Question

Two balls with masses of 2.00 kg and 6.20 kg travel toward each other at speeds of 9.0 m/s and 4.00 m/s, respectively. If the balls have a head-on inelastic collision and the 2.00-kilogram ball recoils with a speed of 8.00 m/s, how much kinetic energy is lost in the collision

Answer 1

`\Delta KE=8.67958\ J`

Step-by-step explanation:

Given:

• mass of the first ball,
  `m_1=2\ kg`

• initial velocity of the first ball,
  `u_1=9\ m.s^(-1)`

• mass of the second ball,
  `m_2=6.2\ kg`

• initial velocity of the second ball,
  `u_2= 4\ m.s^(-1)`

• Final velocity of the first ball,
  `v_1=8\ m.s^(-1)`

Using the law of conservation of linear momentum:

`m_1.u_1+m_2.u_2=m_1.v_1+m_2.v_2`

where:
`v_2=` final velocity of the second ball

`2* 9+6.2* 4=2* 8+6.2* v_2`

`v_2=4.3225\ m.s^(-1)`

Now the kinetic energy lost in the collision:

Δ KE = (Sum of initial individual kinetic energy) - (Sum of individual final kinetic energy)

`\Delta KE=(1)/(2) m_1.u_1^2+(1)/(2) m_2.u_2^2-((1)/(2) m_1.v_1^2+(1)/(2) m_1.v_2^2)`

`\Delta KE=0.5* 2* 9^2+0.5* 6.2* 4^2-(0.5* 2* 8^2+0.5* 6.2* 4.3225^2)`

`\Delta KE=8.67958\ J`