Question

A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate:-

(i) Retardation caused by the wooden block

(ii) Resistance force exerted by the wooden block​

Answer 1

Given :

• Mass of bullet, m = 50 g = 50/1000  kg = 0.05 kg
• Initial velocity of bullet, u = 100 m/s
• Final velocity of bullet, v = 0
• Distance covered by bullet, s = 2 cm = 2/100  m = 0.02 m

To find :

• Retardation caused by the wooden block = ?
• Resistance force exerted by the wooden block = ?

Formulae required :

• Third equation of motion

     2 a s = v² - u²

[ Where a is acceleration, s is distance covered, v is final velocity and u is initial velocity ]

• Expression of Newton's second law

    F = m a

[ Where F is force, m is mass and a is acceleration ]

Solution :

Let, acceleration of bullet be a

then, Using third equation of motion

→ 2 a s = v² - u²

→ 2 a ( 0.02 ) = ( 0 )² - ( 100 )²

→ 0.04 a = - 10000

→ a = - 10000 / 0.04

→ a = -250000  m/s²

Since, acceleration of bullet is -250000 m/s² therefore

• Retardation of bullet would be 250000 m/s².

Now, Let force exerted by bullet on Wooden block be F, then magnitude of Force exerted by wooden block on bullet will be the same as F (and direction would be opposite), since it came to rest. so

Using Newton's second law expression

→ F = m a

→ F = ( 0.05 ) ( -250000 )

→ F = -12500 N

so, Since magnitude of Force exerted by bullet on block is 12500 N therefore,

• Resistive force exerted by wooden block would be 12500 Newtons.

Answer 2

Given :

• mass of bullet= m= 50g
• initial velocity = u =100 m/s
• final velocity = v =0m/s
• distance penetrated =s= 2 cm

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Need to Find:

• Retardation =?
• Resistant force exerted by wooden block =?

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SOLUTION :

given

s= 2cm

⟹s = 2/100 m

⟹ s= 0.02 m

We know from the 3rd equation of motion,

${v}^{2} - {u}^{2} = 2as$

$\implies \: {0}^{2} - {100}^{2} = 2 \times a \times 0.02$

$\implies \: 2 \times a \times 0.02 = - 10000$

$\implies \: a = \dfrac{ - 10000 \: m}{2 \times 0.02 \: {s}^{2} }$

$\implies \: a = - 250000m/ {s}^{2}$

Thus the acceleration is equal to $-250000 m/s^{2}$

• Thus$\red{\bold{retardation=250000m/s^{2}}}$

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Mass = 50g

⟹ Mass= 50/1000 kg

⟹ Mass= 0.05 g

we know,

Force = Mass × Acceleration

⟹Force = 0.05× (-250000) N

⟹ Force = -12500N

• Thus $\red{\bold{12500N\:of\:Force\:}}$ is applied $\red{in \:the \:direction\:opposite}$$\red{to\:motion\:of\:bullet}$