# Homework Help: Help momentum, recoil speed [simple]

1. Nov 3, 2009

### davedays

1. The problem statement, all variables and given/known data
A bullet m= 9.0 g leaves the barrel of a rifle with a speed = 8.0 x 10 ^2 ms-1 (by the may im sure - is this 80 ms-1 ?)

The mass of the rifle = 1.8 kg
The rifle is free to move back
Calculate the speed of recoil

2. Relevant equations
I figured I would need these :
p=mv
F=mv/t

I'm not sure wether that's all I need.

3. The attempt at a solution

Well, first of all I think that the momentum 1 way = the momentum the other way.
so

p1=p2

m1 x v1 = m2 x v2

9 [g] x 80 [ms-1] = 1800 g (1.8 kg right ?) x V2

720 = 1800 V2

V2= 720/1800

V2= 0.4

Is this correct guys ?

2. Nov 3, 2009

### Cryphonus

the way you solve the problem is correct however i think you have a mistake about the speed of the bullet. 8.0x10^2 means the bullet moves with a velocity of 800 m/s.

according to my calculations the answer is 40 m/s....

3. Nov 3, 2009

### davedays

hmm not sure about that .. maybe I have another mistake

m1 x v1 = m2 x v2

9 x 800 = 1800 x v2

7200 = 1800 v2

v2 = 7200 / 1800
v2 = 4

isnt it 4 ms -1 ?

4. Nov 3, 2009

### Pheo1986

yes the method you have used is correct however as cryphonus said, the velocity is 800 m/s. also grams is not the si unit for mass, kilograms is. so you want to use 0.009kg and 1.8kg as the masses.

i got an answer of 4 m/s

5. Nov 3, 2009

### davedays

thanks a lot guys

6. Nov 3, 2009

### Cryphonus

Ok; here is how i solved it; initial momentum is
800 m/s.0.09 kg (mass of the bullet in kilogram);

which is 72 m.kg/s

so if the initial momentum is 72 m.kg/s the final value should be equal to 72 also;

P(final)=1,8kg (mass of the rifle).V(recoilspeed)

72=1,8.V

V=40 m/s ,
it is in the opposite direction of the bullet so if you express the speed of the bullet as a positive value, you have to express the recoil speed as negative

any corrections are welcome....

7. Nov 4, 2009

### Pheo1986

9 g =0.009kg not 0.09kg