Question

A 75 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 620 m/s . Part A What is the recoil speed of the hunter?
Express your answer to two significant figures and include the appropriate units.

Answer 1

To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

m = mass of each object

u = Initial velocity of each object

v = Final velocity of each object

Our values are,

$\text{Mass of hunter} ) m_1 = 75 kg = 75000 g$

$\text{Mass of bullet} = m_2 = 42 g$

$\text{Speed of bullet} = v_2 = 6200 m/s$

$\text{Recoil speed of hunter} =v_1 = ?$

Applying our considerations we have that the above formula would become,

$m_1v_1 + m_2v_2 = 0$

Solving for speed 1,

$v_1 = - \frac{m_2v_2}{m_1 }$

$v_1 = - \frac{42 * 620}{75000 }$

$v_1 = - 0.3472 m/s$

Therefore  the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.

Answer 2

The recoil speed of the hunter standing on the frictionless ice is 0.35m/s².

Given the data in the question;

• Mass of hunter; $m_1 = 75kg$

• Mass of bullet; $m_2 = 42g = 0.042kg$

• Velocity of bullet; $v_2 = 620m/s$

Recoil speed of the hunter; $v_1 =\ ?$

From the law of conservation of momentum:

$MV = mv$

Where m is mass and v is velocity

So

$m_1v_1 = m_2v_2\\\\v_1 = \frac{m_2v_2}{m_1}$

We substitute our given values into the equation

$v_1 = \frac{ 0.042kg \ *\ 620m/s^2}{75kg} \\\\v_1 = \frac{26.04kg.m/s^2}{75kg} \\\\v_1 = 0.35m/s^2$

Therefore, the recoil speed of the hunter standing on the frictionless ice is 0.35m/s².

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