Question

A bullet of mass
mars 7 kg is fined into a
block of metal weighing
metal weighing 7 kg. the block
is frill to move . After the impact, the
velocity
of the bullet and the block is
70 ms What is the initial velocity of
the bullet?​

Answer 1

Answer:

Speed of the bullet before the impact is 700.7 m/s.

Explanation:

Conservation of linear momentum:

The conservation of linear momentum is defined as the initial momentum before the impact and the final momentum after the impact is conserved, if the net work done by the external forces on the system of masses should be zero.

Given,

Mass of the bullet = m_1\ =\ 7\ g\ =\ 0.007\ kgm

1

= 7 g = 0.007 kg

Mass of the block = m_2\ =\ 7\ kgm

2

= 7 kg

Final speed of the block and the bullet = v\ =\ 0.7 m/s.v = 0.7m/s.

Let u be the initial speed of the bullet after the impact. And the block is initially at rest.

From the conservation of linear momentum,

Initial momentum of the system of masses = final momentum of the system of masses.

\begin{gathered}\therefore m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow m_1u\ +\ 0\ +\ (m_1\ +\ m_2)v\\\Rightarrow u\ =\ \dfrac{(m_1\ +\ m_2)v}{m_1}\\\Rightarrow u\ =\ \dfrac{(0.007\ +\ 7)0.7}{0.007}\\\Rightarrow u\ =\ 700.7\ m/s.\end{gathered}

∴m

1

u

1

+ m

2

u

2

= (m

1

+ m

2

)v

⇒m

1

u + 0 + (m

1

+ m

2

)v

⇒u =

m

1

(m

1

+ m

2

)v

⇒u =

0.007

(0.007 + 7)0.7

⇒u = 700.7 m/s.

Hence the initial speed of the bullet before the impact is 700.7 m/s.