Question

6. A bullet of mass 5 g travelling horizontally with a
velocity of 150 m/s strikes a stationary wooden
block and comes to rest in 0.03 s. Calculate the
magnitude of the force exerted by the wooden
block on the bullet.
Ans 25 NT​

Answer 1

Given -

• Mass of bullet = 5 g
• Initial velocity = 150 m/s
• Final velocity = 0 m/s
• Time = 0.03 s

To find -

• Magnitude of force.

Solution -

• v = u + at

Substitute the given values. To find acceleration.

⇛ v = u + at

⇛ 0 = 150 + a × 0.03

⇛ -150 = a × 0.03

⇛ -150/0.03 = a

⇛ 5000 m/s²

• Acceleration = 5000 m/s²

⇛ Mass = 5 g

As we know,

1 g = 0.001 kg

So, Mass = 0.005 kg

• F = ma

Substitute the given values.

⇛ F = 0.005 × 5000

⇛ F = 25 N

》 Force = 25 newton

Answer 2

To Find:-

we should calculate the magnitude of force penetrated by the bullet into the block.

Solution:-

Given That:-

⇒$\sf u = 150m/s$

⇒$\sf v = 0$

⇒$\sf t = 0.03 sec$

⇒$\sf m = 10 g$

⇒$\sf m = 0.005 kg$

We know That :-

S = $\rm \dfrac{( v² - u²)}{2a}$...............(1)

Formula to calculate acceleration

$\sf \implies v = u + at$

a = $\rm \dfrac{( v-u )}{t}$

a = $\rm \dfrac{( 0 - 150)}{0.03}$

$\sf a = -5000 m/s²$...............(2)

Substituting equation ( 2 ) in ( 1 ) gives ,

S = $\rm \dfrac{ 0- (150)²)}{- ( 2 x 5000 )}$...............(1)

S = $\rm \dfrac{22500}{10000}$

$\sf S = 2.25m$

therefore, the distance penetrated by the bullet is 2.25 m

to calculate force ,we know

F = m x a 

⇒ F = 0.005 x 5000

⇒ F = 25 N 

the force exerted by the wooden block on the bullet  is 25 N