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r-ruslan [8.4K]
3 years ago
12

One object (LaTeX: m_1m1) has a mass of 2 kg and moves with a speed of 9 m/s to the right. Another object (LaTeX: m_2m2) has a m

ass of 3 kg and moves to the left with a speed on 3 m/s.
The two objects collide.

After the collision, the first object (LaTeX: m_1m1) moves to the right with a speed of 1.5 m/s. The second object (LaTeX: m_2m2) moves to the right at 2 m/s.

-----

What is the change in the momentum of the first object (LaTeX: m_2m2) during the collision?

Choices were

A)-21

B)-15

C)-3

D)-18

E)0

F)21

G)15
Physics
1 answer:
defon3 years ago
7 0

Answer:

The correct answer is B

Explanation:

The moment of an object is the product of its mass by velocity, it is described by the equation

          p = mv

In this problem they give us the most body and speeds before and after the crash. We must define an environment with both bodies so that the forces during the crash have been internalized and the moment is preserved

         Δp = pf -p₀

         Δp = m v_{f} - m v₀

         Δp = 2 9 -2 1.5

         ΔP = 14.8 m/s

The correct answer is b

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<h3>Final velocity of the block/bullet system</h3>

Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

where;

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v = √(2 x 9.8 x 1.1)

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<h3>Initial velocity of the bullet</h3>

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

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(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)

0.0075u₁ = 4.4428

u₁ = 4.4428/0.0075

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<h3>Initial kinetic energy of the bullet</h3>

K.Ei = ¹/₂m₁u₁²

K.Ei = ¹/₂(0.0075)(592.37)²

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<h3>Final kinetic energy of the block/bullet system</h3>

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²

K.Ef = 10.31 J

<h3>Ratio of final kinetic energy to initial kinetic energy</h3>

= K.Ef/K.Ei x 100%

= (10.31 / 1,315.88) x 100%

= 0.78 %

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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