Question

A man fires a bullet of mass 100 g at a speed of 75 ms–1. The gun is of 3 kg mass. What is the recoil velocity of the gun?

Answer 1

@Callisto and @hartnn @Xishem

Answer 2

oh k so I did like this:

given \(m_1\) = 0.1 kg. and \(m_2 \) = 3 kg.
u = \(75 ms^{-1}\)

recoil velocity = v

0 = 0.1 * 75 +3 v
0 = 7.5 + 3v
-3v = 7.5

v = \(\frac{7.5}{3}=\frac{75}{30}=\frac{15}{6}=2.5 ms^{-1}\)

Answer 3

hence recoil velocity = 2.5 \(ms^{-1}\)

Answer 4

M i right?

if yes then why does the book says the answer as "0.25 m/s" ?

Answer 5

yea~
using the momentum equation
m1v1=m2v2
so 0.1(75)=3v2
v2=7.5/3
=2.5 m/s

probably the answer is wrong ;)

Answer 6

yes you're right

Answer 7

just use conservation of momentum...momentum is not conserved ideally though

Answer 8

hmn so the book is wrong?

Answer 9

for sure..coz i can't see any fault in this calculation...what does your book say?

Answer 10

0.25 m/s

Answer 11

@mathslover your answer is correct.

Answer 12

nope it will be 2.5

Answer 13

thanks a lot all : @ghazi @Kystal @sami-21

Answer 14

:) YW

Answer 15

V = -mv/M

Answer 16

= -0.1*75/3 = -2.5m/s

Answer 17

-ve sign rep the velocity is the opp direction..))

Answer 18

It sure seems like your book is wrong a lot :P.