A bullet of mass 6.5 g that is moving at a speed of 460 m/s embeds itself into a tree, penetrating 24.5 cm into the wood.

Question

What is the magnitude of the average force, in N, exerted by the tree on the bullet? 

Answer 1

Hi Sara,

Looking at Newton's equation F = ma, we will have to find "F", or force. We know the mass "m" since it's given to us. Therefore, there's two steps: we must first calculate acceleration "a" and then we can solve for average force F.

Step 1

We can find acceleration using the kinematics equation V_f² = V₀² + 2a∆s where ∆s is distance. Final velocity V_f is zero since the bullet comes to a stop. Initial velocity is given to us as 460 m/s. ∆s is given to us as 24.5 cm but we must convert this to meters to be consistent – we can input 0.245 m. This gives us a = - 431, 837 m/s². We have a negative sign since the bullet is decelerating.

Step 2

We can use F = ma and multiply m x a to get F. Be very careful to convert mass to kg for this equation to be consistent! 6.5g is 0.0065kg! Multiplying 0.0065 kg x -431,837 m/s² should yield a force of -2,807 N. The reason we have a negative is consistent with the fact that the force is moving in the opposite direction of the bullet in order to slow it down. Since you are looking for just the magnitude and not direction, you can ditch the negative and your answer is 2,807 N.

I hope this helped!