A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an

Question

A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of ?1 = 54.0? at a speed of v1 = 2.80 m/s. It bounces off at an angle of ?2 = 45.0? and a speed of v2 = 2.10 m/s.

a) What is the magnitude of the change in momentum of the billiard ball?

A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of ?1 = 54.0½ at a speed of v1 = 2.80 m/s. It bounces off at an angle of ?2 = 45.0½ and a speed of v2 = 2.10 m/s. a) What is the magnitude of the change in momentum of the billiard ball? b) In which direction does the change of momentum vector point? (Take the x-axis along the cushion and specify your answer in degrees.)

Explanation / Answer

Final momentum of the ball is

            Pfx = - mVf cos 45

                  = -(0.15)(2.1) cos 45

                  = - 0.2227 kgm/s

           Pfy = m Vf sin 45

                 = (0.15)(2.1) sin 45

                 = 0.2227 kgm/s

Initial momentum of the ball is

       Pix = m Vi cos 54

             = (0.15)(2.8) cos 54

              = 0.2469 kgm/s

       Piy = m Vi sin 54

            = (0.15) (2.8) sin 54

             = 0.3398 kg m/s

Change in momentum is

        Px = (-0.2227) - (0.2469 ) = -0.4696 kg m/s

        Py = (0.2227) - (0.3398) = - 0.1171 kgm/s

   magnitude is ? [Px^2 + Py^2 ] = ?[ 0.4696)^2 + (0.1171)^2 ]

                        = 0.484 kg m/s

angle is

        ? = tan^-1 (Py/ Px) = tan^-1 (0.1171/ 0.4696)

                                       = 14 o

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