Question

6. With a 10 Kg of gun, 60 g bullet is fired at a velocity of 160m/s.
The backward thrust that the gun produces on firing will
have the velocity:
(a) 3.2 m/s (b) 1.9 m/s (c) 0.96 m/s (d) 2m/s​

Answer 1

Answer:

(c) 0.96 m/s

Step-by-step explanation:

Given :

$\tt \small{{Mass \: of \: bullet (m_1)= 60 g = 0.06 kg}}$

$\tt \small{Velocity \: of \: bullet (v_1) = 160 m/s}$

$\tt{Mass \: of \: gun \: (m_2) = 10kg }$

$\tt{Recoiled \: \: velocity (v_2) = ?}$

As per the law of conservation of momentum,

$\tt{m_1 \times v_1 = m_2 \times v_2}$

$\tt{0.06 \times 160 = 10 \times v_2}$

$\tt{v_2 = \frac{0.06 \times 160}{10}}$

$\tt {v_2 = 0.06 \times 16}$

$\tt{v_2 = 0.96}$

Thus, The backward thrust that the gun produces on firing will

The backward thrust that the gun produces on firing willhave the velocity 0.96 m/s

Answer : Opt. ( c ) : 0.96 m/s

Answer 2

Answer:

Mass of gun ($\sf{m_{1}}$)= $\tt{10 kg}$

Mass of bullet ($\sf{m_{2}}$)= $\tt{60 g = 0.06 kg}$

Velocity of gun before firing ($\sf{u_{1}}$) = $\tt{0 m/s}$

Velocity of bullet before firing ($\sf{u_{2}}$) = $\tt{0 m/s}$

Velocity of gun after firing = $\sf{v_{1}}$

Velocity of bullet after firing ($\sf{v_{2}}$) = $\tt{160 m/s}$

According to the law of conservation of momentum :

$\tt{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}$

=》 $\tt{0 = 10v_{1} + 0.06(160)}$

=》 $\tt{(-9.6) = 10v_{1}}$

=》 $\tt{v_{1} = (-0.96) m/s}$

Answer: (c) 0.96 m/s backwards