Question

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationary nitrogen nucleus of mass 14m. Find the velocities of the proton and the nitrogen nucleus after the collision.

Answer 1

`V_p = 267.258 m/s`

`V_n = 38.375 m/s`

Step-by-step explanation:

using the law of the conservation of the linear momentum:

`P_i = P_f`

where
`P_i` is the inicial momemtum and
`P_f` is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

`P_i = m(270 m/s)`

`P_f = mV_P + M_nV_n`

where m is the mass of the proton and
`V_p` is the velocity of the proton after the collision,
`M_n` is the mass of the nucleus and
`V_n` is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m
`V_p` + 14m
`V_n`

then, m is cancelated and we have:

270 =
`V_p` +
`14V_n`

This is a elastic collision, so the kinetic energy K is conservated. Then:

`K_i = (1)/(2)MV^2 = (1)/(2)m(270)^2`

and

Kf =
`(1)/(2)mV_p^2 +(1)/(2)(14m)V_n^2`

then,

`(1)/(2)m(270)^2` =
`(1)/(2)mV_p^2 +(1)/(2)(14m)V_n^2`

here we can cancel the m and get:

`(1)/(2)(270)^2` =
`(1)/(2)V_p^2 +(1)/(2)(14)V_n^2`

now, we have two equations and two incognites:

270 =
`V_p` +
`14V_n` (eq. 1)

`(1)/(2)(270)^2` =
`(1)/(2)V_p^2 +(1)/(2)(14)V_n^2`

in the second equation, we have:

36450 =
`(1)/(2)V_p^2 +(1)/(2)(14)V_n^2` (eq. 2)

from this last equation we solve for
`V_n` as:

`V_n = \sqrt{(36450-(1)/(2)V_p^2 )/((1)/(2) ) }`

and replace in the other equation as:

270 =
`V_p +` 14
`\sqrt{(36450-(1)/(2)V_p^2 )/((1)/(2) ) }`

so,

`V_p = -267.258 m/s`

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

`V_n = (270+267.258)/(14)`

`V_n = 38.375 m/s`