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MaRussiya [10]
3 years ago
10

A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal

(x axis) and bounces away at an angle of θf= 45.00 above the horizontal. What is the average force exerted by the wall on the ball if the ball is in contact with the wall for 250.0 ms?
Physics
1 answer:
Mkey [24]3 years ago
3 0

Answer:

   F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

     I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

      cos 45 = vₓ / v

      vₓ = v cos 45

      vₓ = 8 cos 45

      vₓ = 5,657 m / s

We write the moment

Before the crash                          p₀ = m vₓ

After the shock                            p_{f} = -m vₓ

The variation of the moment      Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis           Fₓ t = Δp

       

        Fₓ = 2 m vₓ / t

        Fx = 2 0.450 5.657 / 0.250

        Fx = 20.4 N

We perform the same calculation on the y axis

       sin  45 = vy / v

       vy = v sin 45

       vy = 8 sin 45

       vy = 5,657 m / s

We calculate the initial momentum   po = m v_{y}

Final moment                                      p_{f} = m v_{y}

Variations moment                             Δp = mv_{y} - mv_{y} = 0

Force in the Y-axis                             F_{y} = 0

Therefore the total force is

       F = fx i ^ + Fyj ^

       F = Fx i ^

       F = 20.4 i ^

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A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

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