Answer:
Momentum of 2nd ball is
[tex]P = 31.6 kg m/s[/tex]
direction is given as
[tex]\theta = -37.66 degree[/tex]
Explanation:
As we know that there is no external force on the system of balls so momentum before and after collision will be conserved
So we have
[tex]P_i = 48 \hat i + 0[/tex]
now after collision momentum of two balls is must be same as initial
so we have
[tex]P_i = P_f[/tex]
[tex]48\hat i = (30 cos40 \hat i + 30 sin40\hat j) + (P_{2x}\hat i + P_{2y}\hat j)[/tex]
so we have
[tex]48 = 23 + P_{2x}[/tex]
[tex]P_{2x} = 25 kg m/s[/tex]
for other component we have
[tex]0 = 19.3 + P_{2y}[/tex]
[tex]P_{2y} = -19.3 kg m/s[/tex]
Momentum of 2nd ball is given as
[tex]P = \sqrt{P_2x}^2 + P_{2y}^2}[/tex]
[tex]P = 31.6 kg m/s[/tex]
direction is given as
[tex]tan\theta = \frac{P_{2y}}{P_{2x}}[/tex]
[tex]tan\theta = \frac{-19.3}{25}[/tex]
[tex]\theta = -37.66 degree[/tex]
