Suppose the momentum of a particle of mass moving with velocity has the form

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Special relativity: spacetime and kinematics · Relativistic dynamics and fields in flat spacetime

Relativity and Gravitation, Ch Relativistic momentum, energy and collisions

Problem

Suppose the momentum of a particle of mass

m

moving with velocity

\vec u

has the form

\vec p = f(u)\,m\vec u

, where

f

depends only on the speed

u

and

f(0) = 1

so that the Newtonian result is recovered for slow particles. Test this form on the symmetric glancing collision. Thrower A is at rest in

S

, thrower B is at rest in

S'

, and

S'

moves at

v = 0.800c

along the common

+x

axis. Each throws an identical ball of mass

m

straight across a corridor (along

y

) with speed

u_0

in her own rest frame, with no

x

-velocity in that frame; A's ball goes in

+y

and B's in

-y

. The balls rebound symmetrically, so in her own frame each thrower sees her own ball return with its

y

-velocity reversed and its magnitude unchanged. Transverse distances are the same in both frames, gravity is ignored, the balls are point particles and the collision is instantaneous. You may use the fact that in

S

the

y

-speed of B's ball is

u_0/\gamma(v)

(time dilation) and that its

x

-velocity is

v

. ((a)) Show that conservation of

y

-momentum in

S

requires the total

y

-momentum to vanish, and hence that

f(u_B) = \gamma(v)\,f(u_0)

, where

u_B = \sqrt{v^2 + u_0^2/\gamma(v)^2}

is the speed of B's ball in

S

. ((b)) Take the limit

u_0 \to 0

and deduce

f(v) = \gamma(v)

. Give the numerical value of

f

v = 0.800c

. ((c)) Accepting

\vec p = \gamma(u)m\vec u

and

E = \gamma(u)mc^2

, derive the relation

E^2 - p^2c^2 = m^2c^4

. ((d)) For the case

u_0 = 0.100c

, use

\gamma(u_B) = \gamma(v)\gamma(u_0)

to find the energy of B's ball in

S

in units of

mc^2

, and then use the relation of part (c) to find its momentum in units of

mc

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