Question

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.

Answer 1

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.

Answer 2

The velocity of mass $m$ is $\boxed{0.4v}$ and velocity of mass $4m$ is $\boxed{1.4v}$ 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 $m$.

Initial speed of first body is $2v$.

Mass of second body is $4m$.

Initial speed of second body is $v$.

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:

$\boxed{m_1u_1+m_2u_2=m_1v_1+m_2v_2}$

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

Substitute $m$ for $m_1$, $2v$ for $u_1$, $4m$ for $m_2$ and $v$ for $u_2$ in above equation and simplify.

$6v=v_1+4v_2$  

Rearrange the above equation for value $v_1$ :

$v_1=6v-4v_2$                                                             …… (I)

2.

Equation of conservation of Kinetic energy:

$\boxed{\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2}$

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

Substitute $m$ for $m_1$, $2v$ for $u_1$, $4m$ for $m_2$ and $v$ for $u_2$ in above equation and simplify.

$5v_2^2-12vv_2+7v^2=0$  

Solving the above quadratic equation, we get

$v_2=1.4v$

Put the value of $v_2$ in equation (I).

$v_1=0.4v$

Thus, the velocity of mass $m$ is $\boxed{0.4v}$ and velocity of mass $4m$ is $\boxed{1.4v}$ 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.