Arifle with a mass of m1 = 3.4 kg fires a bullet with a mass of 7.7 g (0.0077 kg). the bullet moves with a velocity of 860 m/s after the rifle is fired. if external forces acting on the rifle can be ignored, what is the magnitude of the recoil velocity of the rifle?

The magnitude of the recoil is v = 1.95 m/s

Explanation:

The momentum of bullet

P = m*v

P = 0.0077kg x 860m/s

P =  6.622 kg m/s

Now using the law of conservation of momentum we have,

P1 + P2 = 0

Now solve to V2 to find the velocity of the rifle so:

m1*v1 + m2*v2 = 0

3.4kg*v1 + 6.622 kg m/s = 0

3.4 kg*v1= -6.622 kg m/s

v1 = 6.622/3.4

v1 = 1.95 m/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//