Question

A mass m, moving at speed 3v, approaches mass, 4m, moving at speed v towards the first mass. They collide elastically head-on. Find expressions for their subsequent speeds.

Answer 1

Do you have any idea by yourself?

Answer 2

Well I was thinking of using m1v1 = m2v2

Answer 3

Bur other than that, no.

Answer 4

Since the collision is inelastic it means that there would be no lose of energy. To find the velocities after the collision it is a good idea to apply both conversation law of linear momentum and kinetic energy which all remain constant during the collision. Let me know if there is something obscure yet?

Answer 5

But the collision is elastic, not inelastic.

Answer 6

*elestic .. sorry.

Answer 7

Actually a collision between two perfectly elastic bodies is such that the final kinetic energy of the system would be the same as the initial kinetic energy of the system.

Answer 8

Well I have a practice problem for this question, with similar numbers and the textbook's solutions as well,

A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V

Answer 9

oops, the speed of 4m was 7.5V

Answer 10

Well, so sill need any more to say?

Answer 11

Yes I just help figuring out how to establish the foundation of the formulas to use.

Answer 12

Well, what is initial and final momentum of the system?

Answer 13

would I use,
.5 mava^2 + .5mbvb^2 = .5mava^2 + .5mbvb^2?

Answer 14

m -->3v
4m-->v

Initial momentum: m(3v) + 4m(v) = 7mv
Final momentum: mV1 + 4mV2
--> 7mv=mV1 + 4mV2
--> 7v= V1 + 4V2

Total conversation of KE:
= m(3v)^2 + 4m(v)^2 = m(V1)^2 + 4m(V2)^2
= 9mv^2 + 4mv^2 = m(V1)^2 + 4m(V2)^2
= 13mv^2 = m(V1)^2 + 4m(V2)^2
= 13v^2 = (V1)^2 + 4(V2)^2
Now we have arrived at two equations:
7v= V1 + 4V2
13v^2 = (V1)^2 + 4(V2)^2

Solve for V1 and V2.. You can substitute V1 from 1st equation into V1 in 2nd equation to obtain V2. Once you get V2 you can solve for V1 using the first eq.

I hope it helps.

Answer 15

*conservation

Answer 16

sorry if I made some typos. Good luck.