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; 16 N(c) – 0.8 ms⁻¹; 8 N(d) –1.6 ms⁻¹; 16 N

Answer 1

see the attachment for the answer

Answer 2

Answer:

0.08 m/s and 16 Newton

Explanation:

$m_1$ = Mass of bullet = 0.01 kg

$m_2$ = Mass of rifle and person = 100 kg

$v_1$ = Final Velocity of bullet = 800 m/s

$v_2$ = Final Velocity of rifle

Since both the bullet and rifle start from rest

$m_1v_1=m_2v_2\\\Rightarrow v_2=\frac{m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.01\times 800}{100}\\\Rightarrow v_2=0.08\ m/s$

The velocity of the person is 0.08 m/s

t = Time taken = 5 s

n = Number of bullets = 10

u = Initial velocity = 0

Impulse

$Ft=m(v_1-u)\ n\\\Rightarrow F=\frac{m_1(v_1-u)}{t}\\\Rightarrow F=\frac{10\times 10^{-3}\times (800-0)}{5}\times 10\\\Rightarrow F=16\ N$

Force on the person is 16 Newton