Question

Theorem of conservation of momentum or momentum:
  two line followers slide on a frictionless surface. The first follower, with a mass of 24.0 gr, initially moves at 2.40 m / s parallel to the x axis, which collides with the second follower, whose mass is 12.0 gr, which is initially at rest, as shown in the figure. After the collision, the first follower moves at 1.03 m / s in a direction that forms an angle β = 31.0 with its initial direction.

A. The final speed of the second follower?
B. The direction of the second follower just after the crash with respect to the positive x-axis?
C. The total kinetic energy and before after the shock and verify if the kinetic energy conservation theorem is fulfilled or not.

Answer 1

# KE,= 9 mu? KE = 2 x 2x10 x 2.4x2 - 69 120103 & ux10 x 1.03 + 2x12x10 s 2018 - 48.74810 35 IKE2ZKEL) enery not consered mi 2.40m/s mia qux10 kg V - 2.40 m/s 2 - 2x varo momentum lesenation in Y direction - Pr = Pj =) o 2 m. V shß + Ma vyama -mi vil simß y may book. hena Vyu – 2x2, 40X &n31 o nou in x drache Er 9.472 mu, - muyas + m2 V y cs f-o Mavybra= mini-mugarp Vycha - (24*2.402 (24x 1.03431) Vagas a = 2001 - 2 0.0349 tand= 88 72; 12 = font ( 17 ) = 89. 20 direction Try = 2,45 mg] valaha 2 Oru399