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A skater embeds a firecracker in a large snowball and pushes it across a frozen pond at 7.6 m/s. The firecracker explodes, breaking the snowball into two equal-mass chunks. One chunk moves off at 9.3 m/s at 19° to the original direction of motion. Discover the speed and direction of the second chunk.

Sagot :

okpalawalter8

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]

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