Question

A bullet moving with a velocity of 10 ms) is brought to rest after penetrating a wooden plank of 4 c thickness. The retardation of the bullet is [ ] a)-1200 ms? b)-1250 ms-2 c) 1200 ms-2 d) 1250 ms-2 ​

Answer 1

Given :-

Initial velocity of the bullet (u = 10 m/s)

Thickness of the wooden plank (s = 4 cm)

It came to rest (v = 0 m/s)

To find :-

Retardation of the bullet .

Solution :‐

Firstly lets find the acceleration of the bullet by using the equations of motion.

Suitable equation:‐

$\: \: \: \clubs \: \boxed{ \bf \: \pink{v {}^{2} - u {}^{2} = 2as }}$

Putting down the values,

Firstly let's convert 4cm - m

$\: \: \bigstar \bf \pink {1m \: = 100cm}$

$\implies \: \bf \: 4cm \: = \dfrac{4}{100} m$

$\implies \: \bf \: 4cm \: = 0.04m$

$\bf \: v {}^{2} - u {}^{2} = 2as$

$\implies \bf \: (0) {}^{2} - (10){}^{2} = 2(a)(0.04)$

$\implies \bf \: 0 - 100 = 0.08a$

$\implies \bf \: - 100 = \dfrac{8a}{100}$

$\implies \bf \: - 100 = \cancel{\dfrac{8a}{100} }$

$\implies \bf \: - 100 = {\dfrac{2a}{25} }$

$\implies \bf \: - 2500 = 2a$

$\implies \bf \: a = \cancel \dfrac{ - 2500}{2}$

$\implies \bf \: a = - 1250 \: ms {}^{ - 2}$

$\bf \pink{So, \: acceleration \: of \: the \: block \: is \: - 1250ms {}^{ - 2} }$

Acceleration is -1250 m/s²

Retardation is 1250 m/s²

So, the correct option is [d]

Know more :-

♣ Equations of motion:-

v = u + at

s = ut + ½ at²

v² - u² = 2as

where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

s = distance covered

Answer 2

Answer:

Initial velocity of the bullet (u = 10 m/s)

Thickness of the wooden plank (s = 4 cm)

It came to rest (v = 0 m/s)

To find :-

Retardation of the bullet .

Solution :‐

Firstly lets find the acceleration of the bullet by using the equations of motion.

Suitable equation:‐

\: \: \: \clubs \: \boxed{ \bf \: \pink{v {}^{2} - u {}^{2} = 2as }}♣

v

2

−u

2

=2as

Putting down the values,

Firstly let's convert 4cm - m

\: \: \bigstar \bf \pink {1m \: = 100cm}★1m=100cm

\implies \: \bf \: 4cm \: = \dfrac{4}{100} m⟹4cm=

100

4

m

\implies \: \bf \: 4cm \: = 0.04m⟹4cm=0.04m

\bf \: v {}^{2} - u {}^{2} = 2asv

2

−u

2

=2as

\implies \bf \: (0) {}^{2} - (10){}^{2} = 2(a)(0.04)⟹(0)

2

−(10)

2

=2(a)(0.04)

\implies \bf \: 0 - 100 = 0.08a⟹0−100=0.08a

\implies \bf \: - 100 = \dfrac{8a}{100}⟹−100=

100

8a

\implies \bf \: - 100 = \cancel{\dfrac{8a}{100} }⟹−100=

100

8a

\implies \bf \: - 100 = {\dfrac{2a}{25} }⟹−100=

25

2a

\implies \bf \: - 2500 = 2a⟹−2500=2a

\implies \bf \: a = \cancel \dfrac{ - 2500}{2}⟹a=

2

−2500

\implies \bf \: a = - 1250 \: ms {}^{ - 2}⟹a=−1250ms

−2

\bf \pink{So, \: acceleration \: of \: the \: block \: is \: - 1250ms {}^{ - 2} }So,accelerationoftheblockis−1250ms

−2

Acceleration is -1250 m/s²

Retardation is 1250 m/s²

So, the correct option is [d]

Know more :-

♣ Equations of motion:-

v = u + at

s = ut + ½ at²

v² - u² = 2as

where,

v = final velocity

u = initial velocity

a = acceleration

t = time taken

s = distance covered