A 100g mass of cookie dough hits inelastically with a 300g mass of cookie dough

Question

A 100g mass of cookie dough hits inelastically with a 300g mass of cookie dough and they stick. The 100g mass was moving to the right, at 40m/s and the 300g mass was moving to the left at 5m/s. What is the final velocity of the massive cookie dough ball? What fraction of energy is lost because of the collision? The combined cookie dough ball from the end of problem one approaches a spring with a spring constant of 400N/m. How much is the spring compressed when the cookie dough ball comes to a stop? The combined cookie dough ball from the end of problem one approaches an inclined plane of 30 degrees . It. moves up the plane without friction. How far up the hill does it go before it comes to a stop? (I'm looking for the distance from the beginning of the slope to the point when it comes to a stop.) The combined cookie dough ball from the end of problem one approaches a flat surface that has a coefficient of kinetic friction of 0.1. How far does it go before it comes to a stop?

Explanation / Answer

Given that the mass of the cookie is m1 = 100 g = 0.1 kg                                                        m2 = 300 g = 0.3 kg Initial velcoity of the mass m1 is u1 = 40 m/s Initial velcoity of the mass m2 is u2 = - 5.0 m/s ------------------------------------------------------------------------------ (a) Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 + m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                            = 6.25 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 0.906           = 90.6 % (b)From conservation of energy            (1/2) kx2 = (1/2))(m1 + m2)v2                    kx2 = (m1 + m2)v2                      x2 = (m1 + m2)v2 / k                      x = (m1 + m2)v2 / k                      x = 0.0625 m                         = 6.25 cm (c) Let L be the horizotnal distance traveled by the combined mass on the inclined plane      Then the vertical height reached by the combined mass is                   h = L sin From conservation of energy                (1/2)(m1 + m2)v2   =( m1 + m2)gh                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2% Given that the mass of the cookie is m1 = 100 g = 0.1 kg                                                        m2 = 300 g = 0.3 kg Initial velcoity of the mass m1 is u1 = 40 m/s Initial velcoity of the mass m2 is u2 = - 5.0 m/s ------------------------------------------------------------------------------ (a) Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 + m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                            = 6.25 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 0.906           = 90.6 % (b)From conservation of energy            (1/2) kx2 = (1/2))(m1 + m2)v2                    kx2 = (m1 + m2)v2                      x2 = (m1 + m2)v2 / k                      x = (m1 + m2)v2 / k                      x = 0.0625 m                         = 6.25 cm (c) Let L be the horizotnal distance traveled by the combined mass on the inclined plane      Then the vertical height reached by the combined mass is                   h = L sin From conservation of energy                (1/2)(m1 + m2)v2   =( m1 + m2)gh                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2% Initial velcoity of the mass m2 is u2 = - 5.0 m/s ------------------------------------------------------------------------------ (a) Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 + m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                            = 6.25 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 0.906           = 90.6 % (b)From conservation of energy            (1/2) kx2 = (1/2))(m1 + m2)v2                    kx2 = (m1 + m2)v2                      x2 = (m1 + m2)v2 / k                      x = (m1 + m2)v2 / k                      x = 0.0625 m                         = 6.25 cm (c) Let L be the horizotnal distance traveled by the combined mass on the inclined plane      Then the vertical height reached by the combined mass is                   h = L sin From conservation of energy                (1/2)(m1 + m2)v2   =( m1 + m2)gh                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%                    kx2 = (m1 + m2)v2                      x2 = (m1 + m2)v2 / k                      x = (m1 + m2)v2 / k                      x = 0.0625 m                         = 6.25 cm (c) Let L be the horizotnal distance traveled by the combined mass on the inclined plane      Then the vertical height reached by the combined mass is                   h = L sin From conservation of energy                (1/2)(m1 + m2)v2   =( m1 + m2)gh                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%                      x2 = (m1 + m2)v2 / k                      x = (m1 + m2)v2 / k                      x = 0.0625 m                         = 6.25 cm (c) Let L be the horizotnal distance traveled by the combined mass on the inclined plane      Then the vertical height reached by the combined mass is                   h = L sin From conservation of energy                (1/2)(m1 + m2)v2   =( m1 + m2)gh                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%                (1/2)(m1 + m2)v2   =( m1 + m2)gLsin                         v2   =2gLsin                        L = v2 /2gsin Then we get   L = 3.985 m (d) if the combined mass traveling in the rough plane, acceleration of the masses is            a = - g From equation of motion              vf2 - v2 = 2as              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%              0 - v2 = 2(-g)s                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%                    s = v2 / 2g                       = 19.93 m (E)      Since the collision in inelastic the two cookies are moves with common velcotiy after collision     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%     From conservation of momentum               m1*u1 - m2*u2 = (m1 + m2)v                         v = (m1*u1 - m2*u2 ) / (m1 + m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                  v = (u1 - u2 ) / 2                                (Since m1 = m2)                                     = 17.5 m/s The fractional loss in kinetic energy is           = 1 - (1/2)(m1 + m2)v2 / [ (1/2)m1*u12 + (1/2)m2*u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 1 - v2 / [ (1/2)u12 + (1/2)u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 1 - 2v2 / [ u12 + u22 ]           = 95.2%

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