Question

A projectile proton with a speed of 1400 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 74

Answer 1

Let u_p = initial velocity of prijectile proton before collision = 1400 m/s

u_t = initial velocity 0f target proton = 0 ( due to at rest)

V_p = final velocity of projectile proton after collision

V_t = final velocity of target proton after collision

From Conservation of momentum,     P_initial = P_final

Momentum along X- direction:

m *u_p + m*u_t = m* V_p cos 74+ m* V_t cos 16                     ( since both are proton so both have same mass)

1400 = V_p cos 74 + V_t cos 16                                ----------------------1

momentum along Y- direction:

0 = m*V_p sin 74 - m* V_t sin 16              ( initial momentun along Y-direction = 0),

( -ve sign due to opposite direction)

V_p sin 74 = V_t sin 16

V_p = 0.286 V_t                             -------------------2

By putting this value in eq. 1

1400 = 0.286 V_t *cos 74 + V_t *cos 16

V_t = 1345.8 m/s

From eq 2

V_p = 384.92 m/s