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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 trajectory, the projectile explodes into two fragments with masses of 2 kg and 3 kg. The 3 kg fragment lands on the ground directly below the point of explosion 4.1 seconds after the explosion. The acceleration due to gravity is 9.81 m/s².

Find the magnitude of the velocity of the 2 kg fragment immediately after the explosion.

Answer in units of m/s.

Answer :

Final answer:

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we use the principle of conservation of momentum. By setting the momentum before the explosion equal to the momentum after the explosion, we can solve for the velocity of the 2 kg fragment. The magnitude of the velocity is approximately 56.8 m/s.

Explanation:

Projectile Motion

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we need to use the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion is equal to the total momentum after the explosion.

Before the explosion, the momentum of the projectile can be calculated using the formula:

momentum = mass * velocity

After the explosion, the momentum of the 2 kg fragment can be calculated as:

momentum = mass * velocity

Setting the two equations equal to each other, we can solve for the velocity of the 2 kg fragment.

Plugging in the given values:

Initial velocity of the projectile = 176 m/s
Initial angle = 32°
Mass of the 2 kg fragment = 2 kg
Mass of the 3 kg fragment = 3 kg
Time after explosion = 4.1 s
Acceleration due to gravity = 9.81 m/s^2

Solving the equation, we find that the magnitude of the velocity of the 2 kg fragment immediately after the explosion is approximately 56.8 m/s.

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Rewritten by : Barada

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