Question

18. 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

@imqwerty can u please help!!

Answer 2

@imqwerty can u please help!!

Answer 3

alright :)
so when the bullet will be like penetrating into the tree stump it will keep slowing down. why? because the tree stump will apply some force on it which will slow it down

alright so initial speed of bullet was was this-> \(u=3 \times 104\) cm/s
we need to take this speed in meters/seconds so we divide it by 100
so
\(u= \Large \frac{3 \times 104}{100}\) m/sec

and we have distance traveled \(S=0.05\) meters

Lets say that it deaccelerates with \(a\) m/sec^2

and it stops at end so final velocity \(v=0\) m/s

now we will use this equation to find "\(a\)" -> \(v^2-u^2=2aS\)

substituting the values we have we get this->

\(0^2- \Large \left(\frac{3 \times 104}{100}\right ) =2 \times a \times 0.05\)
solving we will get \(a=-97.344\) m/s^2

here the negative sign denotes that the bullet is deaccelerating

now we gotta find the force

force is given by this formula-> \(Force=mass \times acceleration\)

here the mass must be in kilograms
and acceleration in meters/sec^2

we got \(a=-97.344\)m/s^2
and mass of bulllet is=2 gram which is-> \(\Large \frac{2}{1000}\) kilograms

now we put these into the force equation we get this->

\(Force= \Large \frac{2}{1000} \times -97.344\)

\(F=-0.194688 ~Newtons\)

here the negative sign of force denotes that it is acting in opposite direction

:)

Answer 4

ok thanx

Answer 5

np :)