Question

A particle with an initial linear momentum of 3.72 kg · m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of 7.44 kg · m/s, directed along the positive y-axis. The final momentum of the first particle is 5.58 kg · m/s, directed 45.0° above the positive x-axis. Find the final momentum of the second particle.

magnitude
direction	above the negative x-axis

Answer 1

___________________________________________

P1i = 3.72 kg.m/s

P1ix = 3.72 kg.m/s

P1iy = 0

P2i =7.44 kg.m/s

P2ix = 0

P2iy = 7.44 kg.m/s

P1f = 5.58 kg.m/s

P1fx = P1f*cos(45)

= 5.58*cos(45)

= 3.94566 kg.m/s

P1fy = P1f*sin(45)

= 5.58*sin(45)

= 3.94566 kg.m/s

Let P2fx and P2fy are the component of momentum of second particle.

Apply, conservation of momentum in x-direction

P1ix + P2ix = P1fx + P2fx

3.72 + 0 = 3.94566 + P2fx

P2fx = - 0.22566 kg.m/s

Apply, conservation of momentum in y-direction

P1iy + P2iy = P1fy + P2fy

0 + 7.44 = 3.94566 + P2fy

P2fy = 3.49434 kg.m/s

so, P2f = sqrt(P2fx^2 + P2fy^2)

= sqrt(0.22566 ^2 + 3.49434^2)

= 3.5016 kg.m/s  

___________________________________________

Direction : theta = tan^-1(P2fy/P2fx)

= tan^-1(-3.49434/0.22566)

= 86.305 degrees aabove the negatve x axis  

___________________________________________

Please rate
if any mistake in this answer please comment i will clarify your doubt . thank you