Question

A 50 kg archer, standing on frictionless ice, shoots a 100 g arrow at speed 100 m/s. what is the recoil speed of the archer?

Answer 1

Answer: The recoil speed of the archer is -0.2 m/s

Explanation:

To calculate the recoil speed of the archer, we use the equation of law of conservation of momentum, which is:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,  

$m_1$ = mass of archer = 50 kg

$u_1$ = Initial velocity of archer = 0 m/s

$v_1$ = Final velocity (recoil speed) of archer = ? m/s

$m_2$ = mass of arrow = 100 g = 0.100 kg   (Conversion factor:  1 kg = 1000 g)

$u_2$ = Initial velocity of arrow = 0 m/s

$v_2$ = Final velocity of arrow = 100 m/s

Putting values in above equation, we get:

$(50\times 0)+(0.100\times 0)=(50\times v_1)+(0.100\times 100)\\\\v_1=\frac{-(0.100\times 100)}{50}=-0.2m/s$

Hence, the recoil speed of the archer is -0.2 m/s