on a smooth surface and fires 10 shorts horizontally, in 5 sec .Each bullet has a mass of 10g with a muzzle velocity of 800 m/s .The final velocity acquired by the person are :-

• (a) -1.6 m/s; 8 N
• (b) -0.08 m/s; 16 N ✔️
• (c) -0.8 m/s ; 8 N
• (d) -1.6 m/s ; 16 N

$\bf{\underline{\pink{Concept:-}}}$

$\bf\purple{{Law\:of\: Conservation\:of\: momentum}}$

law of conservation of momentum states that:-

• If net force is acting on body is zero than momentum of the body system before collision is equal to the momentum of the body in collision.

or,

• According to law of conservation of momentum, total momentum of a system remain constant if no external unbalanced force acts on the system.

$\bf{\underline{\blue{Given:-}}}$

• $\bf\:{m_1}={10}^{-3}\: kg$

• $\bf\:{m_2}=100\:kg$

• $\bf\:{v_1}=800\:m/s$

$\bf{\underline{\green{To\:Find:-}}}$

• $\bf\:{v_2}=\:?$

$\bf{\underline{\orange{Solution:-}}}$

$\bf\implies\:{m_1}{v_1}={m_2}{v_2}$

$\bf\implies\:{v_2}=\dfrac{m_1 v_1}{m_2}$

$\bf\implies\:{v_2}=\dfrac{(10\times\:{10}^{-3})\times\:800}{100}$

$\bf\green{{\implies\:{v_2}=-0.08\:m/s}}$

$\bf\pink{{\therefore}}$ The required velocity is -0.08 m/s.

then,

$\bf\:F=\dfrac{dp}{dt}$

$\bf\implies\:f=\dfrac{10\times\:10\times\:{10}^{-3}\times\:800}{5}$

$\bf\blue{{\implies\:f=16\:N}}$

so, correct option is (b) -0.08 m/s; 16 N