Question

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a frictionless surface and when the carts collide they stick together. What is the ratio of the initial kinetic energy of the two-cart system to the final kinetic energy of the two-cart system?

Answer 1

Answer:1.5

Step-by-step explanation:

Given

mass of first cart
`m_1=6 kg`

initial Velocity
`u_1=3 m/s`

mass of second cart
`m_2=3 kg`

`u_2=0 m/s`

In the absence of External Force we can conserve momentum

`m_1u_1+m_2u_2=(m_1+m_2)v`

`v=(m_1u_1+m_2u_2)/(m_1+m_2)`

`v=(6* 3+3* 0)/(6+3)`

`v=2 m/s`

Final kinetic Energy of two masses

`K.E._2=(1)/(2)(m_1+m_2)v^2`

`K.E._2=(1)/(2)\cdot (3+6)\cdot (2)^2`

`K.E._2=18 J`

Initial Kinetic Energy

`K.E._1=(1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2`

`K.E._1=(1)/(2)6* 3^2+0`

`K.E._1=27 J`

`ratio =(K.E._1)/(K.E._2)=(27)/(18)=1.5`