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dedylja [7]
3 years ago
7

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

t is the recoil speed of the hunter?
Physics
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Where,m_{1} = mass of hunter

m_{2} = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

0+0=70\times v_{1}+0.042\times590

v_{1}=-\dfrac{0.042\times590}{70}

v=-0.354\ m/s

Hence, The recoil velocity is 0.354 m/s.

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