Question

A bullet of mass 120g is fired horizontally into a fixed wooden
block with a speed of 20m/s. The bullet is brought to rest in the
block in 0.1s by a constant resistance Calculate the (1)
magnitude of the resistance (ii) distance moved by the bullet in
the wood​

Answer 1

Answer:

hey mate here is ur ans

Explanation:

m = mass of bullet= 120g

speed of bullet initial u = 20m/s

then final speed become = 0

time of bullet = 0.1 s

acceleration = v-u /t = 0-20 / 0.1 = -200m/s^2

s = ut + 1/2 at^2

s= 20×0.1+1/2 ×(-200) ×(0.1)^2

s = 2-1 = 1m

force of bullet= F = ma = 120×200 = 24000gm/s

resistance of block by 0.1s = 24000×0.1= 2400gm

hope it help u

Answer 2

$\huge \fbox{ \underline \purple{Answer }}$

Let us write the given data first

GIVEN:

✧ Mass of the Bullet(m) = 120 g = 120/1000 = 0.12 kg

✧ Initial speed of the bullet (u) = 20 m/s

✧ Time taken to come to rest (t) = 0.1 sec

✧ Final velocity (v) = 0 m/s

{HINT: They are given as "The bullet is brought to rest" then final velocity is 0}

TO FIND:

(i)   magnitude of the resistance { means acceleration}

(ii) distance moved by the bullet in  the wood​ (s) =?

HERE:

✯ m = mass of the travelling bullet

✯ u  =  initial velocity of the bullet

✯ v  =  final velocity of the bullet

✯ t  =  time travelled by the bullet

✯ a  = magnitude of resistance of the bullet

✯ s  = distance travelled by the bullet

$\huge \fbox{ \underline \green{Solution: }}$

${\huge{\underline{\small{\mathbb{\blue{ (i) Magnitude of Resistance }}}}}}$

✯ v = u + at

✯ a = v - u / t

✯ a = 0 - 20/0.1

✯ a = -200 m/s

$\huge \fbox{ \underline \orange{a = -200 m/s }}$ is the magnitude of resistance of the bullet

✪  Force of bullet(F) = ma = 120×200 = 24000 gm/s

✪  Resistance of block by 0.1 s in grams  = 24000×0.1= 2400 gms

${\huge{\underline{\small{\mathbb{\blue{ (ii) Distance Moved by bullet: }}}}}}$

$\huge \fbox{ \underline \purple{✪ Method 1✪ }}$

✪ v² -u² = 2as

✪ (0)² - (20)² = 2 × -200 × s

✪ -400 = -400 × s

✪ S = 1 metre

$\huge \fbox{ \underline \purple{✪ Method 2✪ }}$

✪ S = ut +$\frac{1}{2}$ at²

✪ S = 20 × 0.1 +$\frac{1}{2}$ × -200 × (0.1)²

✪ S = 2 + -100 × 0.01

✪ S = 2 + (-1)

✪ S = 2-1

✪ S = 1 Metre

$\huge \fbox{ \underline \red{S = 1 m }}$ is the distance travelled by bullet after firing

$\underline{\underline{ \sf \huge \red{✪Extra Bytes✪}}}$

❆ Use the newtons 3 laws of motion accornding to given quantities and find the answers . They are,

❆ v = u + at  

❆ s = ut + 1/2 at²

❆ v² - u² = 2as

❆ Sn = u + $\frac{a}{2}$ (2n -1) used to find time in nth seconds

❆ With these formulas you can apply the given quantities and find the answers

Hope it helps

Thanks for asking:-)