Help with conservation of kinetic energy and momentum

[skyze]

Homework Statement

Alpha particle mass - 4amu
velocity alpha particle - 3.0x10^5m/s

¹⁹⁷Au Nucleus mass - 197amu

Head-on elastic collision. Find the final velocity of both the particles

Homework Equations

Kinetic Energy conserved:
1/2m₁v_1i = 1/2m₁v_1f + 1/2m₂v_2f

Momentum Conserved:

m₁v_1i = m₁v_1f + m₂v_2f

The Attempt at a Solution

For Momentum:

4 x (3 x 10^5) = 4V_1f + 197v_2f
3 x 10^5 = V_1f + 49.25v_2f
V_1f = 3 x 10^5 - 49.25v_2f


For Kinetic Energy

4 x (3.0 x 10^5)² = 4V_1f² + 197v_2f²
3.6 x 10¹¹ = 4V_1f + 197v_2f
9 x 10¹⁰ = V_1f + 49.25v_2f

Subbing in one for the other


9x 10¹⁰ = (300,000 - 49.25v_2f)² + 49.25v_2f


The v_2f and V_1f values I get doesn't work out



Thanks in advance for your help.

 

[Redbelly98]

The Attempt at a Solution

For Momentum:

4 x (3 x 10^5) = 4V_1f + 197v_2f
3 x 10^5 = V_1f + 49.25v_2f
V_1f = 3 x 10^5 - 49.25v_2fFor Kinetic Energy

4 x (3.0 x 10^5)² = 4V_1f² + 197v_2f²
3.6 x 10¹¹ = 4V_1f + 197v_2f
9 x 10¹⁰ = V_1f + 49.25v_2f

Subbing in one for the other 9x 10¹⁰ = (300,000 - 49.25v_2f)² + 49.25v_2f

That last v_2f is really v_2f², right?

The v_2f and V_1f values I get doesn't work out

What values do you get? It looks like the right approach.

p.s. Welcome to Physics Forums!