Question

A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.8 m/s, at an angle of θ = 25.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity (both magnitude and direction) after the collision.

I need a good explanation please.

thank you

Answer 1

x - components

P0 = Pf

m*V_x0 + m*V_x0 = mV_xf + mV_xf

m(5.3) + m(0) = m(4.8*cos 25) + mVf*cos(theta)

V_f*cos(theta) = 0.949 m/s ---------> (1)

y - components

m*V_y0 + m*V_y0 = mV_yf + mV_yf

m(0) + m(0) = m(4.8*sin 25) + mVf*sin(theta)

V_f sin(theta) = -2.02 m/s ----------> (2)

from (1) and (2) equations

V_f cos(theta) + V_f sin(theta) = (0.949) +(-2.02)

(V_f)^2 (cos^2 (theta) + sin^2 (theta) ) = (0.949)^2 + (-2.02)^2

(V_f)^2 = 0.900 + 4.08

V_f = 2.23 m/s

V_f sin(theta) / V_f cos(theta) = -2.02 / 0.949

V_f tan(theta) = -2.02 / 0.949

theta = -64.8 degrees

V_f = 2.23 m/s at 64.8 degrees below the original line of the motion