Question

A bullet of mass 12 grams enters a piece of wood travelling at a speed of 500 m/s. It exits from the piece of wood at a speed of 190 m/s. Calculate the average force exerted by the wood on the bullet if the thickness of the wood is 5 cm.

Answer 1

Task. A bullet of mass $m=12$ grams enters a piece of wood travelling at a speed of $v_{0}=500$ m/s. It exits from the piece of wood at a speed of $v_{1}=190$ m/s. Calculate the average force exerted by the wood on the bullet if the thickness of the wood is $d=5$ cm.

Solution. Assume that the bullet moved with constant acceleration $a$. Then the average force will be equal to $F=ma$.

Let $t$ be the time taken for the bullet to come through the wood. Then

$a=\frac{v_{1}-v_{0}}{t},$

whence

$v_{1}-v_{0}=at.$

On the other hand, since the bullet is moved with constant acceleration, the distance $d$ is equal to

$d=v_{0}t+\frac{at^{2}}{2}=v_{0}t+\frac{t}{2}\cdot at=v_{0}t+\frac{t}{2}\cdot(v_{1}-v_{0})=v_{0}t+\frac{v_{1}t}{2}-\frac{v_{0}t}{2}=\frac{(v_{0}+v_{1})t}{2}.$

Hence

$t=\frac{2d}{v_{0}+v_{1}}.$

Therefore

$a=\frac{v_{1}-v_{0}}{t}=\frac{(v_{1}-v_{0})(v_{0}+v_{1})}{2d}=\frac{v_{1}^{2}-v_{0}^{2}}{2d},$

and the average force is equal to

$F=ma=m\frac{v_{1}^{2}-v_{0}^{2}}{2d}.$

Substituting values we obtain the average force:

$F=0.012*\frac{190^{2}-500^{2}}{2*0.05}=\frac{0.012*(-213900)}{0.1}=-25668N.$

The sign “$-$” means that the direction of force is opposite to the velocity.