A bullet leaves a rifle with a velocity of 100 m/s and the rifle mass 2.5 kg recoils with a velocity of 1 m/s. Find the mass of the bullet. PLEASEEEE HELP ME.👏

Answer 1

according to law of conservation of momentum
MU + mu = MV + mv
Let M = 2.5 kg
m = m kg
initial velocity = 0 ( cause both were at rest before firing)
V = -1 m/s (it acts in opp. direction)
v = 100 m/s

2.5 X 0 + m x 0 = 2.5 x -1 + m X 100
0 = -2.5 + 100m
2.5 = 100m

25/1000 or 0.025 = m <= answer

Answer 2

Please understand this question

Answer 3

The preamble the problem and experience to help out in the past year and the problem.

Answer 4

A bullet leaves a rifle with a velocity of 100 m/s and the rifle of mass 2.5 kg recoils with a velocity of 1 m/s. the mass of the bullet can be determined by the mass of conservation of momentum.

According to the conservation of the momentum is given by \bold{m 1 v 1=m 2 v 2}m1v1=m2v2

Hence by considering the mass of the bullet as m1 and the rifle with the mass of m2 and hence v1 is the velocity of the bullet and the v2 be the velocity of the rifle.

m 1 v 1=m 2 v 2m1v1=m2v2

m 1=\frac{m 2 v 2}{v 1}m1=
v1
m2v2
​


m 1=\frac{2.5 * 1}{100}m1=
100
2.5∗1
​


m1 = 0.025kg

m1 = 25g

Hence the mass of the bullet is generalised and found as 25g.

Mass of bullet is 25g

Answer 5

V=0
V=-1m/s
V=100m/s
2.5*0+ mv=0
=2.5*-1+m*100
0=-2.5+100m
2.5=100m
25/1000 or 0.025=m

*this is the sine of multiply ✖

Answer 6

All the answers above are wrong..
We should divide 25 with 10000.....
then the answer is 0.25kg
but u they all divided 25 from 1000...
hence got wrong answers