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. -0.8 ms^-1 ; 16 N
• B. -0.8 ms^-1 ; 8 N
• C. -1.6 ms^-1 ; 16 N
• D. -1.6 ms^-1 ; 8 N

Answer 1

Answer: (A)

$
P (\text { initial })= P (\text { final })
$
$
0= n \times m \times u +( M - n \times m ) \times v
$
where: $n =10, m =10 g =0.01 kg , u =800 m / s , M =100 kg$
$
0=10 \times 0.01 kg \times 800 m / s +(100 kg -10 \times 0.01 kg ) \times v
$
$
v =-\frac{80 kgm / s }{99.9 kgm / s }
$
$
v =-0.8 m / s
$
Then $: F =\frac{\Delta P }{\Delta t }=\frac{10 \times 0.01 kg \times 800 m / s }{5 s }=16 N$