A 30 g bullet is fired horizontally with velocity 250m/s from a 1.5 kg gun. If the gun is held loosely in hand, its recoil velocity would be_?
Answers
m1 = mass of gun m2 = mass of bullet
formula
m1u1+m2u2=m1v1+m2v2
here initial velocity of both gun and bullet is 0
so m1v1=-m2v2
now 1500g*v1=30*250
v1=30*250/1500
so v1 = 5m/s is the answer negative indicates in opposite direction
thankyou
Answer:
Recoil velocity will be -5ms⁻¹.
Explanation:
Concept:
- Recoil velocity is the speed at which the stock of the rifle impacts your shoulder.
- According to the law of conservation of momentum, the total momentum of the gun and the bullet will be zero after the gun is fired. Let m be the mass of the bullet and v be the velocity on firing the gun; M is the mass of the gun and V be the velocity with which it recoils.
MV+mv=0
Given:
Mass of bullet, m=30g= .03kg( 1Kg=1000g)
Velocity of bullet, v=250ms⁻¹
Mass of gun, M=1.5kg
Recoil velocity of gun =V
Description:
- Before firing, the system (gun+bullet) is at rest, initial velocity =0
- So initial momentum of the system =0
- Final momentum of the system = momentum of bullet + momentum of gun
=mv+MV
= .03×250+1.5×V
=7.5+1.5V
- According to law of conservation of momentum, Final momentum = Initial momentum.
7.5+1.5V=0
1.5V=−7.5
or
V=−7.5/1.5
=−5ms⁻¹
Note: The direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet. That's why the sign of recoil velocity is negative.