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monitta
4 years ago
13

A football player is preparing to punt the ball down the field. He drops the ball from rest and it falls vertically 1.0 m down o

nto his foot. After he kicks it, the ball leaves the foot with a speed of 18.5 m/s at an angle 57° above the horizontal.
What is the magnitude of the impulse delivered to the ball in kilogram meters per second after he kicks it? An American football has a mass of 425 g.

Physics
1 answer:
Doss [256]4 years ago
4 0

Answer: Impulse = 6.37kgm/s

Explanation: please find the attached file for the solution.

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How does Kinetic Energy and the Law of Conservation of Momentum apply to inelastic<br> collisions?
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Answer:

Explanation:

An inelastic collision is one in which the internal kinetic energy changes (it is not conserved) . The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision.

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In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary sec
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Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

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