Question

a 15 gram bullet is fired from a gun of mass 3kg if the speed of bullet just after firing is 300m/s what is recoil speed of gun

Answer 1

Answer:

MB=15g

Mgun=3kg

Ref. image

Since no Extra force fext=0

Momentum is conserved.

MgUg=MBUB

3×Ug=15×10−3×300

Ug=1.5m/s

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Answer 2

Answer:-

-1.5m/s

Explanation:-

Given :

• Mass of bullet: 15g

• Mass of gun: 3kg

• Speed of the bullet after firing: 300m/s

To Find :

• Recoil velocity/speed of the gun

Solution :

$\sf{}Mass\ of\ the\ bullet(m_1)= 15g= 0.015kg\ \ \ \ \ [1g =\dfrac{1}{1000}kg]$

$\sf{}Initial\ velocity\ of\ the\ bullet(u_1)= 0m/s\ \ \ \ \ [As\ it\ was\ in\ rest]$

$\sf{}Final\ velocity\ of\ the\ bullet= 300m/s$

$\sf{}Mass\ of\ the\ gun(m_2)= 3kg$

$\sf{}Initial\ velocity\ of\ the\ gun(u_2)= 0m/s\ \ \ \ \ [As\ it\ was\ in\ rest]$

$\sf{}Final\ velocity\ of\ the\ gun= ?$

Let’s find out Final velocity of the gun by Conversion of momentum

According the Conversion law of of momentum

$\sf{}m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$\sf{}0.015\times 0 + 3\times 0= 0.015\times 300 +3\times v_2$

$\sf{}\implies 0=4.5 +3v_2$

$\sf{}\implies-4.5= 3v_2$

$\sf{}\implies-\dfrac{4.5}{3}= v_2$

$\sf{}\therefore v_2=-1.5m/s$

Therefore,recoil speed of the gun is -1.5m/s