Question

Mass A (2.0 kg) is moving with an initial velocity of 15 m/s in the +x-direction, and it collides with mass B (5.0 kg), initially at rest. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision?

Answer 1

`\Delta K=-160.89\ J`

Step-by-step explanation:

It is given that,

Mass of object A,
`m_A = 2\ kg`

Initial speed of object A,
`u_A=15\ m/s`

Mass of object B,
`m_B = 5\ kg`

Initial speed of object A,
`u_B=0` (at rest)

Le V is the final speed when they lock and move with one common velocity. Using the conservation of momentum to find it.

`m_Au_A+m_Bu_B=(m_A+m_B)V`

`m_Au_A=(m_A+m_B)V`

`V=(m_Au_A)/((m_A+m_B))`

`V=(2* 15)/((2+5))`

V = 4.28 m/s

Initial kinetic energy of the system is :

`K_i=(1)/(2)m_Au_A^2`

`K_i=(1)/(2)* 2* (15)^2`

`K_i=225\ J`

Final kinetic energy of the system is :

`K_f=(1)/(2)(m_A+m_B)V^2`

`K_f=(1)/(2)* (2+5)* (4.28)^2`

`K_f=225\ J`

`K_f=64.11\ J`

Let
`\Delta K` is the change in kinetic energy of the system after the collision. It is given by :

`\Delta K=K_f-K_i`

`\Delta K=64.11-225`

`\Delta K=-160.89\ J`

So, the change in kinetic energy of the system is 160.89 Joules.