Question

A proton with an initial speed of 1.82 ✕ 108 m/s in the +x direction collides elastically with another proton initially at rest. The first proton's velocity after the collision is 1.454 ✕ 108 m/s at an angle of 37.0° with the +x-axis.

What is the velocity (magnitude and direction) of the second proton after the collision? magnitude m/s

direction ? ° counterclockwise from the +x-axis

Answer 1

u1 = 1.82 x 10⁸ m/s ; v1 = 1.454 x 10⁸ m/s at = 37 deg

Let v2 be the vel of second proton and theta2 be the angle.

Let, the ball be moving along +X initially

(a)From the convervation of vertical momentum,

m1 v1 sin1 - m2 v2 sin2= 0

since m1 = m2

v1 sin1 = v2 sin2

v2 sin2 = 1.454 x 10⁸ m/s x sin37 = 0.875 x 10⁸ m/s

v2 sin 2 = 0.875 x 10⁸ m/s (1)

From conservation of horizontal momentum

m1 u1 = m2 v2 cos 2 + m1 v1 cos 1

1.82 x 108 = v2 cos2 + 1.454 x 108 x cos37

v2 cos2 = 0.659 x 10⁸ (2)

Dividing (1)/(2)

sin2 / cos2 = 0.875 x 10⁸ / 0.659 x 10⁸

2 = 53.02

v2 sin 2 = 0.875 x 10⁸ m/s

v2 = 0.875 x 10⁸ / sin53.02 = 1.095 x 10⁸ m/s

Hence, v2 = 1.095 x 108 m/s and = 53.02 deg