A 3.1-kilogram gun initially at rest is free to move. When a 0.015-kilogram bullet leaves the gun with a speed of 500. meters per second, what is the speed of the gun?

Answer 1

This a chance for you to apply the law of conservation of momentum.

That means the momentum of the gun going backward is equal and opposite to the momentum of the bullet going forward. The two momenta cancel out.

Mbullet*Vbullet = -Mgun*Vgun

Solve for Vgun

Answer 2

7.5 m/s

Answer 3

pbefore=pafter
m1v1+m2v2 = m1v’1+m2v’2
(3.1kg)(0m/s)+(0.015kg)(0m/s) =
(3.1kg)( v’1)+(0.015kg)(500.m/s)
0kg∙m/s=(3.1kg)( v’1)+7.5kg∙m/s
-7.5kg∙m/s=(3.1kg)( v’1)
v’1=-2.4m/s (its speed is 2.4 m/s)

Answer 4

u=0
0+0= 3.1 v1 + 0.015(500)m/s
-3.1v1= 7.5
v1= 7.5:(-3.1)

V1=-2.42 in absolute value then the answer will be 2.42

Answer 5

u=0
0+0=3.1V+ 0.015(500)M/S
-3.1V = 7.5
V= -2.42M/S OR 2.42m/s in absolute value

Answer 6

Stephanie is a troll

Answer 7

bababboey

Answer 8

i agree that stephanie is an absolute troll hahahah

Answer 9

not the ten year old speaking haha