A 2.3-kg object traveling at 6.1 m/s collides head-on with a 3.5-kg object traveling in the opposite direction at 4.8 m/s. if the collision is perfectly elastic, what is the final speed of the 2.3-kg object?

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of object 1 = m₁ = 2.3 kg

initial velocity of object 1 = u₁ = 6.1 m/s

mass of object 2 = m₂ = 3.5 kg

initial velocity of object 2 = u₂ = -4.8 m/s

Asked:

final velocity of object 1 = v₁ = ?

Solution:

Firstly , we will use Conservation of Momentum Law as follows:

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

[tex]2.3(6.1) + 3.5(-4.8) = 2.3v_1 + 3.5v_2[/tex]

[tex]-2.77 = 2.3v_1 + 3.5v_2[/tex] → Equation 1

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If the collision is perfectly elastic , then:

[tex]u_1 - u_2 = v_2 - v_1[/tex]

[tex]6.1 - (-4.8) = v_2 - v_1[/tex]

[tex]10.9 = v_2 - v_1[/tex] → Equation 2

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Let's solve the two Equations above:

(Equation 1) - 3.5(Equation 2) ↓

[tex]-2.77 - 3.5(10.9) = 2.3v_1 + 3.5v_2 - 3.5(v_2 - v_1)[/tex]

[tex](-2.77) - 3.5(10.9) = (2.3v_1 + 3.5v_2) - 3.5(v_2 - v_1)[/tex]

[tex]-40.92 = 5.8v_1[/tex]

[tex]v_1 = -40.92 \div 5.8[/tex]

[tex]v_1 \approx -7.1 \texttt{ m/s}[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

alokdubeyvidyaatech alokdubeyvidyaatech · Answer 2 · 2022-03-10T12:09:34+01:00

The velocity of an object depends on its position and time. The final velocity of the object with a mass of 2.3 kg is -7.05 m/s.

What is velocity?

The velocity of an object is the rate of change of its position with respect to time interval.

Given that two objects collide with each other. The masses m1 and m2 of the objects are 2.3 kg and 3.5 kg respectively. The initial velocities u1 and u2 of the objects are 6.1 m/s and -4.8 m/s.

The law of conversion on momentum states that the momentum of the system before the collision and after the collision will be the same. This can be written as,

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

Where v1 and v2 are the final velocities of the objects after the collision.

[tex]2.3 \times 6.1 + 3.5 \times (-4.8) = 2.3v_1 + 3.5 v_2[/tex]

[tex]2.3 v_1 + 3.5v_2 = -2.77[/tex]

If the collision is perfectly elastic and objects are moving in the opposite directions, then the velocity of the system before and after the collision will be the same.

[tex]u_1 - u_2 = v_2 - v_1[/tex]

[tex]6.1 -(-4.8) = v_2-v_1[/tex]

[tex]v_2-v_1 = 10.9[/tex]

[tex]v_2 = 10.9+v_1[/tex]

Now put the value of v_1 in the above equation, we get,

[tex]2.3v_1 +3.5(10.9+v_1) = -2.77[/tex]

[tex]2.3v_1 + 38.15 + 3.5 v_1 +2.77 = 0[/tex]

[tex]5.8v_1 = -40.92[/tex]

[tex]v_1 = -7.05 \;\rm m/s[/tex]

Hence we can conclude that the final velocity of the object with a mass of 2.3 kg is -7.05 m/s.

To know more about the velocity, follow the link given below.

https://brainly.com/question/862972.