Solved: "A rifle with a weight of 30 N..."

High School
Physics
Force, Motion and Energy
Acceleration Due to Gravity

Reggie

Answered question

2020-11-01

A rifle with a weight of 30 N fires a 5.0 g bullet with a speed of 300m/s. (a) Find the recoil speed of the rifle. (b) If a700 N man holds the rilfe firmly against his shoulder, find there coil speed of the man and his rifle.

Answer & Explanation

Nathanael Webber

Skilled2020-11-02Added 117 answers

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

initially both gun and bullet are at rest

m_{1} v_{1} + m_{2} v_{2} = 0

v_{1} = - \frac{m_{2} v_{2}}{m_{1}}

= \frac{.5 \cdot 300 \cdot 9.8}{30} = 49.0 \frac{m}{s}

700 + 30 = 730 N = \frac{730}{9.8 k g}

v_{1} = \frac{.5 \times 300 \times 9.8}{730} = 2.01369 \frac{m}{s}

Jeffrey Jordon

Expert2021-10-12Added 2605 answers

When the starting momentum and the end momentum are identical, there is no change in the momentum of an isolated system. Without interacting with the outside world, the particles communicate within the system. The momentum is conserved and we use the law of conservation of momentum

$p_{f} = p_{i}$

$m_{r} = \frac{w}{g} = \frac{30 N}{9.8 m / s^{2}} = 3.06 k g$

The initial momentum of the system is zero

$p_{f} = 0$

$m_{r} v_{r} + m_{b} v_{b} = 0$

$v_{r} = \frac{- m_{b} v_{b}}{m_{r}}$

Now, we plug the values for $m_{b}, v_{b} a n d m_{r}$ into equation (2) to get $v_{r}$

$v_{r} = \frac{- m_{b} v_{b}}{m_{r}}$ $= \frac{- (0.005 k g) (300 m / s)}{3.06 k g}$ =- 0.49 m/s

The velocity of the rifle is $0.49 m / s$ in the opposite direction of the bullet.

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