Question

i'm having trouble starting this problem and when looking at the equations, i can't quite tell what each of those letters represent(only spent 2 class periods on this stuff.... so any help is appreciated
a 2.0 kg mass is traveling at 5 m/s on a smooth horizontal surface when it collides with and sticks to a stationary 6.0 kg mass. the larger mass is attached to a light spring of force constant 150 N/M, find amplitude of the resulting oscillations of this system.

Answer 1

Firstly you must use Conservation of linear momentum law which states the linear momentum of an isolated system remains the same after the collision.
Linear momentum a body designated as letter p and
p = mv
where m mass of a body
v - velocity.
So your system before collision has the linear momentum
\[2 kg \times 5 m/s = 10 kg \times m/s\]
after the collision the mass of a system increased by 6 kg, but the linear momentum remains the same because your system is isolated it doesn't lose any energy in the form of heat or whatever:
\[10 kg \times m/s = 8 kg \times x m/s\]
x = 1.25 m/s it is the velocity of the system after the collision
Now you have initial velocity of a body of 8 kg attached to a light spring. I think it's a good start to solve the problem.

Answer 2

* Linear momentum of a body is designated as letter p

Answer 3

on the math subject, i posted this same question(before i knew there was a physics section) and i had a similar but different answer. the way the other person did it was to first find the energy of the first mass which is 25 J, then equate that to the second mass because energy is conserved and so .5(k)(A)^2 and i got .5733 anyways i don't know who yours relates to this so it made the whole thing more confusing for me...

Answer 4

ok i think i may have figured it out, first i used W(angular velocity) equals to square root of k/m then after finding the W, i plugged that in the equation W=2pif(frequency)... does that seem correct?

Answer 5

Careful: Energy is NOT conserved in an non-elastic collision such as that one.

Answer 6

Velocity of the system calculated using kinetic energy gives me different result i think it's better to use the law of conservation of linear momentum.

Answer 7

@Vincent-Lyon.Fr yes indeed i forgot about that!

Answer 8

On the other hand, momentum is ALWAYS conserved in ANY type of collision.

Answer 9

oh wow, i never thought to use momentum in this problem, when i saw the oscillation thing, i immediately thought of SHM and such, let me do this again and see what i get

Answer 10

After the collision, then all you know about SHM applies.

Answer 11

ok