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Answer:
[tex]v_2 = 6.406 / cos25.29[/tex]
Explanation:
From the question we are told that
Speed of snow ball 7.6m/s
Speed of chunk 9.3m/s at [tex]19 \textdegree[/tex]
Generally the equation for the conservation of momentum is mathematically given as
Let the snow ball be b
and the chunk b/2
According to conservation of momentum we have
[tex]b_u = (b /2) v_1 cos\theta_1 + (b / 2) v_2cos\theta_2[/tex]
[tex]v_2 cos\theta_2 = 2 * 7.6 -9.3 * cos_19[/tex]
[tex]v_2 cos\theta_2 = 15.2 -8.793 = 6.406[/tex]
Therefore
[tex](b/2)v_1 sin\theta_1 = (b/2) v_2sin\theta_2[/tex]
[tex]v_2 sin\theta_2 = 9.3 * sin_19[/tex]
[tex]v_2 sin\theta_2 = 3.027[/tex]
Mathematically From above equations
[tex]tan\theta_2 = 0.4726[/tex]
[tex]\theta_2 = 25.29 \textdegree[/tex]
Therefore the speed and direction of second chunk
[tex]v_2 = 6.406 / cos25.29[/tex]
[tex]v2 = 7.0855m/s[/tex]