In: Physics
A rifle with a weight of 25 N fires a 5.5 g bullet with a speed of 210 m/s. (a) Find the recoil speed of the rifle. m/s
(b) If a 675 N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle. m/s
Given that rifle with a weight of 25 N
If acceleration due to gravity of earth is g (9.8 m/s2) and m is mass of a body , then weight of a body is given as
W= M*g
we know weight W= Mg =25 N
=> Mrifel*9.8= 25
=> Mrifel= 25/9.8
=> Mrifel = 2.551 kg
Before firing , the rifel and the bullet both are at rest and applying law of conversavation of momentum.
(law of conversavation of momentum is the total sum of momentum before firing and after firing should be equal)
mrifelurife+mbulletubullet = mrife l vrifel +mbullet vbullet
Here ugun , ubullet are velocites of Rifel and bullet before firing and vrifel, vbullet are the velocites of Rifel and bulle after firing
Since Rifel nd bullet are at rest before firing urifel, =0, ubullet =0
mrifel urifel +mbullet ubullet = m rifel vrifel +mbullet vbullet
mrifel v rifel+mbullet vbullet =0 (ugun, =0, ubullet =0)
2.55 * Vrifel + 5.5 * 10-3 *210 =0
Vrifel= -0.452 m/s
Negitive because the rifel moves backwards., Therefore recoil.
2)
Given that rifle with a weight of 675 N
If acceleration due to gravity of earth is g (9.8 m/s2) and m is mass of a body , then weight of a body is given as
W= M*g
we know weight W= Mg = 675 N
=> Mman*9.8= 675
=> Mman= 675/9.8
=> Mman = 68.8 kg
If a man of mass 675 N ( or mass 68.8 kg ) is holding a rifel , both man and rilfel behaves as one body and both are at rest before firing
So again applying Law of conservation Momentum'
So (mrifel +m man )urifel +mbullet ubullet = (m rifel + m man )vrifel +mbullet vbullet
(m rifel + m man )vrifel + refil + mbullet vbullet =0
(68.8+0.0055) vrifel + refil + 5.5 * 10-3 * 210 =0
vrifel + refil =0.016m/s
So the recoil speed of man and bullet reduces to 0.016 m/s
That why the man should hold the rifel tightly to make both rifel and man as one body