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3 years ago

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A projectile 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 tr

ajectory, 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.

1 answer:

Gala2k [10]3 years ago

8 0

Answer:

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

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