Question

a bullet of mass 50 g is fired from a gun of mass 50 kg with a velocity of 200 m/sec its recoil velocity is
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Answer 1

Answer:

Solution :

Since for ( gun + bullet) the system , the external force is zero. Hence by momentum conservation  

Initial momentum = Final momentum  

0=μ+Mv  

Mass of bullet , m=50g=50×10−3kg,u=200m/s  

Mass of gun , M=2kg  

0=50×10−3×200+2v  

v=−5m/s  

where the −ve sign indicates that the velocity of gun is opposite to that of the bullet.

Explanation:

Happy Independence day.!!

Answer 2

Answer

Recoil velocity of gun = - 0.2 m/s

Given

A bullet of mass 50 g is fired from a gun of mass 50 kg with a velocity of 200 m/s

To Find

Recoil velocity of gun

Concept Used

To solve this type of questions we need to apply conservation of momentum , in which initial momentum is equals to final momentum .

$\rm \to Initial\ Momentum=Final\ Momentum\\\\\to \rm P_{_{i}}=P_{_f}$

Momentum : It is defined as product of mass and velocity

Solution

Mass of bullet = 50 g = 0.05 kg

Mass of gun = 50 kg

Velocity of bullet = 200 m/s

Recoil velocity of gun = ? m/s

Initial velocity of both gun and bullet system is zero , since they both are in rest .

Apply conservation of momentum ,

$\to \rm P_{_i}=P_{_f}\\\\\to \rm (m_g+m_b)v_i=(m_g.v_r)+(m_b.v_b)\\\\\to \rm (50+0.05)(0)=(50.v_r)+(0.05\times200)\ [\; \because\ v_i=0\ ]\\\\\to \rm 0=50v_r+10\\\\\to \rm 50v_r=-10\\\\\to \rm v_r=-0.2\ m/s\ \; \pink{\bigstar}$

Recoil velocity of gun = - 0.2 m/s

Note : Negative sign of recoil velocity denotes velocity in opposite direction