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bellamvento
25.01.2020 •
Physics
Aprojectile of mass 5 kg is fired with an initial speed of 176 m/s at an angle of 32◦ with the horizontal. at the top of its trajectory, the projectile explodes into two fragments of masses 2 kg and 3 kg . the 3 kg fragment lands on the ground directly below the point of explosion 4.1 s after the explosion. the acceleration due to gravity is 9.81 m/s 2 . find the magnitude of the velocity of the 2 kg fragment immediatedly after the explosion. answer in units of m/s.
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Ответ:
v1 = 377.98 m/s
Explanation:
m = 5 Kg
v0 = 176 m/s
v0x = v0*Cos 32° = 176 m/s*Cos 32° = 149.256 m/s
m1 = 2 Kg
m2 = 3 Kg
t = 4.1 s
g = 9.81 m/s²
Before the explosion
pix = m*v0x = 5 Kg*149.256 m/s = 746.282 Kgm/s
piy = 0
After the explosion
pfx = m1*v1x
knowing that pix = pfx
we have
746.282 = 2*v1x
v1x = 373.14 m/s
v2y = g*t
pfy = m1*v1y + m2*v2y
pfy = 2*v1y + 3*(9.81*4.1)
pfy = 2*v1y + 120.663
knowing that piy = pfy = 0
we have
0 = 2*v1y + 120.663
v1y = 60.33 m/s
Finally we apply
v1 = √(v1x² + v1y²)
v1 = √(373.14² + 60.33²)
v1 = 377.98 m/s
Ответ:
B
Explanation:
I took the assignment :))