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Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on. Part A Find the subsequent speed of mass m.

Respuesta :

Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.

The velocity of mass [tex]m[/tex] is [tex]\boxed{0.4v}[/tex] and velocity of mass [tex]4m[/tex] is [tex]\boxed{1.4v}[/tex] after elastic collision.

Further explanation:

An elastic collision is a collision where both law of conservation of momentum and law of conservation of energy followed.

Given:

Mass of first body is [tex]m[/tex].

Initial speed of first body is [tex]2v[/tex].

Mass of second body is [tex]4m[/tex].

Initial speed of second body is [tex]v[/tex].

Concept:

To obtain the final velocity of both the body for an elastic collision, equation of conservation given as:

1.

Equation of conservation of momentum:

[tex]\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}[/tex]

Here, [tex]m_1[/tex] is the mass of first body, [tex]u_1[/tex] is the initial speed of first body, [tex]v_1[/tex] is the final speed of first body, [tex]m_2[/tex] is the mass of second body, [tex]u_2[/tex] is the initial speed of second body and [tex]v_2[/tex] is the final speed of second body.

Substitute [tex]m[/tex] for [tex]m_1[/tex], [tex]2v[/tex] for [tex]u_1[/tex], [tex]4m[/tex] for [tex]m_2[/tex] and [tex]v[/tex] for [tex]u_2[/tex] in above equation and simplify.

[tex]6v=v_1+4v_2[/tex]  

Rearrange the above equation for value [tex]v_1[/tex] :

[tex]v_1=6v-4v_2[/tex]                                                             …… (I)

2.

Equation of conservation of Kinetic energy:

[tex]\boxed{\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2}[/tex]

Here, [tex]m_1[/tex] is the mass of first body, [tex]u_1[/tex] is the initial speed of first body, [tex]v_1[/tex] is the final speed of first body, [tex]m_2[/tex] is the mass of second body, [tex]u_2[/tex] is the initial speed of second body and [tex]v_2[/tex] is the final speed of second body.

Substitute [tex]m[/tex] for [tex]m_1[/tex], [tex]2v[/tex] for [tex]u_1[/tex], [tex]4m[/tex] for [tex]m_2[/tex] and [tex]v[/tex] for [tex]u_2[/tex] in above equation and simplify.

[tex]5v_2^2-12vv_2+7v^2=0[/tex]  

Solving the above quadratic equation, we get

[tex]v_2=1.4v[/tex]

Put the value of [tex]v_2[/tex] in equation (I).

[tex]v_1=0.4v[/tex]

Thus, the velocity of mass [tex]m[/tex] is [tex]\boxed{0.4v}[/tex] and velocity of mass [tex]4m[/tex] is [tex]\boxed{1.4v}[/tex] after elastic collision.

Learn more:

1. Volume of gas after expansion: https://brainly.com/question/9979757

2. Principle of conservation of momentum: https://brainly.com/question/9484203

3. Average translational kinetic energy: https://brainly.com/question/9078768

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Energy and conservation

Keywords:

Mass m, speed 2v, mass 4m, speed v, two collide elastically, conservation, momentum, energy, 1.4v, 0.4v, first, second, initial and final.

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