Question

A 0.50 kg ball traveling at 6.0 m/s collides head-on with a 1 kg ball moving in the opposite direction at a velocity of -12 m/s. The 0.50 kg ball moves away at -14 m/s after the collision. FInd the velocity of the second ball. PLEASE EXPLAIN!

Answer 1

in this case you can use the conservation of linear momentum.
before
\[m \times V1 +M \times V2=Pb\]
after
\[m \times V1' +M \times V2'=Pa\]
by conservation of linear momentum
Pb=Pa

Answer 2

i am so confused about everything. can u please start from the beginning and plug the numbers in the equation you wrote

Answer 3

The principle of conservation of momentum says that the momentum of the system must be the same before and after the collision. ok?

Answer 4

oh thanks! i get it

Answer 5

wait so does that mean that the second ball's velocity is 6 m/s too!?

Answer 6

no

Answer 7

what is it then?

Answer 8

\[m \times V1+M \times V2= m \times V1'+M \times V2'\]

Answer 9

m=0.5 kg
M = 1kg
V1=6.0 m/s
V2= -12 m/s
V1'= -14 m/s
V2' unknown

Answer 10

is the answer -2?