Help with conservation of kinetic energy and momentum

AI Thread Summary
The discussion revolves around solving a physics problem involving a head-on elastic collision between an alpha particle and a gold nucleus. The equations for conservation of momentum and kinetic energy are set up, but the user encounters difficulties in obtaining consistent final velocities for both particles. They express confusion about the substitution process and the values derived for the final velocities. Other participants affirm that the approach is correct and encourage sharing specific numerical results for further assistance. The conversation emphasizes the importance of careful calculations and substitutions in solving collision problems.
skyze
Messages
1
Reaction score
0

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:
1/2m1v1i = 1/2m1v1f + 1/2m2v2f

Momentum Conserved:

m1v1i = m1v1f + m2v2f

The Attempt at a Solution




For Momentum:

4 x (3 x 10^5) = 4V1f + 197v2f
3 x 10^5 = V1f + 49.25v2f
V1f = 3 x 10^5 - 49.25v2f


For Kinetic Energy

4 x (3.0 x 10^5)2 = 4V1f2 + 197v2f2
3.6 x 1011 = 4V1f + 197v2f
9 x 1010 = V1f + 49.25v2f

Subbing in one for the other


9x 1010 = (300,000 - 49.25v2f)2 + 49.25v2f


The v2f and V1f values I get doesn't work out



Thanks in advance for your help.
 
Physics news on Phys.org
skyze said:

The Attempt at a Solution

For Momentum:

4 x (3 x 10^5) = 4V1f + 197v2f
3 x 10^5 = V1f + 49.25v2f
V1f = 3 x 10^5 - 49.25v2fFor Kinetic Energy

4 x (3.0 x 10^5)2 = 4V1f2 + 197v2f2
3.6 x 1011 = 4V1f + 197v2f
9 x 1010 = V1f + 49.25v2f

Subbing in one for the other 9x 1010 = (300,000 - 49.25v2f)2 + 49.25v2f
That last v2f is really v2f2, right?

The v2f and V1f values I get doesn't work out
What values do you get? It looks like the right approach.

p.s. Welcome to Physics Forums!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Conservation of momentum - with understanding
Replies
2
Views
2K
Conservation of angular momentum and rotational kinetic energy
Replies
2
Views
1K
Air-Track Carts & Spring (Energy)
Replies
17
Views
6K
Conservation of momentum (wrecking ball hits a stationary object)
Replies
10
Views
3K
Solve Elastic Collision: 2D Homework
Replies
4
Views
2K
Solving Orbital Speed with Energy & Angular Momentum Conservation
Replies
7
Views
2K
Potential and Kinetic energy equations including drag coefficient
Replies
1
Views
1K
Conservation of momentum and energy problem (Please check if my setup is right)
Replies
3
Views
2K
Angle that makes kinetic and potential energy of simple pendulum equal
Replies
2
Views
1K
Kinetic Energy & Momentum of a Ball released from Spring
Replies
6
Views
2K

Hot Threads

Recent Insights

Back
Top