Since the shell was at the highest point, its velocity=0. The chunks moved in horizontal plane. Actually, each velocity is horizontal velocity, therefore, these are the velocities of uniform motions.
v1•t=s1=15, v2•t=s2=25 => v1=0.6•v2,
p1=m1•v1 =m1•0.6•v2,
p2=m2•v2
Taking into account the directions of the chunks motions, let’s find the vecor sum of p1 and p2. The angle that p1 and p2 make is 45+60=105º, => (360 - 2•105) =75º. Using the cosine law,
p12= sqrt{p1²+p2²-2p1•p2•cosα} =
=sqrt{ {(m1•0.6•v2)² +(m2•v2)² -2m1•m2•v2•0.6•v2•cos75º}.
||m1=m2||
p12= m2•v2•sqrt{0.6²+1 - 2•0.6•0.26} =1.02•m2•v2
According to the law of conservation of linear momentum
p3 = p12
p3=m3•v3 =>
m3•v3 =1.02•m2•v2,
v3=1.02•m2•v2/m3
since m2=1 kg, m3=5-1-1=3 kg,
v2=25/t
v3= =1.02•1•v2/3=1.02•25/3•t.
s3=v3•t= 1.02•25•t/3•t=1.02•25/3=8.5 m
A 5.0 kg shell is fired vertically upward. At its max height of 350.0 m, it explodes into 3
pieces. A 1.0 kg chunk lands 15.0 m away, 45
O
due northeast from the launch site. A second 1.0
kg piece lands 25.0 m, 60
O
south of east from the launch site. If all 3 pieces land at the same time
on the (level) ground, where does the third piece land?
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