A firework whose mass is 7kg is launched vertically up into the air. After it reaches its the height of 160m, it blasts and splits into two pieces; pieces A and B. The two pieces fly off and travel horizontally. The blasts of the fireworks lasts 10ms. Piece A (mass of 1.50kg ) touches the ground 300 meters away from the launch position
Determine:
a) the magnitude of the momentum transferred to A due to explosion
b) the magnitude of impulse delivered to A by gravity during the explosion
c) how far B lands from the launch position
To solve this problem, we need to use a few key principles of physics, specifically the conservation of momentum and the equations of motion.
a) To find the magnitude of the momentum transferred to piece A due to the explosion, we need to use the principle of conservation of momentum. The initial momentum of the system is zero since the firework was at rest before the explosion. After the explosion, the momentum will still be conserved, but now distributed between the two pieces.
The equation for conservation of momentum is:
m1 * v1 + m2 * v2 = 0,
where m1 and v1 are the mass and velocity of piece A, and m2 and v2 are the mass and velocity of piece B. Since the explosion happens vertically, piece B will only have a vertical velocity, while piece A will have both vertical and horizontal velocities.
Given:
Mass of piece A (m1) = 1.50 kg
Mass of piece B (m2) = (7 - 1.50) kg = 5.50 kg
Vertical velocity of piece B (v2) = 0 m/s (No vertical velocity)
Vertical velocity of piece A (v1) = ?
From the equation of conservation of momentum, we can solve for the velocity of piece A:
1.50 kg * v1 + 5.50 kg * 0 m/s = 0
Solving for v1, we have:
v1 = 0 / 1.50 kg
v1 = 0 m/s
Therefore, the magnitude of the momentum transferred to piece A due to the explosion is zero since it comes to rest immediately after the explosion.
b) To find the magnitude of the impulse delivered to piece A by gravity during the explosion, we need to use the equation:
Impulse = Change in momentum
Since piece A comes to rest immediately after the explosion, the change in momentum is equal to the initial momentum. Thus, the impulse delivered to piece A is equal to the magnitude of the momentum transferred to piece A.
Therefore, the magnitude of the impulse delivered to piece A by gravity during the explosion is also zero.
c) To find how far piece B lands from the launch position, we need to use the equation of motion for horizontal motion:
Distance = Initial horizontal velocity * Time
Given:
Initial horizontal velocity = ?
Time = 10 ms = 0.01 s
To find the initial horizontal velocity, we need to consider that the horizontal velocity is constant and can be determined from the distance traveled by piece A before reaching the ground.
Given:
Distance traveled by piece A = 300 m
Using the equation,
Distance = Initial horizontal velocity * Time,
we can solve for the initial horizontal velocity:
300 m = Initial horizontal velocity * 0.01 s
Initial horizontal velocity = 300 m / 0.01 s
Initial horizontal velocity = 30,000 m/s
Therefore, piece B will land 30,000 meters away from the launch position.