Question

A 10 g bullet travelling at 200 m/s strikes
And remains embedded in a 2kg . Target which is originlly at rest but free to move .At what speed does the target
Move off ?

Answer 1

$\huge{\boxed{\textsf{Answer :- }}}$

♦ In solving above's Question we will be using the Law of conservation of momentum .

♦ Law of conservation of momentum

>> This law states that Total momentum of a system remains constant if no net external force acts on the system.

♦ Now as given

• Mass of Bullet ($m_1$) = 10 g

• Mass of Target ($m_2$)= 2 Kg

• Initial velocity of Bullet($u_1$) = 200 m/s

• Initial velocity of Target($u_2$) = 0

♦ As the bullet is stuck in the target then the net mass $m_3$ = $m_1 + m_2$

= 2000 g + 10 g

= 2010 g

= 2.01 kg

>> Let the final velocity of Target and the bullet be = "$v$"

♦ Then by using Law of conservation of momentum.

$m_1u_1 + m_2u_2 = m_3v$

$\implies (0.01)(200) + (2)(0) = (2.01)(v)$

$\implies 2 + 0 = (2.01)v$

$\implies 2 = (2.01)v$

$\implies v = \dfrac{2}{2.01}$

$\implies v = 0.995 m/s$

So , the target moves with the speed of 0.995 m/s