Momentum ?

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Proton 1 collides elastically with proton 2 that is initially at rest. Proton 1 has an initial speed u_1=3.50*10^5ms^-1 and makes a glancing collision with proton 2, as shown in Figure 4.20. After the collision, proton 1 moves at an angle theta=37^@ to the horizontal axis, and proton 2 deflects at an angle phi with to the same axis. Find the final speeds of the two protons and the angle phi.

The answer provided are[2.80*10^5ms^-1, 2.11*10^5ms^-1, 53^@], but i want to know how to solve.

1 Answer
Sep 24, 2017

See below.

Explanation:

m_1 vec v_0 = m_1 vec v_1 + m_2 vec v_2

vec v_0 = v_0(1,0)
vec v_1 =v_1 (cos theta,sin theta)
vec v_2 =v_2 (cos phi,-sin phi)

then

{(v_0=v_1costheta+v_2cosphi),(0=v_1sintheta-v_2sinphi):}

and solving for v_1,v_2

{(v_1 =(v_0 Sin phi)/(Cos theta Sin phi + Cos phi Sin theta) ),(v_2 =(v_0 Sin theta)/(Cos theta Sin phi + Cos phi Sin theta)):}

Now the collision is perfectly elastic so

1/2m_1 v_0^2=1/2m_1 v_1^2+1/2m_2 v_2^2

or

((v_0^2Sin theta)/(sin(phi+theta)^2) )(m_1 Sin(2 phi + theta)-m_2 Sin theta)=0

then from

(m_1 Sin(2 phi + theta)-m_2 Sin theta)=0

with m_1 = m_2 we easily determine phi and consequently v_1 and v_2