a bullet of mass 30g is fired from a gun with a velocity 116m/s . if the recoil velocity of the gun is 2m/s , then what is the mass of the gun?
Answers
Given:-
→Mass of the bullet(m)=30g
→Velocity of the bullet(v)=116m/s
→Recoil velocity of the gun(V)=2m/s
To find:-
→Mass of the gun(M)
Solution:-
Here,we can see that the mass of the bullet is given in gram(g). So firstly,we shall convert the mass of the bullet from gram(g) to kilogram(kg).
=>1g=0.001kg
=>30g=30(0.001)
=>0.03kg
By using Law of Conservation of momentum,we get:-
=>M(-V)+mv=0
=>M=mv/V
=>M=0.03×116/2
=>M=1.74kg
Thus,mass of the gun is 1.74kg.
Some Extra Information:-
Conservation of momentum:-
We have studied the equation,
=>F=mv-mu/t
=>Fext=mv-mu/t. (Since,F=Fext)
=>Fext×t=mv-mu
When external force is 0:-
=>0×t=mv-mu
=>mv-mu=0
=>mv=mu
=>Thus,Initial momentum=Final momentum
=>Change in momentum is 0,i.e.
momentum remains conserved.
From above we can infer that "When external force acting on a system is zero,the total momentum of the system remains constant/conserved".
Explanation:
Given:-
→Mass of the bullet(m)=30g
→Velocity of the bullet(v)=116m/s
→Recoil velocity of the gun(V)=2m/s
To find:-
→Mass of the gun(M)
Solution:-
Here,we can see that the mass of the bullet is given in gram(g). So firstly,we shall convert the mass of the bullet from gram(g) to kilogram(kg).
=>1g=0.001kg
=>30g=30(0.001)
=>0.03kg
By using Law of Conservation of momentum,we get:-
=>M(-V)+mv=0
=>M=mv/V
=>M=0.03×116/2
=>M=1.74kg
Thus,mass of the gun is 1.74kg.
Some Extra Information:-
Conservation of momentum:-
We have studied the equation,
=>F=mv-mu/t
=>Fext=mv-mu/t. (Since,F=Fext)
=>Fext×t=mv-mu
When external force is 0:-
=>0×t=mv-mu
=>mv-mu=0
=>mv=mu
=>Thus,Initial momentum=Final momentum
=>Change in momentum is 0,i.e.
momentum remains conserved.
From above we can infer that "When external force acting on a system is zero,the total momentum of the system remains constant/conserved".