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 5s. 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
( 1 ) - 1 . 6 ms ¯¹ : 8 N
( 2 ) - 0 . 08 ms¯¹ :16 N
( 3 ) - 0 . 8 ms¯¹ : 16 N
( 4 ) - 1 . 6 ms¯¹ : 16 N ​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\bf\huge\boxed{-0.80\:m/s,\: 16\:N}$

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$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Given —

→Mass of rifle and the man (M) = 100 kg

→mass of bullet (m) = 10 gm

→t = 5 sec

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By applying $\green{\texttt{Linear conservation of momentum}}$

Pi = Pf

⇨ mu1 + Mu2 = mv1 + Mv2

⇒ 0 + 0 = nmu + (M- nm) v

Here $\green{\texttt{initially velocity is zero }}$

⇒0 = 10 × 10/1000 × 800 + (100- 10× 10× 10¯³) v

On solving

⇒v = - 0.80 m/s

[-ive sign just indicating the direction is $\green{\texttt{ opposite}}$ to the direction of bullets]

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Now By

F = ΔP/Δt [average force]

⇒F = (10×10× 10¯³× 800)/5

⇒F = 16 N

Hence, option (3) is the right answer

It should be - 0.80 ms¯¹ : 16 N

Answer 2

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We have to use the Law of momentum conservation:

P ( initial ) = P ( final )

0 = n · m · u + ( M - n · m ) · v

where: 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

v = - 80 kgm/s / 99.9 kgm/s

v = 0.8 m/s

Then : F = Δ P / Δ t = ( 10 · 0.01 kg · 800 m/s ) : 5 s = 16 N

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

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