A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with asecond particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The finalmomentum of the first particle is 3.00 kg-m/s, directed45.0?Above the positive x-axis. Find the final momen -tum of the second particle.

Explanation:

We shall apply conservation of momentum along x and y axis.

Let the final momentum of second particle be p₁ along x axis and p₂ along y axis.

Considering momentum along x axis

2 + 0 = 3 cos 45 + p₁

p₁ = 2-2.12 = - 0.12 kg m/s

Considering momentum along y axis

4 + 0 = 3 sin 45 + p₂

p₂ = 4-2.12 = 1.88  kg m/s

Final momentum = √ ( p₁² + p₂² )

=√ ( .12² + 1.88² )

= 1.88 approx

are there options? cause you ain’t. real specific

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right