Question

A bullet of mass 50 g travelling with a speed of 15ms penetrates into a

Answer 1

Answer:

a) The bullet will penetrate 0.375 meters into the bag of sand.

b) The average force exerted by the sand is 15 newtons.

Explanation:

The statement is incorrectly written. The correct form is presented below:

A bullet of mass 50 grams travelling with a speed of 15 meters per second penetrates into a bag of sand and is uniformly brought to rest in 0.05 second.

a) How far the bullet will penetrate into the bag of sand?

b) The average force exerted by the sand?

b) The average force exerted by the sand on the bullet ($F$), measured in newtons, is determined by Principle of Linear Momentum Conservation and Impulse Theorem:

$F = \frac{m\cdot (v_{f}-v_{o})}{\Delta t}$ (1)

Where:

$m$ - Mass of the bullet, measured in kilograms.

$v_{o}$, $v_{f}$ - Initial and final speeds of the bullet, measured in meter per second.

$\Delta t$ - Impact time, measured in seconds.

If we know that $m = 0.05\,kg$, $v_{o} = 15\,\frac{m}{s}$, $v_{f} = 0\,\frac{m}{s}$ and $\Delta t = 0.05\,s$, then the average force exerted by the sand is:

$F = \frac{(0.05\,kg)\cdot \left(0\,\frac{m}{s}-15\,\frac{m}{s} \right)}{0.05\,s}$

$F = -15\,N$

The average force exerted by the sand is 15 newtons.

a) The distance travelled by the bullet ($\Delta s$), measured in meters, is determined by application of Principle of Energy Conservation and Work-Energy Theorem:

$\Delta s = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot F }$ (2)

If we know that $m = 0.05\,kg$, $v_{o} = 15\,\frac{m}{s}$, $v_{f} = 0\,\frac{m}{s}$ and $F = -15\,N$, then the distance travelled by the bullet is:

$\Delta s = \frac{(0.05\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(15\,\frac{m}{s} \right)^{2}\right]}{2\cdot (-15\,N)}$

$\Delta s = 0.375\,m$

The bullet will penetrate 0.375 meters into the bag of sand.