Abullet with a mass of 3.0 gram, moving at speed of 300 m/s, hits and embeds itself in the trunk of a tree. the tree exerts an average force of 250 n on the bullet. what is the distance the bullet travels inside the trunk?

Explanation:

We have equation for force

          Force = Mass x Acceleration

          Force = 250 N

         Mass = 3 g = 3 x 10⁻³ kg  

         250 = 3 x 10⁻³ x Acceleration

         Acceleration = 83333.33 m/s²

Deceleration of bullet is 83333.33 m/s².

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s  

Acceleration, a = -83333.33 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 300² + 2 x -83333.33 x s

s = 0.54 m  

Distance the bullet travels inside the trunk is 0.54 m  

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s