Question

a hockey player of mass 82 kg is traveling north with a velocity of 4.1 meters per second he collides with the 76 kg player traveling East at 3.4 meters per second if the two players locked together momentarily in what direction will they be going immediately after the Collision how fast will they be moving

Answer 1

mass of first player = 82 kg

speed of first player = 4.1 m/s (towards North)

mass of second player = 76 kg

speed of second player = 3.4 m/s (towards East)

now the two players will collide and stick together so here since there is no external force on the system of two players so we will say momentum of system of players will remain conserved

So here we will have

$P_{1i} = 82 (4.1 \hat j)$

$P_{2i} = 76 (3.4 \hat i)$

now after collision they both move together with same speed so we will have

$P_{1i} + P_{2i} = (m_1 + m_2)v$

$336.2\hat j + 258.4\hat i = (82 + 76)v$

$v = 1.64\hat i + 2.13 \hat j$

so the magnitude of the velocity after they collide is given as

$v = \sqrt{1.64^2 + 2.13^2}$

$v = 2.69 m/s$

direction of motion is given as

$tan\theta = \frac{2.13}{1.64} = 1.3$

$\theta = 52.4 degree$

so they both will move 52.4 degree North of East after collision