Question

(1) A bullet with mass 50.0 g is fired with an initial velocity of 625 m/s from a gun with a mass 3.50 kg. What is the speed of recoil (in m/s) of the gun?

(2) How many Newtons of force are required to stop a truck 2250 kg truck going 32.0 km/h in 4.00s?

Answer 1

1)
Use conservation of momentum law

Answer 2

m1*u1+m2*u2=m1*v1+m2*v2

Answer 3

what does u stand for?

Answer 4

could you work me through it? i'm not too familiar with that formula

Answer 5

m1=mass off bullet=50g=?kg
m2=mass of gun=3.5kg
u1=initial velocity of bullet=0
u2=initial velocity of gun=0
v1=final velocity of bullet =625m/s
v2=?

Answer 6

v2=recoil speed of gun

Answer 7

No

Answer 8

(0.05 kg)(625 m/s)+(3.5 kg)(0 m/s)= (0.065 kg)(625 m/s)+ (3.5 kg)v2

Answer 9

m1=50g=50/1000kg=.05kg
Right?

Answer 10

i think it should be
(0.05 kg)(0 m/s)+(3.5 kg)(0 m/s)= (0.06 kg)(625 m/s)+ (3.5 kg)v2

Answer 11

.05*0+3.5*0=.05*625+.05*v2
v2=?

Answer 12

Why .06 kg?

Answer 13

oh! i'm sorry

Answer 14

*0.05kg

Answer 15

i got 31.25 = 31.25 + 3.5 V2

Answer 16

no you may be using the wrong equation
the correct one is
05*0+3.5*0=.05*625+.05*v2
as @shamim said

Answer 17

.05*0=0
Right?

Answer 18

so you should get -31.25=3.5v2
and i posted the wrong equation
it should be 05*625+3.5*v2

Answer 19

3.5*9=0
Right?

Answer 20

0+0=0
Right?

Answer 21

V2 = -8.9285 m/s, but since speed is scalar it is just 8.9285

Answer 22

Left hand side of my equation will b 0
Right?

Answer 23

- reveals
v1 nd v2 r oposit in direction

Answer 24

Ur bullet is going to forward nd ur gun is going to backward

Answer 25

how do I solve #2 now?

Answer 26

Who will get medal?????!!!!!

Answer 27

whoever helps me solve both problems

Answer 28

How many Newtons of force are required to stop a truck 2250 kg truck going 32.0 km/h in 4.00s?

Answer 29

U should post ur second question again in another post

Answer 30

ok

Answer 31

thanks for the medal @Summersnow8

Answer 32

@rajat97 no problem