The recoil speed is 2.14 m/s
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the bullet-rifle must be conserved before and after the explosion.
Before the explotions, both are at rest, so the total momentum is zero:
[tex]p=0[/tex]
After the shot, the total momentum is:
[tex]p=mv+MV[/tex]
where :
m = 12.70 g = 0.0127 kg is the mass of the bullet
v = 430 m/s is the velocity of the bullet
M is the mass of the rifle
V is the recoil velocity of the rifle
We know the weight of the rifle, W = 25.0 N, so we can find its mass:
[tex]M=\frac{W}{g}=\frac{25.0}{9.8}=2.55 kg[/tex]
And momentum is conserved, we can equate the initial and final momentum and solve for V:
[tex]0=mv+MV\\V=-\frac{mv}{M}=-\frac{(0.0127)(430)}{2.55}=-2.14 m/s[/tex]
Where the negative sign indicates that the rifle's recoil is in the opposite direction: therefore, the speed of recoil is 2.14 m/s.
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