Question

Subject :— Physics


Question :—
A bullet of mass 20 g moving with a speed of 120 m/s hits a thick muddy wall and
penetrates into it. It takes 0.03 seconds to stop in the wall. Find :—
(a) the acceleration of the

bullet in the wall
(b) the force exerted by the wall on the bullet,
(c) the force exerted by the

bullet on the wall and
(d) the distance covered by the bullet in the wall.

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Answer 1

Question :-

A bullet of mass 20 g moving with a speed of 120 m/s hits a thick muddy wall and

penetrates into it. It takes 0.03 seconds to stop in the wall. Find :—

(a) the acceleration of the

bullet in the wall

(b) the force exerted by the wall on the bullet,

(c) the force exerted by the

bullet on the wall and

(d) the distance covered by the bullet in the wall.

____________________

Solution :-

Given Information :-

• Mass of the bullet ➢ 20g or
• Mass of the bullet in ( kg ) ➢ 0.02kg
• Initial velocity of the bullet ( u ) ➢ 120m/s
• Time taken by the bullet ➢ 0.03 seconds
• Final velocity of the bullet ( v ) ➢ 0m/s (as the bullet penetrates so it's velocity is taken 0m/s)

To Find :-

(a) the acceleration of the bullet in the wall

(b) the force exerted by the wall on the bullet

(c) the force exerted by the bullet on the wall and

(d) the distance covered by the bullet in the wall

Formulae Used :-

• $\sf Acceleration = \frac{(v - u)}{t}$
• $\sf force = mass \times acceleration$
• $\sf distance \: covered = speed \times time \: taken$

Calculation :-

( a ) The acceleration of the bullet in the wall

Substituting the given values in the formula of acceleration, We get,

$\sf Acceleration = \frac{(v - u)}{t}$

$\sf : \implies acceleration = \frac{0 - 120}{0.0}$

$\sf: \implies acceleration = \frac{ - 120}{0.03}$

$\sf: \implies acceleration = \red{ \bf{ - 4000m/ {s}^{2} }} \\ \bf{ \underline{or }}\\ \sf retardation = \red{ \bf{ 4000m/ {s}^{2} }}$

__________

( b ) The force exerted by the wall on the bullet

Substituting the values in the formula of force, We get,

: ⇒Force = mass × acceleration

: ⇒Force exerted by wall on bullet = 0.02kg × 4000m/s²

: ⇒80 Newtons

__________

( c ) the force exerted by the bullet on the wall

Force exerted by the bullet on wall = 80 Newtons. Because every action has an equal and opposite reaction.

__________

( d ) the distance covered by the bullet in the wall

Substituting the values in the formula for Distance, We get,

: ⇒Distance covered = speed × time taken

: ⇒120m/s × 0.03 seconds

: ⇒3.6m

Therefore, the distance covered by the bullet = 3.6m

__________

Final Answers :-

(a) Acceleration = -4000m/s or Retardation = 4000m/s

(b) 80 Newtons

(c) 80 Newtons

(d) 3.6m

____________________

Answer 2

Explanation:

Question :-

A bullet of mass 20 g moving with a speed of 120 m/s hits a thick muddy wall and

penetrates into it. It takes 0.03 seconds to stop in the wall. Find :—

(a) the acceleration of the

bullet in the wall

(b) the force exerted by the wall on the bullet,

(c) the force exerted by the

bullet on the wall and

(d) the distance covered by the bullet in the wall.

____________________

Solution :-

Given Information :-

Mass of the bullet ➢ 20g or

Mass of the bullet in ( kg ) ➢ 0.02kg

Initial velocity of the bullet ( u ) ➢ 120m/s

Time taken by the bullet ➢ 0.03 seconds

Final velocity of the bullet ( v ) ➢ 0m/s (as the bullet penetrates so it's velocity is taken 0m/s)

To Find :-

(a) the acceleration of the bullet in the wall

(b) the force exerted by the wall on the bullet

(c) the force exerted by the bullet on the wall and

(d) the distance covered by the bullet in the wall

Formulae Used :-

\sf Acceleration = \frac{(v - u)}{t}Acceleration=

t

(v−u)

\sf force = mass \times accelerationforce=mass×acceleration

\sf distance \: covered = speed \times time \: takendistancecovered=speed×timetaken

Calculation :-

( a ) The acceleration of the bullet in the wall

Substituting the given values in the formula of acceleration, We get,

\sf Acceleration = \frac{(v - u)}{t}Acceleration=

t

(v−u)

\sf : \implies acceleration = \frac{0 - 120}{0.0}:⟹acceleration=

0.0

0−120

\sf: \implies acceleration = \frac{ - 120}{0.03}:⟹acceleration=

0.03

−120

\begin{gathered} \sf: \implies acceleration = \red{ \bf{ - 4000m/ {s}^{2} }} \\ \bf{ \underline{or }}\\ \sf retardation = \red{ \bf{ 4000m/ {s}^{2} }}\end{gathered}

:⟹acceleration=−4000m/s

2

or

retardation=4000m/s

2

__________

( b ) The force exerted by the wall on the bullet

Substituting the values in the formula of force, We get,

: ⇒Force = mass × acceleration

: ⇒Force exerted by wall on bullet = 0.02kg × 4000m/s²

: ⇒80 Newtons

__________

( c ) the force exerted by the bullet on the wall

Force exerted by the bullet on wall = 80 Newtons. Because every action has an equal and opposite reaction.

__________

( d ) the distance covered by the bullet in the wall

Substituting the values in the formula for Distance, We get,

: ⇒Distance covered = speed × time taken

: ⇒120m/s × 0.03 seconds

: ⇒3.6m

Therefore, the distance covered by the bullet = 3.6m

__________

Final Answers :-

(a) Acceleration = -4000m/s or Retardation = 4000m/s

(b) 80 Newtons

(c) 80 Newtons

(d) 3.6m

____________________