Question

A bullet of mass 50 g moving with an initial velocity of 100 m sl, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it.Calculate : (i) initial momentum of the bullet, (ii) final momentum of the bullet, (iii) retardation caused by the wooden block, and (iv) resistive force
exerted by the wooden block

Ans. (i) 5 kg m s-1, (ii) zero, (iii) 2.5 x 105 m s-2, (iv) 12500 N

I gave the answer also but I need explanation please do fast​

Answer 1

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 :

a] initial momentum of the bullet

b]final momentum of the bullet

c]retardation caused by the wooden block

d] resistive force exerted by the wooden block

Given :

• Mass of the bullet = 50 g
• Initial velocity of the bullet = 100 m/s
• Final velocity = 0 m/s
• Distance travelled = 2 cm

To Find :

• Initial momentum of the bullet
• Final momentum of the bullet
• retardation caused by wooden block
• resistive force exerted by the wooden block

Solution :

a] Intial momentum of a body is given by ,

$\large \boxed {\rm{P_i = mu}}$

Where ,

• m is mass of the body
• u is initial velocity

We have ,

• m = 50g = 0.05 kg
• u = 100 m/s

$: \implies \rm \: P_i= (0.05 \: kg)(100 \: m {s}^{ - 1} ) \\ \\ : \implies \rm \: P_i = 5 \: kg.ms {}^{ - 1}$

∴ The initial momentum of the bullet is 5 kg.ms⁻¹

━━━━━━━━━━━━━━━━━━

b] Final momentum of the bullet

Final momentum of a body is given by ,

$\large \boxed {\rm{P_f = mv}}$

Where ,

• $\sf{P_{f}}$ is Final momentum of the body
• m is mass of the body
• v is final velocity of the body

We have ,

• m = 0.05 kg
• v = 0 m/s

$: \implies \rm \: \: P_f = (0.05 \: kg)(0 \: m {s}^{ - 1} ) \\ \\ : \implies \rm \: P_f = 0 \: kgm {s}^{ - 1}$

∴ The Final momentum of the bullet is 0 kgm/s

━━━━━━━━━━━━━━━━━━

c] Retardation caused by block

we have ,

• v = 0 m/s
• u = 100 m/s
• s = 2 cm = 0.02 m

Now using second equation of motion ,

$\boxed {\rm{v {}^{2} - {u}^{2} = 2as }}$

$: \implies \rm \: (0) {}^{2} - (100) {}^{2} = 2a(0.02) \\ \\ : \implies \rm \: - 10000 = 0.04a \\ \\ : \implies \rm \: a = - \frac{10000}{0.04} \\ \\ : \implies \rm \: a = - 250000 \: m {s}^{ - 2}$

∴ The retardation caused by wooden block is -2,50,000 m/s²

━━━━━━━━━━━━━━━━━━━

d] Resistive force exerted by wooden block

Resistive Force exerted by wooden block is ,

$\large \boxed {\rm{F = ma}}$

We have ,

• m = 0.05 kg
• a = 2,50,000 m/s²

$: \implies \rm \: F = (0.05)( - 250000) \\ \\ : \implies \rm \: F = - 12500 \: N$

∴ The Resistive Force exerted by bullet is -12500 N

Answer 2

Answer:

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 :

a] initial momentum of the bullet

b]final momentum of the bullet

c]retardation caused by the wooden block

d] resistive force exerted by the wooden block

Given :

Mass of the bullet = 50 g

Initial velocity of the bullet = 100 m/s

Final velocity = 0 m/s

Distance travelled = 2 cm

To Find :

Initial momentum of the bullet

Final momentum of the bullet

retardation caused by wooden block

resistive force exerted by the wooden block

Solution :

a] Intial momentum of a body is given by ,

\large \boxed {\rm{P_i = mu}}

P

i

=mu

Where ,

m is mass of the body

u is initial velocity

We have ,

m = 50g = 0.05 kg

u = 100 m/s

\begin{gathered} : \implies \rm \: P_i= (0.05 \: kg)(100 \: m {s}^{ - 1} ) \\ \\ : \implies \rm \: P_i = 5 \: kg.ms {}^{ - 1} \end{gathered}

:⟹P

i

=(0.05kg)(100ms

−1

)

:⟹P

i

=5kg.ms

−1

∴ The initial momentum of the bullet is 5 kg.ms⁻¹

━━━━━━━━━━━━━━━━━━

b] Final momentum of the bullet

Final momentum of a body is given by ,

\large \boxed {\rm{P_f = mv}}

P

f

=mv

Where ,

\sf{P_{f}}P

f

is Final momentum of the body

m is mass of the body

v is final velocity of the body

We have ,

m = 0.05 kg

v = 0 m/s

\begin{gathered} : \implies \rm \: \: P_f = (0.05 \: kg)(0 \: m {s}^{ - 1} ) \\ \\ : \implies \rm \: P_f = 0 \: kgm {s}^{ - 1} \end{gathered}

:⟹P

f

=(0.05kg)(0ms

−1

)

:⟹P

f

=0kgms

−1

∴ The Final momentum of the bullet is 0 kgm/s

━━━━━━━━━━━━━━━━━━

c] Retardation caused by block

we have ,

v = 0 m/s

u = 100 m/s

s = 2 cm = 0.02 m

Now using second equation of motion ,

\boxed {\rm{v {}^{2} - {u}^{2} = 2as }}

v

2

−u

2

=2as

\begin{gathered} : \implies \rm \: (0) {}^{2} - (100) {}^{2} = 2a(0.02) \\ \\ : \implies \rm \: - 10000 = 0.04a \\ \\ : \implies \rm \: a = - \frac{10000}{0.04} \\ \\ : \implies \rm \: a = - 250000 \: m {s}^{ - 2} \end{gathered}

:⟹(0)

2

−(100)

2

=2a(0.02)

:⟹−10000=0.04a

:⟹a=−

0.04

10000

:⟹a=−250000ms

−2

∴ The retardation caused by wooden block is -2,50,000 m/s²

━━━━━━━━━━━━━━━━━━━

d] Resistive force exerted by wooden block

Resistive Force exerted by wooden block is ,

\large \boxed {\rm{F = ma}}

F=ma

We have ,

m = 0.05 kg

a = 2,50,000 m/s²

\begin{gathered} : \implies \rm \: F = (0.05)( - 250000) \\ \\ : \implies \rm \: F = - 12500 \: N\end{gathered}

:⟹F=(0.05)(−250000)

:⟹F=−12500N

∴ The Resistive Force exerted by bullet is -12500 N