Question

A 2.0-gram bullet is shot into a tree stump. It enters at a speed of 3.00 x 104 cm/s and comes to rest after having penetrated 0.05 m in to the stump. What was the average force during the impact? Show all calculations leading to an answer.

Answer 1

$$
r = r_0 + \left( \frac{v+v_0}{2}\right )t
$$


Where \(r_0=0,v_0=3(10)^4~cm/s\).

Solve the above for t and insert into @Irishboy123's equation above for \(t_2-t_1\)

$$
F=m{v-v_0\over t}
$$
Where \(v\) is the final velocity and is equal to zero since the bullet stopped after penetration.

Don't forget to convert cm to meters.