Question

A cart of mass m moves with a speed v on a frictionless air track and collides with an identical cart that is stationary. If the two carts stick together after the collision, what is the final kinetic energy of the system?

Express your answer in terms of the variables m and v.

Answer 1

To solve this problem, we will start by defining our variables, then we will use the linear momentum conservation theorem and end with the definition of the kinetic energy of the system.

The mass of each car is the same, so it can be defined as

m = mass of each cart

The initial velocity of body 2 is zero since part of the rest, while the velocity of body 1 is defined as 'v', at the same time both bodies travel together at the same final velocity after impact so we will have to

$v_1=v$

$v_2=0$

$v_f = \text{same for each object}$

$m_{final} = 2m$

Now using conservation of linear momentum:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

With our values that equation is,

$mv+mv_2=(m+m)v_f$

$mv+m(0)=2mv_f$

$v_f=\frac{v}{2}$

Now the final kinetic energy of the system would be given as

$KE = \frac{1}{2}m_f v_f^2$

$KE = \frac{1}{2}(2m)(\frac{v}{2})^2$

$\therefore \mathbf{KE = \frac{1}{4} mv^2}$

Answer 2

The final kinetic energy of the system after the collision is $\frac{1}{2} mv^2$.

The given parameters:

• Mass of the cart, = m
• Speed of the cart, = v

Apply the principle of conservation of energy to determine the final kinetic energy of the system after the collision;

$K.E_1 + K.E_2 = K.E_f \\\\\frac{1}{2} mv^2 + 0 = K.E_f\\\\K.E_f = \frac{1}{2} mv^2$

Thus, the final kinetic energy of the system after the collision is $\frac{1}{2} mv^2$.

Learn more about conservation of energy here: https://brainly.com/question/166559