A ball (I) having a mass of 0.3 kg and moving at a speed of 4.9 strikes a glancing blow on a stationary ball (

A ball (I) having a mass of 0.3 kg and moving at a speed of 4.9 strikes a glancing blow on a stationary ball (II). After the collision, ball I is moving at right angles to its original direction of motion, with a speed of 4.5 m/s.

a) What is the magnitude of the momentum of ball II after the collision?

b) After the collision it is found that the speed of ball II is 13.3 m/s. What is the mass of ball II?

I am confused, I know this is a collision in two direction. and I found that ball II move in 45degree after collision below the +-x axis.
still cant figure out the momentum amount!

Since there are no net external forces on the two-ball system, linear momentum must be conserved.

Choose a coordinate system. I think you've chosen ball (I) to be moving initially in the +x direction, and to be moving finally in the +y direction, so I'll choose that, too. I assume the initial velocity of ball (I) is measured in m/s.

So, the initial momentum in the x direction is

Pix = Pix_I = (m_I) Vix = (0.3 kg) (4.9 m/s) = 1.47 kg∙m/s

and the initial momentum in the y direction is

Piy = 0 kg∙m/s

After the collision, the total momentum in each direction must be the same. Ball (I) is moving entirely in the +y direction, so the momentum in the x direction must be entirely supplied by ball (II).

Pfx_II = 1.47 kg∙m/s

Ball (I) has a momentum

Pfy_I = (0.3 kg) (4.5 m/s) = 1.35 kg∙m/s

but the total in the y direction must remain zero, so

Pfy_II = -1.35 kg∙m/s

So the magnitude of the momentum of ball (II) is

Pf_II = √[ (Pfx_II)² + (Pfy_II)² ] = √[ (1.47 kg∙m/s)² + (-1.35 kg∙m/s)² ] = 2.00 kg∙m/s

You'll note that the direction is *not* 45° below the x axis. It's

arctan(-1.35 kg∙m/s / 1.47 kg∙m/s) = 42.6°

below the x axis.

Now that we know the magnitude of the momentum, the speed will give us the mass: