Question

Question #1ee48

Answer 1

Given

`m_b->"mass of fullback"=80kg`

`m_o->"mass of opponent"=96kg`

`u_x->"vel of fullback towards East( +ve x-axis)"=5.1" m/s"`

`u_y->"vel of opponent towards North( +ve y-axis)"=3" m/s"`

(a) Here the collision is inelastic as KE is not conserved.

We are to proceed our calculation assuming that the collision is perfectly inelastic and the bodies of two player are entangled after collision.

Let the common velocity the players gained be v and its components along x-axis and y-axis are `v_x and v_y` respectively.

Now applying the law of conservation of linear momentum along x-axis, we can write

`m_bxxu_x+m_o xx0=(m_b+m_o)xxv_x`

`v_x=(m_bxxu_x)/(m_b+m_o)=(80xx5.1)/(80+96)~~2.32m/s`

Again applying the law of conservation of linear momentum along y-axis, we can write

`m_b xx0+m_o xxu_y=(m_b+m_o)xxv_y`

`v_y=(m_o xxu_y)/(m_b+m_o)=(96xx3)/(80+96)~~1.64m/s`

(b) So the common velocity of the players

`v=sqrt(v_x^2+v_y^2)=sqrt(2.32^2+1.64^2)~~2.84m/s`

(c) Loss in mechanical energy

`E_"loss"="Initial KE"-"Final KE"`

`=>E_"loss"=1/2m_bxxu_x^2+1/2m_o xxu_y^2-1/2(m_b+m_o)xxv^2`

`=1/2xx80xx5.1^2+1/2xx96 xx3^2-1/2xx176xx2.84^2`

`~~762.83J`

(d) This loss of mechanical will actually be transferred into equivalent amount of heat energy and will eventually be dissipated in the environment.