Question

Please Help

Answer 1

@Shane_B

Answer 2

@Callisto @Preetha
@mukushla
@Jemurray3

Answer 3

@UnkleRhaukus

Answer 4

@hartnn

Answer 5

apply the conservation of momentum

Answer 6

how?

Answer 7

what is 104 m in ur diagram?

Answer 8

is it the position where bullet hits the ground?

Answer 9

yep

Answer 10

does bullet hit the ball which is resting at vertical post?

Answer 11

cnt say

Answer 12

then how does the ball move...who kicked it?

Answer 13

Oh sorry yes .

Answer 14

along horizontal
(m+M) Vcm = m v1 + m v2

Answer 15

there is no force along horizontal direction so v(cm) along horizontal will move with constant velocity

Answer 16

(m+M) Vcm = m v1 + M v2

Answer 17

before collision Vcm = M V2/(m+M) along horizontal
after collision V cm= (mv1 + M v2)/(m+M)

Answer 18

m= ball, M=bullet

Answer 19

before collision Vcm = M V/(m+M) along horizontal , initial velocity of bullet= v and of ball=0
after collision V cm= (mv1 + M v2)/(m+M)

Answer 20

so M V/(m+M) = (mv1 + M v2)/(m+M)

Answer 21

in fact..we are using conservation of momentum only

Answer 22

so M V=(mv1 + M v2)

Answer 23

you know M and u can calculate v1...using it hits the ground at a distance of 20m

Answer 24

yes ...u also have to calculate the time...so that u can use this t
in v1*t=20m

Answer 25

and then use this formula

Answer 26

right?

Answer 27

hmm I got it now, thanks

Answer 28

so now u can solve this problem by ur own..

Answer 29

yep

Answer 30

good....:)