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?

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?

anonymous · Answer

@TheSmartOne ?

thesmartone · Answer

$$\large\sf Force = \frac{ Mass \times (Final ~Velocity - Initial ~Velocity) }{ Change~in~time }$$

anonymous · Answer

force=unknown; mass = 2.0 grams; final velocity = 0; initial velocity = 3.00x104cm/2; change in time = 0.05 ?

thesmartone · Answer

So they gave us that

$\sf Mass = 2.0 grams\
Initial~Velocity = 3.00 \times 10^4 cm/s \
Final~Velocity = 0$

thesmartone · Answer

I'm stumped on the time part... Hmmm

thesmartone · Answer

Idk if we are using the correct formula either :/

anonymous · Answer

the struggle is real...

thesmartone · Answer

This might help you I think :P https://answers.yahoo.com/question/index?qid=20150128075918AAoAJrs

anonymous · Answer

do you know anyone that might know how to help me solve this?

thesmartone · Answer

You can maybe make a new post and I'll tag some users. They always come to my tag faster than another user since they know me. Haha