A hockey puck B rests on a smooth surface of ice and is struck by a second puck

Question

A hockey puck B rests on a smooth surface of ice and is struck by a second puck A , which was originally traveling at 40.0 { m m}/{ m s} and which is deflected 30.0^circ from its original direction. (See the figure below .) Puck B acquires a velocity at a 45.0^circ { m angle} to the original direction of A . The pucks have the same mass.
Compute the speed of puck A after the collision.Compute the speed of puck B after the collision.What fraction of the original kinetic energy of puck A dissipates during the collision?

Explanation / Answer

Given that both A and B has same masses let it be m initial velocity of A = 40 m / s initial velocity of B = 0 m /s total momentum before collision = ( m * 40 ) + ( m * 0 )                                                  = 40 m     - ------- ( 1 ) after collision : velocity of A along horizontal direction = u cos30 = 0.866 u velocity of A along vertical direction = usin 30 = 0.5 u velocity of B along horizontal direction = vcos 45 = 0.707 v velocity of B along vertical direction =  - v sin 45 = - 0.707 v   since it isin - ve axis total momentum along vertical direction = 0 So , m ( 0.5 u ) + m ( - 0.707 v ) =0               0.5 u = 0.707 v                     u = 1.414 v    ------------( 2 ) total momentum along horizontal direction = m ( 0.866 u ) + m( 0.707 v )                                                                =0.866 * m * 1.414 v + 0.707 m v                                                               = 1.224 m v + 0.707 mv                                                                =1.931 mv      ------------------( 3 ) momentum is conserved     So,    eq ( 2 ) = eq ( 3 )                                                      40 m = 1.931 mv                                                          v = 40 / 1.931                         velocity of B                = 20.7 m / s              u = 1.414 * v     since fro ( 2)                  = 1.414 * 20. velocity of A   = 29.26 m /s kinetic energy before collision = ( 1/ 2 ) m ( 40 ) 2                                             = 0.5 * m * 1600                                              =800m           ----------------( 2 ) kinetic energy after collision = ( 1/ 2 ) m u 2 +( 1/ 2 ) m v 2                                          = 0.5 * m * ( 29.26 ) 2 + 0.5 * m * ( 20.7 ) 2                                          = 428.3 m + 214.2 m                                          = 642.5 m loss in KE = KE before collision - KE aftercollision                = 800 m - 642. 5 m                = 157.45 m fraction of dissipated KE = ( 157.45 m /  800 m ) *100                                            = 0.196 * 100                                      = 19.6 %˜20% Given that both A and B has same masses let it be m initial velocity of A = 40 m / s initial velocity of B = 0 m /s total momentum before collision = ( m * 40 ) + ( m * 0 )                                                  = 40 m     - ------- ( 1 ) after collision : velocity of A along horizontal direction = u cos30 = 0.866 u velocity of A along vertical direction = usin 30 = 0.5 u velocity of B along horizontal direction = vcos 45 = 0.707 v velocity of B along vertical direction =  - v sin 45 = - 0.707 v   since it isin - ve axis total momentum along vertical direction = 0 So , m ( 0.5 u ) + m ( - 0.707 v ) =0               0.5 u = 0.707 v                     u = 1.414 v    ------------( 2 ) total momentum along horizontal direction = m ( 0.866 u ) + m( 0.707 v )                                                                =0.866 * m * 1.414 v + 0.707 m v                                                               = 1.224 m v + 0.707 mv                                                                =1.931 mv      ------------------( 3 ) momentum is conserved     So,    eq ( 2 ) = eq ( 3 )                                                      40 m = 1.931 mv                                                          v = 40 / 1.931                         velocity of B                = 20.7 m / s              u = 1.414 * v     since fro ( 2)                  = 1.414 * 20. velocity of A   = 29.26 m /s kinetic energy before collision = ( 1/ 2 ) m ( 40 ) 2                                             = 0.5 * m * 1600                                              =800m           ----------------( 2 ) kinetic energy after collision = ( 1/ 2 ) m u 2 +( 1/ 2 ) m v 2                                          = 0.5 * m * ( 29.26 ) 2 + 0.5 * m * ( 20.7 ) 2                                          = 428.3 m + 214.2 m                                          = 642.5 m loss in KE = KE before collision - KE aftercollision                = 800 m - 642. 5 m                = 157.45 m fraction of dissipated KE = ( 157.45 m /  800 m ) *100                                            = 0.196 * 100                                      = 19.6 %˜20% Given that both A and B has same masses let it be m initial velocity of A = 40 m / s initial velocity of B = 0 m /s total momentum before collision = ( m * 40 ) + ( m * 0 )                                                  = 40 m     - ------- ( 1 ) after collision : velocity of A along horizontal direction = u cos30 = 0.866 u velocity of A along vertical direction = usin 30 = 0.5 u velocity of B along horizontal direction = vcos 45 = 0.707 v velocity of B along vertical direction =  - v sin 45 = - 0.707 v   since it isin - ve axis total momentum along vertical direction = 0 So , m ( 0.5 u ) + m ( - 0.707 v ) =0               0.5 u = 0.707 v                     u = 1.414 v    ------------( 2 ) total momentum along horizontal direction = m ( 0.866 u ) + m( 0.707 v )                                                                =0.866 * m * 1.414 v + 0.707 m v                                                               = 1.224 m v + 0.707 mv                                                                =1.931 mv      ------------------( 3 ) momentum is conserved     So,    eq ( 2 ) = eq ( 3 )                                                      40 m = 1.931 mv                                                          v = 40 / 1.931                         velocity of B                = 20.7 m / s              u = 1.414 * v     since fro ( 2)                  = 1.414 * 20. velocity of A   = 29.26 m /s kinetic energy before collision = ( 1/ 2 ) m ( 40 ) 2                                             = 0.5 * m * 1600                                              =800m           ----------------( 2 ) kinetic energy after collision = ( 1/ 2 ) m u 2 +( 1/ 2 ) m v 2                                          = 0.5 * m * ( 29.26 ) 2 + 0.5 * m * ( 20.7 ) 2                                          = 428.3 m + 214.2 m                                          = 642.5 m loss in KE = KE before collision - KE aftercollision                = 800 m - 642. 5 m                = 157.45 m fraction of dissipated KE = ( 157.45 m /  800 m ) *100                                            = 0.196 * 100                                      = 19.6 %˜20% velocity of B along vertical direction =  - v sin 45 = - 0.707 v   since it isin - ve axis total momentum along vertical direction = 0 So , m ( 0.5 u ) + m ( - 0.707 v ) =0               0.5 u = 0.707 v                     u = 1.414 v    ------------( 2 ) total momentum along horizontal direction = m ( 0.866 u ) + m( 0.707 v )                                                                =0.866 * m * 1.414 v + 0.707 m v                                                               = 1.224 m v + 0.707 mv                                                                =1.931 mv      ------------------( 3 ) momentum is conserved     So,    eq ( 2 ) = eq ( 3 )                                                      40 m = 1.931 mv                                                          v = 40 / 1.931                         velocity of B                = 20.7 m / s              u = 1.414 * v     since fro ( 2)                  = 1.414 * 20. velocity of A   = 29.26 m /s kinetic energy before collision = ( 1/ 2 ) m ( 40 ) 2                                             = 0.5 * m * 1600                                              =800m           ----------------( 2 ) kinetic energy after collision = ( 1/ 2 ) m u 2 +( 1/ 2 ) m v 2                                          = 0.5 * m * ( 29.26 ) 2 + 0.5 * m * ( 20.7 ) 2                                          = 428.3 m + 214.2 m                                          = 642.5 m loss in KE = KE before collision - KE aftercollision                = 800 m - 642. 5 m                = 157.45 m fraction of dissipated KE = ( 157.45 m /  800 m ) *100                                            = 0.196 * 100                                      = 19.6 %˜20%

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